input: What is the significance of level curves in the study of functions of multiple variables?

answers: Level curves are significant because they provide a visual representation of the function's values at different heights, helping to understand the function's behavior in a plane.

context:
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Advanced Calculus Demystified

David Bachman

**Library of Congress Cataloging-in-Publication Data**

Bachman, David.

Advanced calculus demystified / David Bachman.

p. cm. – (Demystified series)

Includes index.

1. Calculus. I. Title.

QA303.2.B23 2007

515–dc22

**Advanced Calculus Demystified**

Copyright © 2007 by The McGraw-Hill Companies. All rights reserved. Except as permitted under the Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of publisher.

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_To Stacy_
**ABOUT THE AUTHOR**

**David Bachman, Ph.D.** is an Assistant Professor of Mathematics at Pitzer College, in Claremont, California. His Ph.D. is from the University of Texas at Austin, and he has taught at Portland State University, The University of Illinois at Chicago, as well as California Polytechnic State University at San Luis Obispo. Dr. Bachman has authored one other textbook, as well as 11 research papers in low-dimensional topology that have appeared in top peer-reviewed journals.
CONTENTS |

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|

Preface

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|

Acknowledgments

**CHAPTER 1** | **Functions of Multiple Variables**

|

1.1 Functions

|

1.2 Three Dimensions

|

1.3 Introduction to Graphing

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1.4 Graphing Level Curves

|

1.5 Putting It All Together

|

1.6 Functions of Three Variables

|

1.7 Parameterized Curves

|

Quiz

**CHAPTER 2** | **Fundamentals of Advanced Calculus**

|

2.1 Limits of Functions of Multiple Variables

|

2.2 Continuity

|

Quiz

**CHAPTER 3** | **Derivatives**

|

3.1 Partial Derivatives

|

3.2 Composition and the Chain Rule

|

3.3 Second Partials

|

Quiz

**CHAPTER 4** | **Integration**

|

4.1 Integrals over Rectangular Domains

|

4.2 Integrals over Nonrectangular Domains

|

4.3 Computing Volume with Triple Integrals

|

Quiz

**CHAPTER 5** | **Cylindrical and Spherical Coordinates**

|

5.1 Cylindrical Coordinates

|

5.2 Graphing Cylindrical Equations

|

5.3 Spherical Coordinates

|

5.4 Graphing Spherical Equations

|

Quiz

**CHAPTER 6** | **Parameterizations**

|

6.1 Parameterized Surfaces

|

6.2 The Importance of the Domain

|

6.3 This Stuff Can Be Hard!

|

6.4 Parameterized Areas and Volumes

|

Quiz

**CHAPTER 7** | **Vectors and Gradients**

|

7.1 Introduction to Vectors

|

7.2 Dot Products

|

7.3 Gradient Vectors and Directional Derivatives

|

7.4 Maxima, Minima, and Saddles

|

7.5 Application: Optimization Problems

|

7.6 LaGrange Multipliers

|

7.7 Determinants

|

7.8 The Cross Product

|

Quiz

**CHAPTER 8** | **Calculus with Parameterizations**

|

8.1 Differentiating Parameterizations

|

8.2 Arc Length

|

8.3 Line Integrals

|

8.4 Surface Area

|

8.5 Surface Integrals

|

8.6 Volume

|

8.7 Change of Variables

|

Quiz

**CHAPTER 9** | **Vector Fields and Derivatives**

|

9.1 Definition

|

9.2 Gradients, Revisited

|

9.3 Divergence

|

9.4 Curl

|

Quiz

**CHAPTER 10** | **Integrating Vector Fields**

|

10.1 Line Integrals

|

10.2 Surface Integrals

|

Quiz

**CHAPTER 11** | **Integration Theorems**

|

11.1 Path Independence

|

11.2 Green's Theorem on Rectangular Domains

|

11.3 Green's Theorem over More General Domains

|

11.4 Stokes' Theorem

|

11.5 Geometric Interpretation of Curl

|

11.6 Gauss' Theorem

|

11.7 Geometric Interpretation of Divergence

|

Quiz

|

**Final Exam**

|

**Answers to Problems**

|

**Index**
**PREFACE**

In the first year of calculus we study limits, derivatives, and integrals of functions with a single input, and a single output. The transition to _advanced calculus_ is made when we generalize the notion of "function" to something which may have multiple inputs and multiple outputs. In this more general context limits, derivatives, and integrals take on new meanings and have new geometric interpretations. For example, in first-year calculus the derivative represents the slope of a tangent line at a specified point. When dealing with functions of multiple variables there may be many tangent lines at a point, so there will be many possible ways to differentiate.

The emphasis of this book is on developing enough familiarity with the material to solve difficult problems. Rigorous proofs are kept to a minimum. I have included numerous detailed examples so that you may see how the concepts really work. All exercises have detailed solutions that you can find at the end of the book. I regard these exercises, along with their solutions, to be an integral part of the material.

The present work is suitable for use as a stand-alone text, or as a companion to any standard book on the topic. This material is usually covered as part of a standard calculus sequence, coming just after the first full year. Names of college classes that cover this material vary greatly. Possibilities include _advanced calculus, multivariable calculus_ , and _vector calculus_. At schools with semesters the class may be called _Calculus III_. At quarter schools it may be _Calculus IV_.

The best way to use this book is to read the material in each section and then try the exercises. If there is any exercise you don't get, make sure you study the solution carefully. At the end of each chapter you will find a quiz to test your understanding. These short quizzes are written to be similar to one that you may encounter in a classroom, and are intended to take 20–30 minutes. They are not meant to test every idea presented in the chapter. The best way to use them is to study the chapter until you feel confident that you can handle anything that may be asked, and then try the quiz. You should have a good idea of how you did on it after looking at the answers. At the end of the text there is a final exam similar to one which you would find at the conclusion of a college class. It should take about two hours to complete. Use it as you do the quizzes. Study all of the material in the book until you feel confident, and then try it.

Advanced calculus is an exciting subject that opens up a world of mathematics. It is the gateway to linear algebra and differential equations, as well as more advanced mathematical subjects like analysis, differential geometry, and topology. It is essential for an understanding of physics, lying at the heart of electro-magnetics, fluid flow, and relativity. It is constantly finding new use in other fields of science and engineering. I hope that the exciting nature of this material is conveyed here.
**ACKNOWLEDGMENTS**

The author thanks the technical editor, Steven G. Krantz, for his helpful comments.
CHAPTER 1 |

---|---

**Functions of Multiple Variables**

**1.1 Functions**

The most common mental model of a _function_ is a machine. When you put some input in to the machine, you will always get the same output. Most of first year calculus dealt with functions where the input was a single real number and the output was a single real number. The study of advanced calculus begins by modifying this idea. For example, suppose your "function machine" took _two_ real numbers as its input, and returned a single real output? We illustrate this idea with an example.

**EXAMPLE 1-1**

Consider the function

_f_ ( _x_ , _y_ ) _= x_ 2 \+ _y_ 2

For each value of _x_ and _y_ there is one value of _f_ ( _x_ , _y_ ). For example, if _x_ = 2 and _y_ = 3 then

_f_ (2, 3) = 22 \+ 32 = 13

One can construct a table of input and output values for _f_ ( _x_ , _y_ ) as follows:

**Problem 1** _Evaluate the function at the indicated point_.

__. _f_ ( _x_ , _y_ ) = _x_ 2 \+ _y_ 3; ( _x_ , _y_ ) = (3, 2)

_. _

_. _

Unfortunately, plugging in random points does not give much enlightenment as to the behavior of a function. Perhaps a more visual model would help....

**1.2 Three Dimensions**

In the previous section we saw that plugging random points in to a function of two variables gave almost no enlightening information about the function itself. A far superior way to get a handle on a particular function is to picture its _graph_. We'll get to this in the next section. First, we have to say a few words about where such a graph exists.

Recall the steps required to graph a function of a single variable, like _g_ ( _x_ ) = 3 _x_. First, you set the function equal to a new variable, _y_. Then you plot all the points ( _x_ , _y_ ) where the equation _y_ = _g_ ( _x_ ) is true. So, for example, you would not plot (0, 2) because 0 ≠ 3 · 2. But you would plot (2, 6) because 6 = 3 · 2.

The same steps are required to plot a function of two variables, like _f_ ( _x_ , _y_ ). First, you set the function equal to a new variable, _z_. Then you plot all of the points ( _x_ , _y_ , _z_ ) where the function _z_ = _f_ ( _x_ , _y_ ) is true. So we are forced to discuss what it means to plot a point with three coordinates, like ( _x_ , _y_ , _z_ ).

Coordinate systems will play a crucial role in this book, so although most readers will have seen this, it is worth spending some time here. To plot a point with two coordinates such as ( _x_ , _y_ ) = (2, 3) the first step is to draw two perpendicular axes and label them _x_ and _y_. Then locate a point 2 units from the origin on the _x_ -axis and draw a vertical line. Next, locate a point 3 units from the origin on the _y_ -axis and draw a horizontal line. Finally, the point (2, 3) is at the intersection of the two lines you have drawn.

To plot a point with three coordinates the steps are just a bit more complicated. Let's plot the point ( _x_ , _y_ , _z_ ) = (2, 3, 2). First, draw _three_ mutually perpendicular axes. You will immediately notice that this is impossible to do on a sheet of paper. The best you can do is two perpendicular axes, and a third at some angle to the other two (see Figure 1-1). With practice you will start to see this third axis as a perspective rendition of a line coming out of the page. When viewed this way it will seem like it is perpendicular.

**Figure 1-1** Three mutually perpendicular axes, drawn in perspective

Notice the way in which we labeled the axes in Figure 1-1. This is a convention, i.e., something that mathematicians have just agreed to always do. The way to remember it is by the _right hand rule_. What you want is to be able to position your right hand so that your thumb is pointing along the _z_ -axis and your other fingers sweep from the _x_ -axis to the _y_ -axis when you make a fist. If the axes are labeled consistent with this then we say you are using a _right handed coordinate system_.

OK, let's now plot the point (2, 3, 2). First, locate a point 2 units from the origin on the _x_ -axis. Now picture a _plane_ which goes through this point, and is perpendicular to the _x_ -axis. Repeat this for a point 3 units from the origin on the _y_ -axis, and a point 2 units from the origin on the _z_ -axis. Finally, the point (2, 3, 2) is at the intersection of the three planes you are picturing.

Given the point ( _x_ , _y_ , _z_ ) one can "see" the quantities _x_ , _y_ , and _z_ as in Figure 1-2. The quantity _z_ , for example, is the distance from the point to the _xy_ -plane.

**Figure 1-2** Plotting the point (2, 3, 2)

**Problem 2** _Which of the following coordinate systems are right handed?_

**Problem 3** _Plot the following points on one set of axes:_

_1._ (1, 1, 1)

_2._ (1, −1, 1)

_3._ (−1, 1, −1)

**1.3 Introduction to Graphing**

We now turn back to the problem of visualizing a function of multiple variables. To graph the function _f_ ( _x_ , _y_ ) we set it equal to _z_ and plot all of the points where the equation _z_ = _f_ ( _x_ , _y_ ) is true. Let's start with an easy example.

**EXAMPLE 1-2**

Suppose _f_ ( _x_ , _y_ ) = 0. That is, _f_ ( _x_ , _y_ ) is the function that always returns the number 0, no matter what values of _x_ and _y_ are fed to it. The graph of _z_ = _f_ ( _x_ , _y_ ) = 0 is then the set of all points ( _x_ , _y_ , _z_ ) where _z =_ 0. This is just the _xy_ -plane.

Similarly, now consider the function _g_ ( _x_ , _y_ ) = 2. The graph is the set of all points where _z_ = _g_ ( _x, y_ ) = 2. This is a plane parallel to the _xy_ -plane at height 2.

We first learn to graph functions of a single variable by plotting individual points, and then playing "connect-the-dots." Unfortunately this method doesn't work so well in three dimensions (especially when you are trying to depict three dimensions on a piece of paper). A better strategy is to slice up the graph by various planes. This gives you several curves that you can plot. The final graph is then obtained by assembling these curves.

The easiest slices to see are given by each of the coordinate planes. We illustrate this in the next example.

**EXAMPLE 1-3**

Let's look at the function _f_ ( _x_ , _y_ ) = _x_ \+ 2 _y_. To graph it we must decide which points ( _x_ , _y_ , _z_ ) make the equation _z_ = _x_ \+ 2 _y_ true. The _xz_ -plane is the set of all points where _y =_ 0. So to see the intersection of the graph of _f_ ( _x_ , _y_ ) and the _xz_ -plane we just set _y_ = 0 in the equation _z_ = _x_ \+ 2 _y_. This gives the equation _z_ = _x_ , which is a line of slope 1, passing through the origin.

Similarly, to see the intersection with the _yz_ -plane we just set _x_ = 0. This gives us the equation _z_ = 2 _y_ , which is a line of slope 2, passing through the origin.

Finally, we get the intersection with the _xy_ -plane. We must set _z =_ 0, which gives us the equation 0 = _x_ \+ 2 _y_. This can be rewritten as   We conclude this is a line with slope

The final challenge is to put all of this information together on one set of axes. See Figure 1-3. We see three lines, in each of the three coordinate planes. The graph of _f_ ( _x_ , _y_ ) is then some shape that meets each coordinate plane in the required line. Your first guess for the shape is probably a plane. This turns out to be correct. We'll see more evidence for it in the next section.

**Figure 1-3** The intersection of the graph of _x_ \+ 2 _y_ with each coordinate plane is a line through the origin

**Problem 4** _Sketch the intersections of the graphs of the following functions with each of the coordinate planes_.

_. _2 _x_ \+ 3 _y_

_. x_2 \+ _y_

_. x_2 \+ _y_ 2

_. _2 _x_ 2 \+ _y_ 2

_. _

_. x_2 − _y_ 2

**1.4 Graphing Level Curves**

It's fairly easy to plot the intersection of a graph with each coordinate plane, but this still doesn't always give a very good idea of its shape. The next easiest thing to do is sketch some level curves. These are nothing more than the intersection of the graph with horizontal planes at various heights. We often sketch a "bird's eye view" of these curves to get an initial feeling for the shape of a graph.

**EXAMPLE 1-4**

Suppose _f_ ( _x_ , _y_ ) = _x_ 2 \+ _y_ 2. To get the intersection of the graph with a plane at height 4, say, we just have to figure out which points in ℝ3 satisfy _z_ = _x_ 2 \+ _y_ 2 and _z_ = 4. Combining these equations gives 4 = _x_ 2 \+ _y_ 2, which we recognize as the equation of a circle of radius 2. We can now sketch a view of this intersection from above, and it will look like a circle in the _xy_ -plane. See Figure 1-4.

**Figure 1-4** (a) The intersection of _z_ = _x_ 2 \+ _y_ 2 with a plane at height 4. (b) A top view of the intersection

The reason why we often draw level curves in the _xy_ -plane as if we were looking down from above is that it is easier when there are many of them. We sketch several such curves for _z_ = _x_ 2 \+ _y_ 2 in Figure 1-5.

**Figure 1-5** Several level curves of _z_ = _x_ 2 \+ _y_ 2

You have no doubt seen level curves before, although they are rarely as simple as in Figure 1-5. For example, in Figure 1-6 we see a topographic map. The lines indicate constant elevation. In other words, these lines are the level curves for the function which gives elevation. In Figure 1-7 we have shown a weather map, with level curves indicating lines of constant temperature. You may see similar maps in a good weather report where level curves represent lines of constant pressure.

**Figure 1-6** A topographic map

**Figure 1-7** A weather map shows level curves

**EXAMPLE 1-5**

We now let _f_ ( _x_ , _y_ ) = _xy_. The intersection with the _xz_ -plane is found by setting _y_ = 0, giving us the function _z_ = 0. This just means the graph will include the _x_ -axis. Similarly, setting _x_ = 0 gives us _z_ = 0 as well, so the graph will include the _y_ -axis. Things get more interesting when we plot the level curves. Let's set _z_ = _n_ , where _n_ is an integer. Solving for _y_ then gives us   This is a hyperbola in the first and third quadrant for _n_ > 0, and a hyperbola in the second and fourth quadrant when _n_ < 0. We sketch this in Figure 1-8.

**Figure 1-8** Level curves of _z_ = _xy_

**Problem 5** _Sketch several level curves for the following functions_.

_. _2 _x_ **_+_** 3 _y_

_. x_2 \+ _y_

_. _

_. x_2 – _y_ 2

**Problem 6** _The level curves for the following functions are all circles. Describe the difference between how the circles are arranged_.

_. x_2 \+ _y_ 2

_. _

_. _

_. _sin( _x_ 2 \+ _y_ 2)

**1.5 Putting It All Together**

We have now amassed enough tools to get a good feeling for what the graphs of various functions look like. Putting it all together can be quite a challenge. We illustrate this with an example.

**EXAMPLE 1-6**

Let _f_ ( _x_ , _y_ ) = _x_ 2 \+ _y_ 2. In Problem 4 you found that the intersections with the xz- and yz-coordinate planes were parabolas. In Example 1-4 we saw that the level curves were circles. We put all of this information together in Figure 1-9(a). Figure 1-9(b) depicts the entire surface which is the graph. This figure is called a _paraboloid_.

**Figure 1-9** Sketching the paraboloid _z_ = _x_ 2 \+ _y_ 2

Graph sketching is complicated enough that a second example may be in order.

**EXAMPLE 1-7**

In Figure 1-10 we put together the level curves of _f_ ( _x_ , _y_ ) = _xy_ , found in Example 1-5, to form its graph. The three-dimensional shape formed is called a _saddle_.

**Figure 1-10** Several level curves of _z_ = _xy_ piece together to form a saddle

**Problem 7** _Use your answers toProblems 4 and  to sketch the graphs of the following functions:_

_. _2 _x_ **_+_** 3 _y_

_. x_2 \+ _y_

_. _2 _x_ 2 \+ _y_ 2

_. _

_. x_2 − _y_ 2

**1.6 Functions of Three Variables**

There is no reason to stop at functions with two inputs and one output. We can also consider functions with three inputs and one output.

**EXAMPLE 1-8**

Suppose

_f_ ( _x_ , _y_ , _z_ ) = _x_ \+ _xy_ \+ _yz_ 2

Then _f_ (1, 1, 1) = 3 and _f_ (0, 1, 2) = 4.

To graph such a function we would need to set it equal to some fourth variable, say w, and draw a picture in a space where there are four perpendicular axes, _x_ , _y_ , _z_ , and _w_. No one can visualize such a space, so we will just have to give up on graphing such functions. But all hope is not lost. We can still describe surfaces in three dimensions that are the level sets of such functions. This is not quite as good as having a graph, but it still helps give one a feel for the behavior of the function.

**EXAMPLE 1-9**

Suppose

_f_ ( _x_ , _y_ , _z_ ) = _x_ 2 \+ _y_ 2 \+ _z_ 2

To plot level sets we set _f_ ( _x_ , _y_ , _z_ ) equal to various integers and sketch the surface described by the resulting equation. For example, when _f_ ( _x_ , _y_ , _z_ ) = 1 we have

1 = _x_ 2 \+ _y_ 2 \+ _z_ 2

This is precisely the equation of a sphere of radius 1. In general the level set corresponding to _f_ ( _x_ , _y_ , _z_ ) = _n_ will be a sphere of radius

**Problem 8** _Sketch the level set corresponding to f_ ( _x_ , _y_ , _z_ ) = 1 _for the following functions_.

_. f_ ( _x_ , _y_ , _z_ ) = _x_ 2 \+ _y_ 2 − _z_ 2

__. _f_ ( _x_ , _y_ , _z_ ) = _x_ 2 − _y_ 2 − _z_ 2

**1.7 Parameterized Curves**

In the previous sections of this chapter we studied functions which had multiple inputs, but one output. Here we examine the opposite scenario: functions with one input and multiple outputs. The input variable is referred to as the _parameter_ , and is best thought of as time. For this reason we often use the variable _t_ , so that in general such a function might look like

_c_ ( _t_ ) = ( _f_ ( _t_ ), _g_ ( _t_ ))

If we fix a value of _t_ and plot the two outputs we get a point in the plane. As _t_ varies this point moves, tracing out a curve, _C_. We would then say _C_ is a curve that is _parameterized by c_ ( _t_ ).

**EXAMPLE 1-10**

Suppose _c_ ( _t_ ) = (cos _t_ , sin _t_ ).Then c(0) = (1, 0) and   If we continue to plot points we see that _c_ ( _t_ ) traces out a circle of radius 1. Indeed, since

cos2 _t_ \+ sin2 _t_ = 1

the coordinates of _c_ ( _t_ ) satisfy _x_ 2 \+ _y_ 2 = 1, the equation of a circle of radius 1. In Figure 1-11 we plot the circle traced out by _c_ ( _t_ ), along with additional information which tells us what value of _t_ yields selected point of the curve.

**Figure 1-11** The function _c_ ( _t_ ) = (cos _t_ , sin _t_ ) parameterizes a circle of radius 1

**EXAMPLE 1-11**

The function _c_ ( _t_ ) = (cos _t_ 2, sin _t_ 2) also parameterizes a circle or radius 1, like the parameterization given in Example 1-10. The difference between the two parameterizations can be seen by comparing the spacing of the marked points in Figure 1-11 with those of Figure 1-12. If we think of _t_ as time, then the parameterization depicted in Figure 1-12 represents a point moving around the circle faster and faster.

**Figure 1-12** The function _c_ ( _t_ ) = (cos _t_ 2, sin _t_ 2) parameterizes a circle of radius 1 in a different way

**EXAMPLE 1-12**

Now let _c_ ( _t_ ) = ( _t_ cos _t_ , _t_ sin _t_ ). Plotting several points shows that _c_ ( _t_ ) parameterizes a curve that spirals out from the origin, as in Figure 1-13.

**Figure 1-13** The function _c_ ( _t_ ) = ( _t_ cos _t_ , _t_ sin _t_ ) parameterizes a spiral

Parameterizations can also describe curves in three-dimensional space, as in the next example.

**EXAMPLE 1-13**

Let _c_ ( _t_ ) = (cos _t_ , sin _t_ , _t_ ). If the third coordinate were not there then this would describe a point moving around a circle. Now as _t_ increases the height off of the _xy_ -plane, i.e., the _z_ -coordinate, also increases. The result is a spiral, as in Figure 1-14.

**Figure 1-14** The function _c_ ( _t_ ) = (cos _t_ , sin _t_ , _t_ ) parameterizes a curve that spirals around the _z_ -axis

**Problem 9** _Sketch the curves parameterized by the following:_

_. _( _t_ , _t_ )

_. _( _t_ , _t_ 2)

_. _( _t_ 2, _t_ )

_. _( _t_ 2, _t_ 3)

_. _(cos 2 _t_ , sin 3 _t_ )

**Problem 10** _The functions given inExamples 1-10 and 1-11 parameterize the same circle in different ways. Describe the difference between the two parameterizations for negative values of t_.

**Problem 11** _Find a parameterization for the graph of the function y_ = _f_ ( _x_ ).

**Problem 12** _Describe the difference between the following parameterized curves:_

_. c_( _t_ ) = (cos _t_ , sin _t_ , _t_ 2)

_. _

_. c_( _t_ ) = ( _t_ cos _t_ , _t_ sin _t_ , _t_ )

**Quiz**

**Problem 13**

_. Determine if the coordinate system pictured is left or right handed_.

_. Let_

_a. Sketch the intersections of the graph of f_ ( _x_ , _y_ ) _with the xy-plane, the xz-plane, and the yz-plane_.

_b. Sketch the level curves for f_ ( _x_ , _y_ ).

_c. Sketch the graph of f_ ( _x_ , _y_ ).

_. Sketch the curve parameterized by c_( _t_ ) = (2 cos _t_ , 3 sin _t_ ).
CHAPTER 2 |

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Fundamentals of Advanced Calculus

2.1 Limits of Functions of Multiple Variables

The study of calculus begins in earnest with the concept of a limit. Without this one cannot define derivatives or integrals. Here we undertake the study of limits of functions of multiple variables.

Recall that we say   if you can make _f_ ( _x_ ) stay as close to _L_ as you like by restricting _x_ to be _close enough_ to a. Just how close "close enough" is depends on how close you want _f_ ( _x_ ) to be to _L_.

Intuitively, if   we think of the values of _f_ ( _x_ ) as getting closer and closer to _L_ as the value of _x_ gets closer and closer to a. A key point is that it should not matter _how_ the values of _x_ are approaching a. For example, the function

does not have a limit as _x_ → 0. This is because as _x_ approaches 0 from the right the values of the function _f_ ( _x_ ) approach 1, while the values of _f_ ( _x_ ) approach −1 as _x_ approaches 0 from the left.

The definition of limit for functions of multiple variables is very similar. We say

if you can make _f_ ( _x_ , _y_ ) stay as close to _L_ as you like by restricting ( _x_ , _y_ ) to be _close enough_ to ( _a_ , _b_ ). Again, just how close "close enough" is depends on how close you want _f_ ( _x_ , _y_ ) to be to _L_.

Once again, the most useful way to think about this definition is to think of the values of _f_ ( _x_ , _y_ ) as getting closer and closer to _L_ as the point ( _x_ , _y_ ) gets closer and closer to the point ( _a_ , _b_ ).The difficulty is that there are now an infinite number of directions by which one can approach ( _a_ , _b_ ).

**EXAMPLE 2-1**

Suppose _f_ ( _x_ , _y_ ) is given by

We consider

First, let's see what happens as ( _x_ , _y_ ) approaches (0, 0) along the _x_ -axis. For all such points we know _y_ = 0, and so

Now consider what happens as ( _x_ , _y_ ) approaches (0, 0) along the y-axis. For all such points we have _x_ = 0, and so

We conclude the values of _f_ ( _x_ , _y_ ) approach different numbers if we let ( _x_ , _y_ ) approach (0, 0) from different directions. Thus we say   does not exist.

Showing that a limit does not exist can be very difficult. Just because you can find multiple ways to come at ( _a_ , _b_ ) so that the values of _f_ ( _x_ , _y_ ) approach the same number _L_ does not necessarily mean   There might be some way to approach ( _a_ , _b_ ) that you haven't tried that gives a different number. This is the key to the definition of limit. We say the function has a limit only when the values of _f_ ( _x_ , _y_ ) approach the same number no matter how ( _x_ , _y_ ) approaches ( _a_ , _b_ ). We illustrate this in the next two examples.

**EXAMPLE 2-2**

Suppose _f_ ( _x_ , _y_ ) is given by

As we let ( _x_ , _y_ ) approach (0, 0) along the _x_ -axis (where _y_ = 0) we have

Similarly, as we let ( _x_ , _y_ ) approach (0, 0) along the _y_ -axis (where _x_ = 0) we have

But if we let ( _x_ , _y_ ) approach (0, 0) along the line _y_ = _x_ we have

So once again we find   does not exist.

Our third example is the trickiest.

**EXAMPLE 2-3**

Let

As ( _x_ , _y_ ) approaches (0, 0) along the _x_ -axis (where _y_ = 0) we have

As ( _x_ , _y_ ) approaches (0, 0) along the _y_ -axis (where _x_ = 0) we have

If we approach (0, 0) along the line _y_ = _x_ we get

As _x_ approaches 0 we have

So far it is looking like perhaps

since, as ( _x_ , _y_ ) approaches (0, 0) along the _x_ -axis, the _y_ -axis, and the line _y_ = _x_ , the values of _f_ ( _x_ , _y_ ) approach 0. But what happens if we let ( _x_ , _y_ ) approach (0, 0) along the _curve y_ = _x_ 2? In this case

We can evaluate the limit of this function as _x_ approaches 0 by dividing the numerator and denominator by _x_ 4.

So again the limit does not exist.

**Problem 14** _Show that the following limits do not exist:_

_1_.

_2_.

_3_.

_4_.

2.2 Continuity

We say a function _f_ ( _x_ , _y_ ) is _continuous_ at ( _a_ , _b_ ) if its limit as ( _x_ , _y_ ) approaches ( _a_ , _b_ ) equals its value there. In symbols we write

Most functions you can easily write down are continuous at every point of their domain. Hence, what you want to avoid are points outside of the domain, where you may have

1. Division by zero.

2. Square roots of negatives.

3. Logs of nonpositive numbers.

4. Tangents of odd multiples of

In each of these situations the function does not even exist, in which case it is certainly not continuous. But even if the function exists it may not have a limit. And even if the function exists, and the limits exist, they may not be equal.

**EXAMPLE 2-4**

Suppose

There is no zero in the denominator when ( _x_ , _y_ ) = (1, 1), so _f_ ( _x_ , _y_ ) is continuous at (1, 1).

**EXAMPLE 2-5**

Evaluate

There are no values of _x_ and _y_ that will make the denominator 0, so the function is continuous everywhere. Since the value of a continuous function equals its limit, we can evaluate the above simply by plugging in (0, 0).

**Problem 15** _Find the domain of the following functions:_

__.

|
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__.

|

__.

|

**Problem 16** _Consider the function_

_Is f_ ( _x_ , _y_ ) _continuous at_ (0, 0)?

Quiz

**Problem 17**

_. Show that the function_

_does not have a limit as_ ( _x_ , _y_ ) → (0, 0).

_. Is the function_

_continuous at_ (0, 0)?

_. Find the domain of the function_

CHAPTER 3 |

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**Derivatives**

**3.1 Partial Derivatives**

What shall we mean by the _derivative_ of _f_ ( _x_ , _y_ ) at a point ( _x_ 0, _y_ 0)? Just as in one variable calculus, the answer is the slope of a tangent line. The problem with this is that there are multiple tangent lines one can draw to the graph of _z_ = _f_ ( _x_ , _y_ ) at any given point. Which one shall we pick to represent the derivative? The answer is another question: " _Which_ derivative?" We will see that at any given point there are lots of possible derivatives; one for each tangent line.

Another way to think about this is as follows. Suppose we are at the point ( _x_ 0, _y_ 0) and we start moving. While we do this we keep track of the quantity _f_ ( _x_ , _y_ ). The rate of change that we observe is the derivative, but the answer may depend on which direction we are traveling.

Suppose, for example, that we are observing the function _f_ ( _x_ , _y_ ) _= x_ 2 _y_ , while moving through the point (1, 1) with unit speed. Suppose further that we are traveling parallel to the _x_ -axis, so that our _y_ -coordinate is always one. We would like to know the observed rate of change of _f_ ( _x_ , _y_ ). Since the _y_ -coordinate is always one the values of _f_ ( _x_ , _y_ ) that we observe are always determined by our _x_ -coordinate: f ( _x_ , 1) = _x_ 2. The rate of change of this function is given by its derivative: 2 _x_. Finally, when _x_ = 1 this is the number 2.

The above is a particularly easy computation. Given any function, if you are traveling in a direction which is parallel to the _x_ -axis then your _y_ -coordinate is fixed. Plugging this number in for _y_ then gives a function of just _x_ , which we can differentiate. Here's another example.

**EXAMPLE 3-1**

We compute the rate of change of _f_ ( _x_ , _y_ ) = _x_ 3 _y_ 3 at the point (1, 2), when we are traveling parallel to the _x_ -axis. During our travels the value of _y_ stays fixed at 2. Hence, the values of the function we are observing are determined by our _x_ -coordinate: _f_ ( _x_ , 2) _=_ 8 _x_ 3.The derivative of this function is then 24 _x_ 2, which takes on the value 24 when _x_ is one.

What if we wanted to repeat our computations, with different values of _y_? It would be helpful to keep the letter " _y_ " in our computations, and plug in the value at the very end. Notice that when we plugged in a number for _y_ it became a constant, and was treated as such when we differentiated with respect to _x_. If we leave the letter _y_ in our computations we can still treat it as a constant.

**EXAMPLE 3-2**

Let _f_ ( _x_ , _y_ ) = _x_ \+ _xy_ \+ _y_ 2. We wish to treat _y_ as a constant, just as if we had plugged in a number for it, and take the derivative with respect to _x_. Recall that the derivative of a sum of functions is the sum of the derivatives. So we will discuss the derivatives of each of the terms of _x_ \+ _xy_ \+ _y_ 2 individually.

There is no occurrence of _y_ in the first term, so it is particularly easy. Its derivative is just one.

The second term is a little trickier. Since we are treating _y_ as a constant this is of the form _const·x_. The derivative of such a function is just _const_. So the derivative of _xy_ is just _y_.

Finally, the quantity _y_ 2 is also a constant. The derivative of a constant is zero. Hence our answer is 1 + _y_.

Notice in the above example that if we thought of _x_ as constant, and _y_ as the variable, then the derivative would have been very different. We need some notation to tell us what is changing and what is being kept constant. We use the symbols   to represent the _partial derivative with respect to x_. This means _x_ is considered a variable, every other letter is a constant, and we differentiate. Similarly, the notation   is the _partial derivative with respect to y_.

**EXAMPLE 3-3**

Again we let _f_ ( _x_ , _y_ ) = _x_ \+ _xy_ \+ _y_ 2. We compute

There are various other ways to think of partial derivatives that are useful. One is completely algebraic. Recall that the derivative with respect to _x_ of a function _f_ ( _x_ ) of one variable is defined as the limit

The partial with respect to _x_ is defined similarly. Just remember that _y_ is kept constant, so that _f_ ( _x_ , _y_ ) really becomes a function of just _x_. We then apply the above definition to get:

The partial with respect to _y_ is defined similarly

There is yet another way to think about the partial derivative. We began this section by claiming that the derivative will still represent the slope of a tangent line. In the following figure we see the graph of an equation _z_ = _f_ ( _x_ , _y_ ). Through the point ( _x_ 0, _y_ 0) in the _xy_ -plane there is also drawn a vertical plane parallel to the _xz_ -plane. The intersection of this vertical plane with the graph is a curve. The slope of the tangent line to this curve is exactly the value of the partial derivative with respect to _x_ at ( _x_ 0, _y_ 0). To see the partial derivative with respect to _y_ we would have a similar picture, where the vertical plane is parallel to the _yz_ -plane.

**Problem 18** _Compute_   _and_   _for the following functions_.

__. _x_ \+ _xy_

__. _x_ ln _y_

__.

**Problem 19** _Compute_   and   _for the following functions_.

__. _x_ 2 _y_ 3

__.

**Problem 20** _For the function_ f ( _x_ , _y_ ) = − _x_ \+ _xy_ 2 − _y_ 2 _find all places where both_   _are zero_.

**3.2 Composition and the Chain Rule**

**3.2.1 COMPOSITION WITH PARAMETERIZED CURVES**

Suppose we have a parameterized curve _ϕ_ ( _t_ ) _=_ ( _x_ ( _t_ ), _y_ ( _t_ )) in the plane. That is, for a given value of _t_ we are given the numbers _x_ ( _t_ ) and _y_ ( _t_ ), which we visualize as a point in the plane. We can also take these two numbers and plug them in to a function f ( _x_ , _y_ ). The result is the composition _f_ ( _ϕ_ ( _t_ )). Notice that only one number goes in to this function, and only one number comes out.

**EXAMPLE 3-4**

Let _f_ ( _x_ , _y_ ) = _x_ 2 \+ _y_ 2. Let _ϕ_ ( _t_ ) = ( _t_ cos _t_ , _t_ sin _t_ ). Then

_f_ ( _ϕ_ ( _t_ )) = _f_ ( _t_ cos _t_ , _t_ sin _t_ ) = _t_ 2 cos2 _t_ \+ _t_ 2 sin2 _t = t_ 2

There are various ways to visualize the composition. One is to imagine the graph of _z_ = _f_ ( _x_ , _y_ ) in three dimensions. Then draw the parameterized curve _ϕ_ ( _t_ ) in the _xy_ -plane, and imagine a vertical piece of paper curled up so that it sits on this curve. Now mark where the paper intersects the graph of _f_ ( _x_ , _y_ ), and unroll it. The result is the graph of _f_ ( _ϕ_ ( _t_ )).

Since _f_ ( _ϕ_ ( _t_ )) is a function of one variable we can talk about its derivative just as if we were in a first term calculus class. But what we want to do here is relate it to the derivatives of _f_ ( _x_ , _y_ ), _x_ ( _t_ ), and _y_ ( _t_ ). The formula we end up with is the multivariable calculus version of the "chain rule"

**EXAMPLE 3-5**

Let _f_ ( _x_ , _y_ ) and _ϕ_ ( _t_ ) be defined as in the previous example. We wish to compute   when   Note first that

To use the chain rule we must compute the partial derivatives of _f_ ( _x_ , _y_ ) at this point. First note that

Thus,

We also need the derivatives of _x_ ( _t_ ) and _y_ ( _t_ ) when

Thus,

Finally, we have

Now that we have gone through the pain of using the chain rule to compute the derivative, it should be pointed out that the same answer could have been found _much_ faster directly. We know from the previous example that _f_ ( _ϕ_ ( _t_ )) _= t_ 2. So the derivative function is 2 _t_. Plugging in   then immediately gives

**Problem 21** _Let ϕ_ ( _t_ ) = (2 _t_ , _t_ 2). _Suppose you don_ ' _t know what f_ ( _x_ , _y_ ) _is, but you know_   _and_   _Compute_

**Problem 22** _Suppose you don_ ' _t know what ϕ_ ( _t_ ) _=_ ( _x_ ( _t_ ), _y_ ( _t_ )) _is, but you know ϕ_ (2) = (1, 3), _x_ ′(2) _= −_ 2, _and y_ ′(2) = 1 _. Let f_ ( _x_ , _y_ ) _= x_ 2 _y. Compute_

**3.2.2** **COMPOSITION OF FUNCTIONS OF MULTIPLE VARIABLES**

We now look at the idea of composition with more complicated types of functions, as in our next example.

**EXAMPLE 3-6**

Let _x_ ( _u_ , _v_ ) = _uv_ and _y_ ( _u_ , _v_ ) = _u_ 2 \+ _v_ 2. Suppose _f_ ( _x_ , _y_ ) = _x_ \+ _y_. Then we may form the composition _f_ ( _x_ ( _u_ , _v_ ), _y_ ( _u_ , _v_ )) as follows.

_f_ ( _x_ ( _u_ , _v_ ), _y_ ( _u_ , _v_ )) = _x_ ( _u_ , _v_ ) + _y_ ( _u_ , _v_ ) = _uv_ \+ _u_ 2 \+ _v_ 2

Notice that the result is a function whose input is a pair of numbers, _u_ and v, and whose output is a single number. Hence, we may talk about the partial derivatives, with respect to _u_ and _v_ , of the function given by composition. The result is given by another variant of the chain rule. As in the previous example, we will begin with the functions _x_ ( _u_ , _v_ ), _y_ ( _u_ , _v_ ), and _f_ ( _x_ , _y_ ). The following formulas give the partial derivatives of the composition _f_ ( _x_ ( _u_ , _v_ ), _y_ ( _u_ , _v_ )).

At this point it is mathematically meaningless to think of terms like "∂ _x_ " as entities in themselves that can be canceled, but thinking this way may help you remember the above formulas. Note that the formula for the partial derivative of _f_ with respect to _u_ , for example, has two terms. If you cancel the "quantity" ∂ _x_ in the first term, and ∂ _y_ in the second, you are left with   in both.

**EXAMPLE 3-7**

Let _f_ ( _x_ , _y_ ) _= x_ 2 \+ _xy_. Suppose we don't know _x_ ( _u_ , _v_ ) or _y_ ( _u_ , _v_ ), but we know _x_ (1, 2) _=_ 3, _y_ (1, 3) _= −_ 2,   and   Then we may use the chain rule to compute   To do this we will need to know   at the point ( _x_ (1, 2), _y_ (1, 2)) = (3, −2).

We now employ the chain rule:

**Problem 23** _The following table lists values for a function f_ ( _x_ , _y_ ) _and its partial derivatives_.

_Let x_ ( _u_ , _v_ ) _= uv and y_ ( _u_ , _v_ ) = _u_ \+ _v_ 2. _Find the following partial derivatives at the indicated points_.

__.

__.

**Problem 24** _Let f_ ( _x_ , _y_ ) = sin( _x + y_ ), _x_ ( _u_ , _v_ ) = _u_ \+ _v_ , _and y_ ( _u_ , _v_ ) = _u_ − _v. Find the following quantities when_

__. _f_ ( _x_ , _y_ )

__.

__.

**3.3 Second Partials**

In the previous sections we saw that the partial derivatives of a function _f_ ( _x_ , _y_ ) are also functions of _x_ and _y_. We can therefore take the derivative again with respect to either variable.

**EXAMPLE 3-8**

Let _f_ ( _x_ , _y_ ) _= x_ 2 _y_ 3. The partial derivatives are   and   We can take the partial derivative again of both of these functions with respect to either variable:

As is customary, we adopt the following shorthand notations for the second derivatives:

The quantities   are called the _mixed partials_. The above example illustrates an amazing fact: Under reasonable conditions the mixed partials are always equal! The "reasonable conditions" are only that the mixed partials exist and are themselves continuous functions. The proof of this goes back to the limit definition of the partial derivative.

**Problem 25** _Find all second partial derivatives of the following functions_.

__. _f_ ( _x_ , _y_ ) = _xy_

__. _f_ ( _x_ , _y_ ) = _x_ 2 − _y_ 2

__. _f_ ( _x_ , _y_ ) _=_ sin( _xy_ 2)

**Quiz**

**Problem 26** _Let f_ ( _x_ , _y_ ) _= x_ 2 _y_ \+ _x_ 3 _y_ 2.

__. _Find_

__. _If ϕ_ ( _t_ ) = ( _t_ 2, _t_ − 1), _then what is f_ ( _ϕ_ ( _t_ )) _?_

__. _Suppose you don't know what ψ_ ( _t_ ) _=_ ( _x_ ( _t_ ), _y_ ( _t_ )) _is, but you know ψ_ (2) _=_ (1, 1),   _and_   _Find the derivative of f_ ( _ψ_ ( _t_ )) _when t_ = 2.

__. _Suppose x and y are functions of u and v_ , _x_ ( _u_ , _v_ ) = _u_ 2 \+ _v_ , _and y_ (1, 1) = 1. _What would_   _have to be when_ ( _u_ , _v_ ) = (1, 1), _if_
CHAPTER 4 |

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**Integration**

**4.1 Integrals over Rectangular Domains**

The integral of a function of one variable gives the area under the graph and above an interval on the _x_ -axis called the _domain of integration_. For functions of two variables the graph is a surface. We will interpret the act of integration as that of finding the volume below the surface, and above some region in the _xy_ -plane. Eventually, we will examine how to do this with very general-shaped regions. We begin by looking at rectangles.

Let _R_ be the region of the _xy_ -plane pictured above. Suppose we want to find the volume of the region in  3 which sits above _R_ , and below the graph of _z_ = _f_ ( _x_ , _y_ ), as in the following figure. We do this by following these steps.

1. Begin by choosing a grid of points {( _x i_, _y j_)} in _R_ , so that the horizontal and vertical spacings between adjacent points are Δ _x_ and Δ _y_ , respectively.

2. Connect these grid points to break up _R_ into rectangles. Note that the area of each rectangle is Δ _x_ Δ _y_.

3. We now draw a box of height _f_ ( _x_ _i_ , _y_ _j_ ) above each such rectangle to get a figure which approximates the desired volume. The volume of each box is its length × width × height, which is _f_ ( _x _i__ , _y _j__ )Δ _x_ Δ _y_.

4. We now add up the volumes of all of these boxes to obtain the quantity

5. Finally, we repeat this process indefinitely, each time using a grid with Δ _x_ and Δ _y_ smaller and smaller. The corresponding figures that we get approximate the desired volume more and more. In the limit we get what we're after, which we denote as

Basic properties of summation and limits allow us to rearrange the above equation as follows:

The quantity in the brackets   is exactly the definition of what you get from _f_ ( _x_ , _y_ ) when you fix _y_ and integrate _x_ , as in the following example.

**EXAMPLE 4-1**

Let _f_ ( _x_ , _y_ ) = _x_ 2 \+ _y_ 3.If we fix _y_ = 3 then this is the function _f_ ( _x_ ) _= x_ 2 \+ 27. We may now integrate this function over some range of values of _x_ , such as [0, 2]:

If we don't substitute the value 3 for _y_ , but we still think of _y_ as a constant, very little changes:

In the above example, notice that when we think of _y_ as a constant and we integrate with respect to _x_ , our answer is a function of _y_. We may now integrate this new function with respect to y. This is precisely what we are instructed to do in the limit we obtained for

**EXAMPLE 4-2**

Let _f_ ( _x_ , _y_ ) = _xy_ 2. Suppose _Q_ is the rectangle with vertices at (1, 1), (2, 1), (1, 4), and (2, 4). To find the volume under the graph of _f_ and above this rectangle we wish to compute

We work inside the brackets first. To do this integral we must pretend _y_ is a constant.

We now have

Note that in the definition of   we could have rearranged the terms so that the integral with respect to _y_ was inside the brackets. The fact that the answer does not depend on what order you integrate is called _Fubini's Theorem_.

**EXAMPLE 4-3**

Let _S_ be the rectangle with vertices at (1, 0), (2, 0), (1, 1), and (2, 1). We integrate the function   We compute the integral of _f_ ( _x_ , _y_ ) over _S_ by doing the integral with respect to _y_ first:

To save space the integral   is often written as

**Problem 27** _Compute the following:_

__.

|
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__.

|

__.

|

**Problem 28** _Let R be the rectangle in the xy-plane with corners at_ (−1,−1), (−1, 0), (2,−1), _and_ (2, 0). _Find the volume below the graph of z_ = _x_ 3 _y and above R_.

**Problem 29** _Let R be the rectangle with vertices at_ (0, 0), (1, 0), (0, 1), _and_ (1, 1). _Find a formula for_

**Problem 30** _Let V be the volume below the graph of_  , _and above the rectangle with corners at_ (0, 0), (2, 0), (0, 3), _and_ (2, 3). _Find the area of the intersection of V with the vertical plane through the point_ (1, 1) _which is parallel to the xz-plane_.

**4.2 Integrals over Nonrectangular Domains**

In the previous section we saw how to compute the volume under the graph of a function and above a rectangle. But what if we are interested in the volume which lies above a nonrectangular area? For example, suppose now that _R_ is the region of the _xy_ -plane that is bounded by the graph of _y_ = _g_ ( _x_ ), the _x_ -axis, and the line _x_ = 1. We would like to find the volume of the figure _V_ which lies below the graph of _z_ = _f_ ( _x_ , _y_ ) and above the region _R_.

In the previous section we computed volume by breaking up the region in question into small boxes. Our strategy here is to cut it into "slabs" parallel to the _yz_ -plane. To specify the location of each slab we must give a value of _x_. Hence, the volume of each slab, calculated as an integral with respect to _y_ , is a function of _x_. Adding the slabs up is just like integrating with respect to _x_.

We follow these steps:

1. Choose points { _x_ _i_ } in the interval [0, 1].

2. Compute the area _A_ ( _x_ _i_ ) of the intersection of _V_ with the plane _x = x i_. We do this by plugging _xi_ in for the variable _x_ in the function _f_ ( _x_ , _y_ ) and then integrating with respect to _y_. Notice that _y_ ranges from 0 to _g_ ( _xi_ ) on this domain. So the requisite area is given by

3. The volume of a thin slab is then given by

4. Adding up the volumes of all of the slabs thus gives

5. Finally, the desired volume is obtained from this quantity by choosing smaller and smaller slabs. This is equivalent to taking a limit as Δ _x_ tends toward 0.

The key to understanding this formula is the limits of integration. Each slab is parallel to the _y_ -axis. The area of the side of the slab is thus computed as an integral with respect to _y_. But the range of values that _y_ can take on depends on what the value of _x_ is. Hence, the limits of integration of the inner integral depend on _x_.

**EXAMPLE 4-4**

Let _R_ be the region in the _xy_ -plane bounded by the graph of _y_ = _x_ 2, the _x_ -axis, and the line _x =_ 1. We compute the volume below the graph of _z_ = _xy_ 2 and above _R_ as follows:

**EXAMPLE 4-5**

We now use the above ideas to compute a more complicated volume. Let _Q_ be the region of the _xy_ -plane bounded by the graphs of _y_ = _x_ 2 and _y_ = 1 − _x_ 2. We wish to determine the volume that lies below the graph of _z_ = _x_ 3 \+ _y_ and above Q.

The above figure depicts the region _Q_. Notice that for a fixed value _x_ 0 of _x_ the values of _y_ range from _x_ 2 to 1 − _x_ 2. This tells us the limits of integration for the inner integral. Now notice that the smallest and largest possible values of _x_ are where the graphs of _x_ 2 and 1 **−** _x_ 2 coincide. To find this we solve

_x_ 2 = 1 − _x_ 2

Which implies

2 _x_ 2 = 1

And hence,

This tells us the limits of integration for the outer integral.We now compute the desired volume

Sometimes the problem is set up so that integrating with respect to _x_ first is more natural, as the following example illustrates.

**EXAMPLE 4-6**

Let _S_ be the region of the _xy_ -plane bounded by the graph of _y_ = _x_ 2, the _y_ -axis, and the lines _y =_ 1 and _y =_ 2. We will integrate the function _f_ ( _x_ , _y_ ) _= x_ over this domain.

The domain _S_ is pictured in the following figure. Notice that for a fixed value _y_ 0 of _y_ the range of values that _x_ can take on goes from 0 to   This tells us the limits of integration for the inner integral. The smallest value _y_ 0 can be is 1 and the largest value is 2. These are the limits for the outer integral.

We now compute

Often you can set up an integral in two ways: one so that _dx_ comes first, and one so that _dy_ comes first. Sometimes you'll find that only one of these ways makes the problem accessible. We illustrate this in the next example.

**EXAMPLE 4-7**

Let _P_ be the region of the _xy_ -plane bounded by the graph of _y = x_ , the _x_ -axis, and the line _x =_ 1. We wish to integrate the function   over _P_.

The region _P_ is depicted above twice. Notice from the figure on the left that if we fix a value y0 of _y_ then _x_ can range from y0 to 1. The smallest possible value for _y_ 0 is 0 and the biggest is 1. This tells us the limits of integration when we set up the integral with _dx_ first.

If, on the other hand, we fix a value _x_ 0 of _x_ then _y_ can range from 0 to _x_ , as in the figure on the right. The smallest possibility for _x_ is 0 and the largest possibility is 1. This tells us that the integral can also be set up as

It is only possible to do the second of these integrals, which we compute as follows:

**Problem 31** _Evaluate:_

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.

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**Problem 32** _Let T be the region of the xy-plane bounded by the graph of y = x_ 2 − _x_ − 2 _and the x-axis. Integrate the function f_ ( _x_ , _y_ ) = _x_ 2 _over T_.

**Problem 33** _Let R be the region of the positive quadrant of the xy-plane bounded by y = x and y = x_ 3. _Set up two different integrals for f_ ( _x_ , _y_ ) _over R_.

**Problem 34** _Evaluate the following integral by switching the order of integration:_

**4.3 Computing Volume with Triple Integrals**

The cylindrical figure pictured below has a base with area _A_ and has height 1. The volume is therefore the product of these quantities, namely _A ·_ 1 = _A_. But the top surface of this figure is the graph of the equation _z =_ 1. The integral of this function gives the volume below the graph, which we just found out was A. In other words, if you want to find the area of a region _R_ of the _xy_ -plane, you just need to compute the integral of the function _f_ ( _x_ , _y_ ) _=_ 1 over _R_ :

**EXAMPLE 4-8**

Suppose _R_ is the region below the graph of _y = g_ ( _x_ ), above the _x_ -axis, and between the lines _x = a_ and _x = b_. Then we can find the area of _R_ by evaluating the double integral:

But this just reduces to

which should, of course, be familiar from first term calculus as the area under the graph of _y = g_ ( _x_ ).

Just as one can compute area by evaluating the double integral of the function _f_ ( _x_ , _y_ ) _=_ 1, one can also compute volume by a _triple integral_ of the function _f_ ( _x_ , _y_ , _z_ ) _=_ 1. The tricky part usually involves finding the limits of integration.

**EXAMPLE 4-9**

Let _V_ be the region of  3 bounded by the planes _y_ = _x_ , _z_ = _x_ \+ _y_ , _z_ = 0, _y_ = 0, and _x_ = 1, as pictured below.

We set up a triple integral to compute the volume of _V_. The order of integration will be _dz dy dx_. To find the limits for the innermost integral (the one with respect to z)we fix _x_ and _y_ and observe that _z_ can range from 0 to _x_ \+ _y_. For the second integration we just fix _x_ , and observe that _y_ can vary from 0 to _x_. Finally, the smallest value that _x_ can take on is 0 and the largest is 1, determining the limits of the outermost integral. We now compute

**Problem 35** _Use a triple integral to find the volume of the solid bounded by the graph of z =_ 1 − _y_ 2, _and the planes z =_ 0, _x =_ 0, _and x =_ 1.

**Problem 36** _Find the volume of the solid bounded by the graph of z_ = 1 − _x_ − _y_ 2, _and the coordinate planes z =_ 0 _and x =_ 0.

**Problem 37** _Set up a triple integral to find the volume which lies below the paraboloid z =_ 1 − _x_ 2 − _y_ 2 _and above the xy-plane_.

**Problem 38** _Write a triple integral which will compute the volume bounded by the sphere x_ 2 \+ _y_ 2 \+ _z_ 2 = 1.

**QUIZ**

**Problem 39**

_. Evaluate the following integrals:_

_a_.

_b_.

_. Set up an integral for the volume which lies between the cone   and the xy-plane_.
CHAPTER 5 |

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**Cylindrical and Spherical Coordinates**

**5.1 Cylindrical Coordinates**

A _coordinate system_ is a systematic way of locating a point in (some) space by specifying a few numbers. The coordinate system you are most familiar with is called _rectangular coordinates_.In  3 this works by specifying _x_ , _y_ , and _z_ , which represent distances along the _x_ -, _y_ -, and _z_ -axes, respectively.

Rectangular coordinates are extremely cumbersome when trying to define certain common shapes, such as cylinders or spheres. A second way to determine the location of a point is to adapt polar coordinates to three dimensions. In this case we give a location in the _xy_ -plane by a value of _r_ and _θ_ , and then the height off of the _xy_ -plane by a value of _z_.

Converting from cylindrical coordinates to rectangular coordinates is as easy as it is in polar coordinates. The _z_ coordinate is exactly the same in both systems. To get _x_ and _y_ you just use the formulas

**EXAMPLE 5-1**

We locate the point _r_ = 2,   and _z_ = 7 in rectangular coordinates:

So the desired point is at

**Problem 40** _Write the rectangular coordinates of the following points_.

__.

__.

__.

**5.2 Graphing Cylindrical Equations**

To get a real feel for cylindrical coordinates it is helpful to look at the graphs of a few simple equations. The graph of something like _z_ = 3 in cylindrical coordinates is the same as the graph of _z_ = 3 in rectangular coordinates (it's the same z!): a horizontal plane that hits the _z_ -axis at 3. The next two examples are a bit more interesting.

**EXAMPLE 5-2**

Consider the cylindrical equation _r_ = 3. This is the set of all points in  3 which are exactly 3 units away from the _z_ -axis. The shape is a cylinder of radius 3 centered on the _z_ -axis.

**EXAMPLE 5-3**

We now examine the cylindrical equation   This describes the set of all points that make an angle of   with the _xz_ -plane. The shape is a plane through the _z_ -axis that is halfway between the _xz_ -plane and the _yz_ -plane.

More complicated equations involve relationships between _r, θ_ , and _z_ , as in the next example.

**EXAMPLE 5-4**

Consider the equation _z = r_ 2. To visualize the graph we are being instructed to find all of the points in  3 whose _z_ -coordinate is the square of their distance to the _z_ -axis. The most relevant feature of the equation, however, is the lack of the coordinate _θ_. This tells us that no matter what _θ_ is we will see the same picture.

Following this observation we consider the half plane which meets the _z_ -axis at some arbitrary angle. Take this plane out of the picture and look at it head-on, as in the figure below. As you move left and right in this plane it is the quantity _r_ that is changing. As you move up and down it is _z_ that changes. Hence, it makes sense to label our axes _r_ and _z_ in this plane.

In the half plane at the right we have depicted the graph of _z_ = _r_ 2, a familiar parabola. Now, the key to understanding the complete picture is to place this parabola in _every_ half plane at the left. The result is the surface that you get when you rotate the graph of _y_ = _x_ 2 (in  2 **)** around the _y_ -axis. It is called a _paraboloid_ , and is depicted below (see also Figure 1-9 for a computer-generated picture).

**Problem 41** _Describe the shape of the following cylindrical equations_.

__. _θ_ = 0

__.

__. _z_ = 0

__. _z = r_

__. _z_ 2 \+ _r_ 2 = 1

**Problem 42** _Sketch the graphs of the following equations_.

__.

__. _r_ 2 − _z_ 2 = 1

__. _z_ 2 − _r_ 2 = 1

**5.3 Spherical Coordinates**

Cylindrical coordinates are very convenient for describing cylinders and other surfaces of revolution. In this section we explore _spherical coordinates_ , which are useful for describing many other shapes.

To locate a point using spherical coordinates you need to know three quantities: _ρ_ (pronounced "Row"), _θ_ , and _ϕ_ (pronounced "Fee"). The quantity _θ_ is precisely the same as in cylindrical coordinates. _ρ_ is the distance from the origin. _ϕ_ is the angle with the _z_ -axis, as pictured below.

Given the spherical coordinates of a point we would like to be able to convert to rectangular coordinates. The easiest quantity to find is _z_. Consider the triangle in the figure below. The side adjacent to the angle _ϕ_ is precisely _z_. The hypotenuse is _ρ_. So   Solving for _z_ then gives

_z_ = _ρ_ cos _ϕ_

To find _x_ and _y_ it is helpful to first find the quantity _r_ from cylindrical coordinates. Going back to the triangle in the above figure, we see that the side opposite the angle _ϕ_ is precisely _r_. Hence,   and thus

_r_ = _ρ_ sin _ϕ_

We know from the previous section that _x_ = _r_ cos _θ_ and _y_ = _r_ sin _θ_. Substituting for _r_ gives

**EXAMPLE 5-5**

We find the rectangular coordinates of the point where

**Problem 43** _Convert the following to rectangular coordinates_.

__.

__.

__.

**5.4 Graphing Spherical Equations**

We now explore the graphs of several equations in spherical coordinates.

**EXAMPLE 5-6**

The simplest equation to understand is something like _ρ_ = 2. This describes the set of all points that are precisely 2 units away from the origin. In other words, this is a sphere of radius 2.

If _θ_ is a constant then the equation is the same in spherical coordinates and cylindrical coordinates; a plane through the z-axis. The previous example shows that when _ρ_ is constant we get a sphere. In the next example we see what happens when _ϕ_ is held constant.

**EXAMPLE 5-7**

Consider the equation  . This describes the points which make an angle of   with the _z_ -axis. This is a circular cone, centered on the _z_ -axis.

As with cylindrical coordinates, things get more difficult when equations involve relationships between multiple coordinates.

**EXAMPLE 5-8**

We graph the equation _ρ_ = _ϕ_ .Notice that the coordinate _θ_ does not appear in the equation. This tells us that the intersection of the graph with every half plane incident to the _z_ -axis is the same. (Recall that we encountered a similar type of graph when we looked at some cylindrical equations.) This, in turn, tells us that the graph is a surface of revolution.

In the following figure we have sketched one half plane. In this half plane we have indicated how a spiral is the graph of _ρ_ = _ϕ_.

To form the complete graph we must now take this spiral and rotate it about the _z_ -axis.

In the previous example we saw that when _θ_ did not appear in a spherical equation the graph was a surface of revolution. In the next example we explore what happens when _ϕ_ does not appear in a spherical equation.

**EXAMPLE 5-9**

We now explore the graph of _ρ_ = _θ_. We do this again by investigating various half planes incident to the _z_ -axis. This time the pictures are not all the same. When _θ_ = 0, for example, we are looking at the graph of _ρ_ = 0, which is just a point. When   we are looking at   which is half of a circle of radius   As _θ_ increases we see larger and larger half circles incident to the _z_ -axis, as in the figure below.

Putting it all together is challenging. We have done this for you in the computer-generated picture below. The result is reminiscent of a sea shell.

**Problem 44** _Describe the graphs of the following spherical equations_.

__. _ρ_ = 0

__.

__. _ϕ_ = _π_

__. _ρ_ sin _ϕ_ = 2

__. _ρ_ cos _ϕ_ = 2

**Problem 45** _Sketch the graphs of the following spherical equations_.

__. _ρ_ = sin _ϕ_

__. _ρ_ = cos _θ_

**Quiz**

**Problem 46**

__. _Give rectangular coordinates for the following points:_

_a_. _The point with cylindrical coordinates r_ = 1,   _and z_ = 2.

_b_. _The point with spherical coordinates ρ_ = 2,

__. _Sketch the graphs of the following:_

_a_. _The cylindrical equation r_ = cos _θ_.

_b_. _The spherical equation θ = ϕ_.
CHAPTER 6 |

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**Parameterizations**

**6.1 Parameterized Surfaces**

In Chapter 1 we encountered parameterized curves. These were functions that had one input (the _parameter_ ) and multiple outputs. One visualizes such curves by drawing the range of the function. In this section we introduce _parameterized surfaces_. These are functions which have two inputs and multiple (usually three) outputs. Again, we visualize such functions by picturing their range. This is most often some surface in  3.

**EXAMPLE 6-1**

We explore the parameterization given by

_ψ_ ( _u_ , _v_ ) = ( _u_ , _v_ , _u_ 2 \+ _v_ 2)

Note that for each point in the range of _ψ_ the _z_ -coordinate is obtained from the _x_ and _y_ coordinates by squaring them and adding. In other words, if a point is in the range of _ψ_ then it is also on the graph of _z_ = _x_ 2 \+ _y_ 2. This is the paraboloid depicted in Figure 1-9.

**EXAMPLE 6-2**

We now look at a slight modification of the previous example.

_ψ_ ( _u_ , _v_ ) = (2 _u_ , _v_ , _u_ 2 \+ _v_ 2)

Note that this parameterization is obtained from the previous one by doubling the _x_ -coordinate of every point. This has the effect of stretching the graph in the _x_ -direction. Intersections with horizontal planes will thus be ellipses, as opposed to circles as in the previous example.

Example 6-1 illustrated another important idea. Suppose we want to represent the graph of a function _z_ = _f_ ( _x_ , _y_ ) as a parameterized surface. Then we may simply write

_ψ_ ( _u_ , _v_ ) = ( _u_ , _v_ , _f_ ( _u_ , _v_ ))

Note that the _z_ -coordinate of every point in the range is obtained from the _x-_ and _y_ -coordinates by plugging them into _f_. Hence, each point in the range is indeed on the graph of _f_.

It is not much more difficult to represent graphs in other coordinate systems as parameterized surfaces. The trick is to always translate to rectangular coordinates.

**EXAMPLE 6-3**

Suppose we want to write the graph of the cylindrical equation _z_ = _r_ 2 as a parameterized surface. First, we write down the translation from cylindrical to rectangular coordinates:

Now we substitute _r_ 2 for _z_ :

Hence, a parameterization may be written as follows:

_ψ_ ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _ψ_ , _r_ 2)

**EXAMPLE 6-4**

Suppose we want to find a parameterization for the graph of the spherical equation

_ρ_ = cos _ϕ_

First we write down the translation from spherical to rectangular coordinates:

Now we substitute cos _ϕ_ for _ρ_ :

This gives us the parameterization

_ψ_ ( _θ_ , _ϕ_ ) = (cos _ϕ_ sin _ϕ_ cos _θ_ , cos _ϕ_ sin _ϕ_ sin _θ_ , cos _ϕ_ cos _ϕ_ )

**Problem 47** _Find parameterizations for the graphs of the following equations_.

__. _z_ = _x_ 2

__. _r_ = _θ_ 2

__. _ρ_ = _θ_ 2

**Problem 48** _The following are parameterizations of the graphs of rectangular, cylindrical, or spherical equations. Find these equations_.

__. _ϕ_ ( _ρ_ , _θ_ ) = ( _ρ_ sin _θ_ cos _θ_ , _ρ_ sin2 _θ_ , _ρ_ cos _θ_ )

__. _ϕ_ ( _x_ , _z_ ) = ( _x_ , _x_ \+ _xz_ , _z_ )

__. _ϕ_ ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , sin _r_ )

**6.2 The Importance of the Domain**

When you are trying to specify a particular shape, the domain of the parameterization can be just as important as the function. We illustrate this with an example.

**EXAMPLE 6-5**

Suppose we want to parameterize the part of a sphere of radius 1 that lies in the first octant (i.e., the region of  3 where _x_ , _y_ , _z_ ≥ 0). The equation of a sphere is particularly simple in spherical coordinates, so we start here. Such an equation is _ρ_ = 1. Writing down the translation to rectangular coordinates and substituting 1 for _ρ_ then yields the parameterization

_ψ_ ( _θ_ , _ϕ_ ) = (sin _ϕ_ cos _θ_ , sin _ϕ_ sin _θ_ , cos _ϕ_ )

But how do we make sure we just get the part of the sphere we want? The answer is to restrict the values of _θ_ and _ϕ_ that can be plugged into _ϕ_ :

0 ≤ _θ_ ≤  , 0 ≤ _ϕ_ ≤

These ranges constitute the domain of _ϕ_.

**EXAMPLE 6-6**

We parameterize the portion of the graph of _z_ = _r_ 2 that lies inside the cylinder _x_ 2 \+ _y_ 2 = 1. The equation whose graph we are interested in is given in cylindrical coordinates, so it is natural to start there. The restriction on the domain can also be naturally expressed in cylindrical coordinates, so we do not anticipate any problems. In Example 6-3, we found a parameterization for the desired surface:

_ψ_ ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _r_ 2)

The points of  3 that lie inside the cylinder _x_ 2 \+ _y_ 2 = 1 all have _r_ ≤ 1. Hence, we restrict the domain of our parameterization to

0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ 2 _π_

**Problem 49** _The function_

_ψ_ ( _θ_ , _ϕ_ ) = (sin _ϕ_ cos _θ_ , sin _ϕ_ sin _θ_ , cos _ϕ_ )

_parameterizes a sphere of radius 1. How would you restrict the values of θ and ϕ to just get the portion of the sphere where_

__. _x_ , _y_ ≥ 0

__. _x_ ≤ 0, _y_ ≥ 0, _z_ ≤ 0

**Problem 50** _Describe the difference between the shapes with the following parameterizations_.

__.

__. ψ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _r_ ); 0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ 2 _π_

__. ψ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _r_ ); 0 ≤ _r_ ≤ 1, 0 ≤ θ ≤ _π_

**6.3 This Stuff Can Be Hard!**

Parameterizations are extremely powerful tools that can describe very complicated surfaces. But with such great power can come great difficulty. There may be no obvious way to look at a complicated parameterization and know what shape it describes. For example, consider the parameterization

No expert would be able to look at an equation so complicated and immediately know what it looks like. This is where computers become an invaluable tool. Plugging this equation into a standard graphing program reveals the following beautiful picture.

Despite the apparent difficulty in understanding such complicated functions, there are a few tricks that can help. These are basic ways in which you can take a parameterization and modify it to suit your needs. For example, if you multiply some coordinate by a constant, this has the effect of stretching the parameterized shape by that factor. If you add a constant to some coordinate the shape is moved in that direction.

**EXAMPLE 6-7**

Consider the following parameterization:

ψ( _θ_ , _ϕ_ ) = (2 sin _ϕ_ cos _θ_ \+ 1, 3 sin _ϕ_ sin _θ_ − 2, cos _ϕ_ )

0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _ϕ_ ≤ _π_

This looks much like a parameterization of a sphere from spherical coordinates. But the _x_ -coordinate was multiplied by 2, and then 1 was added to it. Similarly, the _y_ -coordinate was multiplied by 3 and then 2 was subtracted from it. Thus the parameterized shape is an ellipsoid with center at (1, −2, 0).

**Problem 51** _Describe how the following parameterizations are related to graphs of functions in rectangular, cylindrical, or spherical coordinates_.

__. ψ( _y_ , _z_ ) = ( _y_ 2 \+ _z_ , 2 _y_ , _z_ − 1)

__. ψ( _x_ , _y_ ) = ( _y_ , _x_ , _x_ \+ sin _y_ )

__. ψ( _θ_ , _z_ ) = (2 cos _θ_ , 3 sin _θ_ \+ 1, _z_ − 1)

**Problem 52** _Parameterize the surface whose level curves are the ellipses shown below_.

**6.4 Parameterized Areas and Volumes**

Earlier in the text we looked at parameterized curves in  2 and  3. This chapter began with parameterized surfaces in  3. Now we look at parameterized areas in  2 and parameterized volumes in  3. Often these types of parameterizations begin with writing down the translation from some coordinate system to rectangular coordinates, and then restricting the domain.

**EXAMPLE 6-8**

The translation from polar coordinates to rectangular coordinates is given by

We can use this to parameterize the area that is both inside the unit circle and in the first quadrant:

ψ( _r_ , θ) = ( _r_ cos _θ_ , _r_ sin _θ_ )

0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤

Note that it is the restriction on the domain that guarantees that the correct area is parameterized.

**EXAMPLE 6-9**

Let _R_ be the region below the graph of _y_ = _f_ ( _x_ ), above the _x_ -axis, and between the lines _x_ = _a_ and _x_ = _b_. Then _R_ is parameterized by

ψ( _x_ , _t_ ) = ( _x_ , _t f_ ( _x_ ))

_a_ ≤ _x_ ≤ _b_ , 0 ≤ _t_ ≤ 1

In three dimensions one usually begins with rectangular, cylindrical, or spherical coordinates.

**EXAMPLE 6-10**

Consider the parameterization

ψ( _r_ , _θ_ , _z_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _z_ )

1 ≤ _r_ ≤ 2, 0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _z_ ≤ 1

The function itself is just the translation from cylindrical to rectangular coordinates. The restriction on the domain means that we have only parameterized the region between cylinders of radii 1 and 2, and between the planes _z_ = 0 and _z_ = 1.

Of course, as in the previous section, one can distort and move shapes by multiplying and adding constants to each coordinate.

**EXAMPLE 6-11**

We examine the parameterization

ψ( _ρ_ , _θ_ , _ϕ_ ) = (2 _ρ_ sin _ϕ_ cos _θ_ , _ρ_ sin _ϕ_ sin _θ_ , _ρ_ cos _ϕ_ )

0 ≤ _ρ_ ≤ 1, 0 ≤ _θ_ ≤  , 0 ≤ _ϕ_ ≤

Without the 2 in the first component of the parameterization this would be one-eighth of a solid ball. The 2 just stretches it in the _x_ -direction, making it one-eighth of an ellipsoid.

**Problem 53** _Sketch the areas of the plane parameterized by the following_.

__. ψ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ ), 0 ≤ _r_ ≤ 1, 0 ≤ _θ_ , ≤

__. ψ( _r_ , _θ_ ) = ( _r_ cos _θ_ , 2 _r_ sin _θ_ ), 1 ≤ _r_ ≤ 2, 0 ≤ _θ_ , ≤ _π_

**Problem 54** _Describe the volumes given by the following parameterizations_.

__. ψ( _ρ_ , _θ_ , _ϕ_ ) = ( _ρ_ sin _ϕ_ cos _θ_ , _ρ_ sin _ϕ_ sin _θ_ , _ρ_ cos _ϕ_ ), 1 ≤ _ρ_ ≤ 2, 0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _ϕ_ ≤

__. ψ( _r_ , _θ_ , _z_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , 2 _z_ \+ 1), 0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _z_ ≤ 1

**Problem 55** _Find a parameterization for the volume which lies below the cone z_ = _r_ , _above the xy-plane, and inside the cylinder x_ 2 \+ _y_ 2 = 1.

**Quiz**

**Problem 56**

__. _Find a parameterization for the portion of the cylinder of radius 2, centered on the z-axis, which lies below the graph of z_ = 2 _r and above the xy-plane_.

__. _Sketch the region in_  3 _parameterized by the following:_

_ψ_ ( _ρ_ , _θ_ , _ϕ_ ) = (2 _ρ_ sin _ϕ_ cos _θ_ , _ρ_ sin _ϕ_ sin _θ_ , _ρ_ cos _ϕ_ )

_where_ 0 ≤ _ρ_ ≤ 1, 0 ≤ _θ_ ≤  , _and_ 0 ≤ _ϕ_ ≤ _π_.
CHAPTER 7 |

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**Vectors and Gradients**

**7.1 Introduction to Vectors**

Mathematically, there is very little difference between a vector and a point. In two dimensions, for example, a point is really just a pair of numbers. We visualize this pair as a dot in a plane. A two-dimensional vector is similar. Technically, it is also just a pair of numbers. But this time we visualize it as an arrow (Figure 7-1). This arrow does not have a preferred location. The two numbers that specify it only give its _magnitude_ (i.e., size) and _direction_. This can be convenient, since certain algebraic operations with vectors can be seen visually by moving them around.

Figure 7-1 The point (1, 2) and the vector 〈1, 2〉

Some differences between points and vectors are purely cosmetic. For example, the numbers that you give to specify a point are called its _coordinates_. The numbers that specify a vector are its _components_. The main difference between points and vectors is that we can do algebra with vectors. This is a topic studied in detail in a class on _linear algebra_. We will need some of the ideas from linear algebra here.

The first algebraic operation we will study with vectors is addition. This is very easy: to add two vectors we simply add its components.

**EXAMPLE 7-1**

We compute the sum of the vectors 〈2, 3〉 and 〈−1, 2〉 as follows:

〈2, 3〉 + 〈−1, 2〉 = 〈2 − 1, 3 + 2〉 = 〈1, 5〉

We can visualize the operation of addition as follows. To add a vector _V_ to a vector _W_ we slide _W_ so that its "tail" (the base of the arrow) coincides with the "tip" of _V_. The result of _V + W_ is then the vector whose tail is at the tail of _V_ and whose tip is at the tip of _W_.

Negatives are equally easy. If _V_ is a vector then we let _−V_ denote the vector whose components are the negatives of the components of _V_. The picture, of course, is a vector which has the same length as _V_ , but points in the exact opposite direction.

**EXAMPLE 7-2**

If _V_ = 〈2, 3〉 then _−V_ = 〈−2, −3〉.

Finally, we define subtraction by combining addition and negatives:

_V_ − _W_ = _V_ \+ (− _W_ )

The result of subtraction is therefore exactly what you would expect: the vector whose components are the difference between those of _V_ and _W_.

The _magnitude_ (or _length_ ) of a vector is easily computed by the Pythagorean Theorem. We denote the magnitude of a vector _V_ by the absolute value bars | · |. Hence,

**EXAMPLE 7-3**

A vector whose length is one is often called a _unit vector_.

Another common operation is multiplying a vector by a number. This is referred to as _scalar multiplication_. Like addition and subtraction, this is done componentwise. That is, the product of a scalar (i.e., a number) _c_ and a vector _V_ is a vector whose components are _c_ times the components of _V_. That is,

_c_ 〈 _a_ , _b_ 〉 = 〈 _ca_ , _cb_ 〉

Geometrically, scalar multiplication produces a vector which points in the same direction, but whose magnitude has been altered by the given scale factor. For example, if we multiply a vector by the scalar 2, then we get a new vector, pointing in the same direction, which is twice as long.

**EXAMPLE 7-4**

Suppose we would like to find a unit vector that points in the same direction as 〈3, 4〉. The trick is to multiply it by some scalar, and arrange it so that the resulting vector has length one. In other words, we would like to find a number _c_ such that

| _c_ 〈3, 4〉 | = 1

A quick computation gives us

Hence, 5 _c_ = 1, and so _c_ =   We now multiply the vector 〈3, 4〉 by this number to get the answer

**Problem 57** _For the following vectors V and W compute V + W, − V, and V − W_.

__. _V_ = 〈1, 6〉, _W_ = 〈6, 1〉

__. _V_ = 〈0, 0〉, _W_ = 〈1, 2〉

__. _V_ = 〈−1, 1〉, _W_ = 〈1, −2〉

**Problem 58** _Find the magnitude of the following vectors_.

__. 〈2, 3〉

__. 〈1, 3〉

**Problem 59** _Find a unit vector that points in the same direction as_ 〈2, 1〉.

**Problem 60** _Find a unit vector that is perpendicular to the vector_ 〈5, 12〉.

**7.2 Dot Products**

After we have mastered the basic operations of addition and subtraction we move on to multiplication. Unfortunately, we do not have anything quite as straightforward. It is tempting to define the product of _V_ and _W_ as the vector whose components are the products of the components of _V_ and _W_. This particular operation, however, does not turn out to be terribly useful.

Instead we define the product of two vectors _V_ and _W_ to be the sum of the products of the components. This means the result of multiplication is a single number, as opposed to a vector. This product is called the _dot product_ , since it is always denoted with a dot. In symbols, we write

〈 _a_ , _b_ 〉 · 〈 _c_ , _d_ 〉 = _ac_ \+ _bd_

**EXAMPLE 7-5**

〈2, 3〉 · 〈3, 4〉 = 6 + 12 = 18

The usefulness of the dot product comes from the fact that it has a nice geometric interpretation. Suppose the angle between vectors _V_ and _W_ is _θ_. Then,

_V_ · _W_ = | _V_ || _W_ | cos _θ_

This follows from some tricky trigonometry as follows. Consider the following triangle:

Now recall the _Law of Cosines_ , which says that if a triangle has side lengths _a, b_ , and _c_ , and the angle between _a_ and _b_ is _θ_ , then

_c_ 2 = _a_ 2 \+ _b_ 2 − 2 _ab_ cos _θ_

If we apply this to the above triangle we get

| _V_ − _W_ |2 = | _V_ |2 \+ | _W_ |2 − 2| _V_ || _W_ | cos _θ_

which we can rearrange to

Now suppose _V_ = 〈 _v_ 1, _v_ 2〉 and _W_ = 〈 _w_ 1, _w_ 2〉. Then,

And so

_V_ · _W_ = _v_ 1 _w_ 1 \+ _v_ 2 _w_ 2 = | _V_ || _W_ | cos _θ_

In retrospect, this formula is very reasonable. Consider two extreme cases:

_Case 1_. Suppose we take the dot product of _V_ with itself. By definition,

But in this case _θ_ = 0, so cos _θ_ = 1. Hence,

_Case 2_. The other extreme case is when _V_ and _W_ are perpendicular. The slope of the line containing _V_ is   Similarly, the slope of the line containing _W_ is   If these lines are perpendicular then their slopes are negative reciprocals, so   Cross multiplying and bringing everything to one side of the equation then gives _v_ 1 _w_ 1 \+ _v_ 2 _w_ 2 = 0. But the quantity on the left is precisely the definition of _V_ · _W_. To verify the cosine formula just note that if _V_ and _W_ are perpendicular then cos _θ_ = 0, so | _V_ || _W_ | cos _θ_ = 0 as well.

**EXAMPLE 7-6**

The cosine of the angle between vectors 〈2, 3〉 and 〈−1, 2〉 can be found using the dot product

Hence,

**EXAMPLE 7-7**

Since the cosine of the angle between two vectors is zero if and only if the vectors are perpendicular, the dot product can be used to detect this. For example,

〈2, 3〉 · 〈−6, 4〉 = −12 + 12 = 0

so these vectors must be perpendicular.

**Problem 61** _Compute the dot product of the following pairs of vectors_.

__. _V_ = 〈2, 4〉, _W_ = 〈−3, 1〉

__. _V_ = 〈0, 7〉, _W_ = 〈5, 2〉

__. _V_ = 〈−2, −1〉, _W_ = 〈6, −3〉

**Problem 62** _Compute the cosine of the angle between the following pairs of vectors_.

__. _V_ = 〈2, 4〉, _W_ = 〈−3, 1〉

__. _V_ = 〈−2, −1〉, _W_ = 〈6, −3〉

**Problem 63** _Suppose V and W are vectors of length 3. What are the smallest and largest possible values for V_ · _W? What is the smallest possible value of |V_ · _W|?_

**Problem 64** _Use the dot product to decide if each of the following pairs of vectors are perpendicular_.

__. _V_ = 〈2, 4〉, _W_ = (−4, 8)

__. _V_ = 〈2, 4〉, _W_ = (8, −4)

__. _V_ = 〈−2, −1〉, _W_ = 〈−3, 6〉

__. _V_ = 〈−2, −1〉, _W_ = 〈6, −3〉

**7.3 Gradient Vectors and Directional Derivatives**

In this section we return to the concept of differentiation of functions of two variables. Recall that the partial derivative   was the rate of change of the function _f_ ( _x_ , _y_ ) if you walk in the positive _x_ -direction with unit speed. Similarly, the partial derivative   is the rate of change if you walk in the positive _y_ -direction with unit speed. We are left with the obvious question, "What is the rate of change of _f_ ( _x_ , _y_ ) if you walk in some other direction?"

Let _V_ = 〈 _a_ , _b_ 〉 represent a vector which points in the direction of travel. The length of _V_ will be one, to reflect the fact that we are walking with unit speed. We already know two things:

It is reasonable, then, to expect

**EXAMPLE 7-8**

Suppose _f_ ( _x_ , _y_ ) = _xy_ 2. Then the rate of change of _f_ ( _x_ , _y_ ), at the point (2, 1), in the direction   is

The rate of change of _f_ ( _x_ , _y_ ), in the direction of _V_ , is called a _directional derivative_ , and is denoted as ∇ _V_ _f_ ( _x_ , _y_ ). Hence, we have the formula

But notice that the right side of this equation can be rewritten as a dot product:

We can further shorten this equation by coming up with new notation for   Henceforth we will call this vector the _gradient_ of _f_ , and denote it as ∇ _f_. The equation for ∇ _V_ _f_ then becomes

∇ _V_ _f_ = _V_ · ∇ _f_

**EXAMPLE 7-9**

The gradient of the function _f_ ( _x_ , _y_ ) = _x_ 2 _y_ 3 at the point ( _x_ , _y_ ) is

∇ _f_ ( _x_ , _y_ ) = 〈2 _xy_ 3, 3 _x_ 2 _y_ 2〉

The gradient at the point (1, 1), then, would be ∇ _f_ (1, 1) = 〈2, 3〉.

Recall that the rate of change of a function is also the slope of a tangent line to its graph, as long as you are traveling with speed one. Here's a nice way to think about the situation. Suppose you are climbing a mountain, and you have a good trail map in your hands. Let ( _x_ , _y_ ) be your coordinates when you locate yourself on the map. The function _f_ ( _x_ , _y_ ) is your elevation at that point. Now turn your body to face the direction _V_ (on your map). If you sight up or down so that your gaze just grazes the mountainside then you are looking along the tangent line whose slope is given by ∇ _V_ _f_ .

Recall that the dot product is given by the product of the magnitudes of the two vectors, times the cosine of the angle between them. If we fix the point we are at then ∇ _f_ is a fixed vector. If | _V_ | = 1, then the only way to change ∇ _V_ _f_ = _V_ · ∇ _f_ is to change the angle between _V_ and ∇ _f_. The largest this quantity can be is when the value of cos _θ_ is largest. This happens when _θ_ = 0. We conclude that the largest value of ∇ _V_ _f_ is given by

∇ _V_ _f_ = _V_ · ∇ _f_ = | _V_ ||∇ _f_ | cos _θ_ = |∇ _f_ |

and that this value is attained when _V_ (the direction we are facing) coincides with the direction of ∇ _f_. In other words, if you are on the mountainside and you want to face directly uphill you should point yourself in the direction of the gradient vector. When you do this and sight along the mountainside the slope you see is the magnitude of the gradient vector.

**EXAMPLE 7-10**

Let _f_ ( _x_ , _y_ ) = _e_ _xy_ 2. At the point ( _x_ , _y_ ) the gradient vector is

So, at the point (2, 3) we have ∇ _f_ (2, 3) = 〈9 _e_ 18, 12 _e_ 18〉. The largest slope of any tangent line to the graph at the point (2, 3) is then given by

What if you were standing on the mountainside and wanted to face the direction you would have to travel to keep your elevation constant? In other words, how would you find the direction of your level curve? If you were facing such a direction you would be looking along a horizontal line, i.e., a line whose slope is zero. The only way for ∇ _V_ _f_ = _V_ · ∇ _f_ to be zero is for _V_ and ∇ _f_ to be perpendicular.

**EXAMPLE 7-11**

Suppose again _f_ ( _x_ , _y_ ) = _e_ _xy_ 2. In the previous example we saw ∇ _f_ (2, 3) = 〈9 _e_ 18, 12 _e_ 18〉. A vector which points in a direction perpendicular to this would be 〈4, −3〉 (Check this!). Hence, this vector is tangent to a level curve at the point (2, 3).

**Problem 65** _For each of the functions f_ ( _x_ , _y_ ) _below answer the following questions:_

• _Find the gradient vector_ ∇ _f_ ( _x_ , _y_ ).

• _Calculate the rate of change of f_ ( _x_ , _y_ ) _at the point_ (1, 1) _in the direction_

• _Find a unit vector that points in the direction of the maximum rate of change at the point_ (1, 1).

• _Find the largest slope of any tangent line at_ (1, 1).

• _Find a unit vector that lies in a line tangent to a level curve through_ (1, 1).

__. _x_ ln _y_

__. 2 _x_ \+ 3 _y_

__. _x_ 2 _y_ \+ _xy_ 3

**7.4 Maxima, Minima, and Saddles**

At a local maximum or a local minimum of a graph the tangent plane is horizontal. Another way to say this is that the slope of every tangent line to a local maximum or minimum is zero. But recall that the slope of a tangent line at ( _x_ , _y_ ) in direction _V_ is given by _V_ · ∇ _f_. The only way for this to be zero for every possible _V_ is if ∇ _f_ is the zero vector.

**EXAMPLE 7-12**

We show that _f_ ( _x_ , _y_ ) = _x_ 2 _y_ 3 cannot have a maximum or minimum at any point where _x_ and _y_ are nonzero. The gradient is

∇ _f_ ( _x_ , _y_ ) = 〈2 _xy_ 3, 3 _x_ 2 _y_ 2〉

If _x_ and _y_ are nonzero then this is not the zero vector. This tells us that there is a direction where the slopes of tangent lines are nonzero, and hence we cannot be at a local maximum or minimum.

Unfortunately, just because the gradient is the zero vector it does not necessarily mean that there is a local maximum or minimum.

**EXAMPLE 7-13**

The gradient of _f_ ( _x_ , _y_ ) = _x_ 2 \+ _y_ 2 and _g_ ( _x_ , _y_ ) = _x_ 2 − _y_ 2 is 〈0, 0〉 at the origin. The function _f_ ( _x_ , _y_ ) has a minimum there, while _g_ ( _x_ , _y_ ) has a saddle.

In a first-semester calculus class you learned to detect local maxima and minima by a second-derivative test. We would like to do the same thing here. The problem, of course, is that there are _four_ second partial derivatives! To keep track of all this information we often write them in a _matrix_ , as follows:

Now we examine this matrix for several functions whose graph is familiar. Each of these has a gradient vector equal to zero at the origin.

1. _x_ 2 _\+ y_ 2. This function has a local minimum at the origin. The matrix of second partials is

2. − _x_ 2 − _y_ 2. This function has a local maximum at the origin. The matrix of second partials is

3. _x_ 2 − _y_ 2. This function has a saddle at the origin. The matrix of second partials is

4. _xy_. This function also has a saddle at the origin. The matrix of second partials is

The first two give us a clue as to the quantity we would like to look at. Consider the product of the upper-left and lower-right entries of the matrix. For the maximum and minimum above this quantity is positive and for the first of the above saddles it is negative. However, this alone would not be enough to distinguish maxima and minima from saddles, as the second of the saddles shows. To compensate we must subtract the product of the upper-right and lower-left entries, yielding the formula

However, since the mixed partials are equal we can shorten this to

This is indeed the right quantity to look at, in the sense that if it is greater than zero you have a maximum or minimum, and if it is less than zero you have a saddle. Unfortunately, if it is zero you have no information; you may be at a maximum, minimum, saddle, or something much more bizarre. Nonetheless, we will single this out as our first test.

**Test 1**

 . Suppose that at some point ( _x_ 0, _y_ 0) we have ∇ _f_ ( _x_ 0, _y_ 0) = 〈0, 0〉.

_D_ ( _x_ 0, _y_ 0) > 0 ⇒ _f_ ( _x_ , _y_ ) has a local max or min at ( _x_ 0, _y_ 0).

_D_ ( _x_ 0, _y_ 0) < 0 ⇒ _f_ ( _x_ , _y_ ) has a saddle at ( _x_ 0, _y_ 0).

_D_ ( _x_ 0, _y_ 0) = 0 ⇒ No information about _f_ ( _x_ 0, _y_ 0).

**EXAMPLE 7-14**

We find the location of all saddles of the function _f_ ( _x_ , _y_ ) = _x_ 2 \+ 2 _xy_ \+ 3 _y_ 3. First, we will need to narrow down the possibilities by finding the critical points. To do this we find the gradient.

∇ _f_ = 〈2 _x_ \+ 2 _y_ , 2 _x_ \+ 9 _y_ 2〉

Setting this equal to the vector 〈0, 0〉 gives us the system of equations

The first equation tells us that _x_ = − _y_. Plugging this into the second equation then gives

−2 _y_ \+ 9 _y_ 2 = 0

Either _y_ = 0 (and hence _x_ = 0) or we can divide both sides of this equation by _y_ to get

−2 + 9 _y_ = 0

Solving then gives us _y_ =   Hence, we have critical points at (0, 0) and

To determine which of these are saddles we compute the matrix of second partials:

And so

_D_ ( _x_ , _y_ ) = 36 _y_ − 4

We now check each critical point:

When _D_ ( _x_ 0, _y_ 0) > 0 it would be nice to have a second test to determine whether ( _x_ 0, _y_ 0) is a local maximum or a local minimum. Such a test can be easily guessed from our prototypical examples, _f_ ( _x_ , _y_ ) = _x_ 2 \+ _y_ 2 and _f_ ( _x_ , _y_ ) = − _x_ 2 − _y_ 2. Notice that in both cases   and   have the same sign. When this sign is positive we have a local minimum and when it is negative we have a local maximum. This is precisely our second test.

**Test 2**

 . Suppose that at some point ( _x_ 0, _y_ 0) we have ∇ _f_ ( _x_ 0, _y_ 0) = 〈0, 0〉 and _D_ ( _x_ 0, _y_ 0) > 0.

**EXAMPLE 7-15**

Recall the function _f_ ( _x_ , _y_ ) = _x_ 2 \+ 2 _xy_ \+ 3 _y_ 3 from the previous example. We found critical points at (0, 0) and   and determined that at   there was a local maximum or a local minimum. To determine which we need only look at   Since this was 2, and 2 > 0, we conclude that at this critical point there is a local minimum.

It is important to keep in mind that if _D_ ( _x_ 0, _y_ 0) = 0 then we have no information about the nature of _f_ ( _x_ 0, _y_ 0). We illustrate this in the next example.

**EXAMPLE 7-16**

Consider the following functions:

1. _f_ ( _x_ , _y_ ) = _x_ 4 \+ _y_ 4

2. _f_ ( _x_ , _y_ ) = − _x_ 4 − _y_ 4

3. _f_ ( _x_ , _y_ ) = _x_ 2 _y_ 2

In each case the only critical point is at (0, 0) and _D_ (0, 0) = 0. But at (0, 0) in the first case there is a local minimum, in the second there is a local maximum, and in the third there is a saddle.

**Problem 66** _Find the local maxima, minima, and saddles of the following functions_.

__. _xy_ \+ 2 _x_ − 3 _y_ − 6

__. _x_ 3 − _xy_ \+ _y_ 2

**Problem 67** _For the function_ sin( _x_ \+ _y_ ) _show that D_ ( _x_ , _y_ ) = 0 _at every point_ ( _x_ , _y_ ). _Does this function have maxima, minima, or saddles?_

**Problem 68** _If, for some point_ ( _x_ 0, _y_ 0) _you know D_ ( _x_ 0, _y_ 0) > 0 _and_   ( _x_ 0, _y_ 0) > 0 _then show that_  ( _x_ 0, _y_ 0) > 0. ( _This tells us that you may use _ ( _x_ 0, _y_ 0) _instead of_   ( _x_ 0, _y_ 0) _when distinguishing maxima from minima_.)

**7.5 Application: Optimization Problems**

An important application of the mathematics of the previous section is to optimization problems. You encountered several such problems in a first semester calculus class. You can now handle much more complicated situations. We illustrate this with an example.

**EXAMPLE 7-17**

We find the largest volume of a box with an open top, and surface area 100 m2. First, we let the dimensions of the box be _a, b_ , and _c_ , with _c_ the height. Then the surface area is given by

_S. A_. = 100 = _ab_ \+ 2 _ac_ \+ 2 _bc_

Solving this equation for _c_ gives

The volume, of course, is the quantity _abc_. Substituting for _c_ then gives us

Thinking of volume, then, as a function _V_ ( _a_ , _b_ ) we now search for critical points. The gradient is a bit tedious to compute, but with some work you can show it simplifies to

Each component of this vector can only be zero if the numerators are zero, so we get two equations:

0 = −2 _a_ 2 _b_ 2 \+ 200 _b_ 2 − 4 _ab_ 3

0 = −2 _a_ 2 _b_ 2 \+ 200 _a_ 2 − 4 _a_ 3 _b_

We may safely assume that neither _a_ nor _b_ are zero. Dividing the first equation by 2 _b_ 2 and the second by 2 _a_ 2 then gives

0 = − _a_ 2 \+ 100 − 2 _ab_

0 = − _b_ 2 \+ 100 − 2 _ab_

(7-1)

Solving both equations for 2 _ab_ and setting them equal to each other gives us

− _a_ 2 \+ 100 = − _b_ 2 \+ 100

and thus (since we are only interested in positive values of _a_ and _b_ ) we may conclude _a_ = _b_. Combining this with Equation 7-1 tells us

0 = − _a_ 2 \+ 100 − 2 _a_ 2

Solving this for _a_ reveals

The fact that _a_ = _b_ then gives us

Finally, our expression for _c_ in terms of _a_ and _b_ above yields

The volume is thus

**7.6 LaGrange Multipliers**

Suppose you wanted to look for the maximum value of _f_ ( _x_ ) =   on the interval [−2, 3]. This function has two critical points, and the one at _x_ = −1 is a local maximum. But to conclude that _f_ (−1) =   is the maximum value of _f_ ( _x_ ) on the indicated domain would be incorrect. This is because the maximum is attained at one of the endpoints of the domain, namely at _x_ = 3. At this point _f_ (3) = 6.

The same kind of thing can happen with functions _f_ ( _x_ , _y_ ) of two variables. Using techniques of the previous section we can look for, and analyze, critical points in the interior of the domain _D_ of _f_. But if we are interested in the maximum value that _f_ attains on _D_ we may have to look at the points around the boundary of _D_ , as in Figure 7-2. In the one-variable case this was easy, since it was just a matter of checking two points. When there are multiple variables there are an infinite number of points around the boundary of _D_ , so we need some technique other than just plugging values into _f_.

**Figure 7-2** The function _f_ ( _x_ , _y_ ) has a local maximum at the point _q_ , but the absolute maximum on the domain _D_ is the point _p_ on the boundary

In Figure 7-3 we see the contours for the function _f_ ( _x_ , _y_ ) of Figure 7-2, superimposed on the domain _D_. As is customary, we will denote the points around the boundary of _D_ as ∂ _D_. (Don't get this confused with a partial derivative.) Notice that the maximum and minimum values of _f_ on ∂ _D_ occur at the places where the level curves of _f_ are tangent to ∂ _D_.

**Figure 7-3** The contours of _f_ ( _x_ , _y_ ) and the domain _D_. The absolute maximum of _f_ on _D_ is at the point _p_ where a level curve is tangent to ∂ _D_

The key fact that we will use is that the vectors ∇ _f_ are always perpendicular to the level curves of _f_. So, at a point _p_ = ( _x_ 0, _y_ 0) where a level curve of _f_ is tangent to ∂ _D_ we will see ∇ _f_ ( _x_ 0, _y_ 0) perpendicular to ∂ _D_.

The next step is to find a new function, _g_ ( _x_ , _y_ ), for which ∂ _D_ is a level curve. The vector ∇ _g_ is always perpendicular to all of its level curves. In particular, ∇ _g_ ( _x_ 0, _y_ 0) will be perpendicular to ∂ _D_ at ( _x_ 0, _y_ 0). See Figure 7-4.

**Figure 7-4** The vector ∇ _f_ is a scalar multiple of the vector ∇ _g_

Since ∇ _f_ ( _x_ 0, _y_ 0) and ∇ _g_ ( _x_ 0, _y_ 0) are perpendicular to the same curve they must be parallel to each other. If vectors _V_ and _W_ are parallel, then one is a scalar multiple of the other. In other words, there is a constant λ such that _V_ = _λW_. Putting all this together, we find that if a maximum (or minimum) of _f_ occurs at a point ( _x_ 0, _y_ 0) of ∂ _D_ then there must be a constant λ such that ∇ _f_ ( _x_ 0, _y_ 0) = λ∇ _g_ ( _x_ 0, _y_ 0). Using this information we get a finite number of candidate points at which to look for maxima and minima.

**EXAMPLE 7-18**

Let _f_ ( _x_ , _y_ ) = 2 _x_ 2 \+ 3 _y_ 3, and suppose _D_ is the set of points ( _x_ , _y_ ) in  2 satisfying _x_ 2 \+ _y_ 2 ≤ 1. We will look for the largest value attained by _f_ ( _x_ , _y_ ) on ∂D.

To use the "method of Lagrange multipliers" the first step is to find a function _g_ ( _x_ , _y_ ) for which ∂ _D_ is a level curve. This is very easy: _g_ ( _x_ , _y_ ) = _x_ 2 \+ _y_ 2. Then the level curve _g_ (x, y) = 1 is the curve _x_ 2 \+ _y_ 2 = 1, which is precisely the boundary of _D_.

The next step is to look for points ( _x_ 0, _y_ 0), and values of λ, such that ∇ _f_ ( _x_ 0, _y_ 0) = λ∇ _g_ ( _x_ 0, _y_ 0). We thus compute

So if 〈4 _x_ , 9 _y_ 2〉 = λ 〈2 _x_ , 2 _y_ 〉 we have 4 _x_ = λ2 _x_ and 9 _y_ 2 = λ2 _y_. The first of these equations tells us that either _x_ = 0 or λ = 2.

_Case 1.x_ = 0. The boundary of _D_ is the set of points satisfying _x_ 2 \+ _y_ 2 = 1. So if _x_ = 0 then _y_ = ±1. This means that we must check the points (0, 1) and (0, −1) for maxima and minima.

_Case 2. λ_ = 2. Combining this with the equations 9 _y_ 2 = λ2 _y_ then tells us 9 _y_ 2 = 4 _y_. So again there are two cases: either _y_ = 0 or 9 _y_ = 4, in which case _y_ =  .

_Subcase A. y_ = 0. Similar to Case 1 above, the boundary of _D_ is the points satisfying _x_ 2 \+ _y_ 2 = 1. If _y_ = 0 then _x_ = ±1. So we get two more points to check: (1, 0) and (−1, 0).

_Subcase B. y_ =   Plugging this into _x_ 2 \+ _y_ 2 = 1 gives us   Solving for _x_ then gives   Thus we get two more points:   and

We now have six points at which to look for maxima and minima of _f_. We do this simply by plugging them in:

So the maximum value of _f_ on ∂ _D_ is at (0, 1) and the minimum value is at (0, −1).

**Problem 69** _Let f_ ( _x_ , _y_ ) = _x_ 2 \+ _y_ 2 \+ 2 _y_ − 1 _and_

_. Find the locations of all critical points of f_ ( _x_ , _y_ ) _on D_.

_. Find the absolute minimum of f_ ( _x_ , _y_ ) _on D_.

**7.7 Determinants**

In the previous chapter we saw that the dot product was a way to take two vectors and produce a meaningful number. In two dimensions there is another such way. Suppose _V_ = 〈a, b〉 and _W_ = 〈c, d〉 are vectors and consider the parallelogram formed by two parallel copies of each.

We wish to determine the area of this parallelogram. Notice in the figure below that there are four triangles and two small rectangles inside a larger rectangle. The area of the desired parallelogram is found by subtracting these areas.

The area of the large parallelogram is ( _a_ \+ _c_ )( _b_ \+ _d_ ). The area of each small rectangle is _bc_. The area of two of the triangles is  _ab_ and the other two is  _cd_. Thus the desired area is

We often organize the information contained in a set of vectors by writing them in a matrix. The area of the parallelogram between them is then the _determinant_ of the matrix. We write this as follows:

Notice that there is a bit more information in the determinant than just the area, since our answer may be negative. The sign of the answer tells us the order of _V_ and _W_. If we go counterclockwise to get from _V_ to _W_ then the sign of the determinant is positive. Otherwise it is negative. For this reason we call the value of the determinant the _signed area of the parallelogram spanned by V and W_.

**EXAMPLE 7-19**

We compute the (signed) area of the parallelogram spanned by the vectors 〈1, 2〉 and 〈4, 3〉.

In three dimensions there is also a determinant, but it is slightly more complicated. If we take four parallel copies of three vectors then they span a 3-dimensional figure called a _parallelepiped_. You should think of this as a cube that has been stretched, skewed, and rotated. The determinant of the matrix that is comprised of these vectors is then the volume of this figure.

A matrix containing the vectors _U_ = 〈 _a_ , _b_ , _c_ 〉, _V_ = 〈 _d, e, f_ 〉, and _W_ = 〈 _g_ , _h, i_ 〉 would be written like this

To write the determinant of this matrix it is helpful to copy the entire matrix next to the original one. Then add up all the "right-slanting" products and subtract all the "left-slanting" products.

Once again this answer may be negative, indicating the way in which _U_ , _V_ , and _W_ are situated. The sign is positive if, when we sweep the fingers of our right hand from _U_ to _V_ , our thumb points in the direction of _W_. Once again, this is the "right-hand rule," that we encountered earlier.

A second way to remember how to compute the determinant of a 3 × 3 matrix is to reduce it to computing the determinants of three 2 × 2 matrices:

Notice that each term on the left is preceded by one element from the top row of the matrix, with alternating signs. The 2 × 2 matrix in each term is the one you get from the original matrix by deleting the corresponding row and column. So, for example, the first term is preceded by _a_ , since that is the first element of the top row. The 2 × 2 matrix in the first term is the one you get by eliminating the row and column containing _a_.

**EXAMPLE 7-20**

We compute the (signed) volume of the parallelepiped spanned by 〈1, 2, 3〉, 〈1, −1, 1〉, and 〈0, 2, 2〉.

**Problem 70** _Find the (signed) area of the parallelogram spanned by_

__. 〈1, 3〉 _and_ 〈−1, 2〉

__. 〈1, 6〉 _and_ 〈1, 1〉

_. _ 〈2, 3〉 _and_ 〈6, 9〉

**Problem 71** _Suppose V and W are parallel vectors in_  2. _Let M be the matrix whose rows are V and W. Show that the determinant of M is zero_.

**Problem 72** _Calculate the_ ( _signed_ ) _area of the parallelepiped spanned by_

__. 〈1, 2, 3〉, 〈1, 0, 2〉, and 〈−2, 2, −3〉

__. 〈0, 1, 3〉, 〈−1, 2, 1〉, and 〈2, 0, −1〉

**7.8 The Cross Product**

A challenging problem is to find the area of a parallelogram defined by two vectors in three dimensions. If the vectors are _V_ = 〈a, b, c〉 and _W_ = 〈d, e, f〉 then this area is given by

The general form of this equation looks a lot like the Pythagorean Theorem: some quantity is the square root of the sum of squares. We can employ this observation to make the above formula a bit more compact:

Area( _V, W_ ) = |〈 _bf_ − _ec, cd_ − _af, ae_ − _bd_ 〉|

Note that we started with two vectors, _V_ and _W_ , and ended up with a third that was somehow related. This gives us a new kind of product, called the _cross product:_

_V_ × _W_ = 〈 _bf_ − _ec, cd_ − _af, ae_ − _bd_ )

Recall that the dot product was a way to take two vectors and multiply them, with the result being a scalar. The cross product, by comparison, is a way to multiply and get a third vector.

The astute reader will notice that each component of the cross product is the determinant of a 2 × 2 matrix. We may thus write its formula like this:

It is traditional to define the vectors **i** = 〈1, 0, 0〉, **j** = 〈0, 1, 0〉, and **k** = 〈0, 0, 1〉. Using this we may rewrite the definition of the cross product as

We introduced a negative sign in the second term and switched the order of the columns of the second matrix. Now our formula looks just like a determinant:

This last formula is probably the easiest way to remember how to compute the cross product of two vectors.

**EXAMPLE 7-21**

We compute the area of the parallelogram spanned by the vectors 〈1, 1, 0〉 and 〈1, 2, 3〉. To do this we first compute the cross product:

The desired area is now the magnitude of this vector:

We have already seen that the magnitude of the cross product has some geometric significance. What about its direction? To get a clue we compute _V_ · _(V_ × _W_ ). Suppose _V_ = 〈 _a, b, c_ 〉 and _W_ = 〈 _d, e, f_ 〉. Then, we compute:

From this we may conclude that _V_ × _W_ is perpendicular to _V_. A similar computation shows _W_ · ( _V_ × _W_ ) = 0, and hence _V_ × _W_ is perpendicular to _W_ as well. We conclude _V_ × _W_ is perpendicular to the _plane_ containing _V_ and _W_.

**EXAMPLE 7-22**

We find a unit vector that is perpendicular to both 〈1, 1, 0〉 and 〈1, 2, 3〉. From the previous problem we know that the cross product of these two vectors is the vector 〈3, −3, 1〉, and that the magnitude of this vector is   To get a unit vector which points in the same direction we just divide by the magnitude:

**Problem 73** _Find the area of the parallelogram spanned by the vectors_

__. 〈1, 2, 3〉 _and_ 〈−1, 0, 1〉

__. 〈1, 1, 0〉 _and_ 〈1, 0, 1〉

__. 〈1, 2, 3〉 _and_ 〈3, 1, 2〉

**Problem 74** _Find a unit vector which is perpendicular to each of the following pairs of vectors_.

__. 〈1, 2, 0〉 _and_ 〈1, 1, 1〉

__. 〈1, 1, 2〉 _and_ 〈1, 1, 1〉

**Problem 75** _Show that V_ × _W_ = − _W_ × _V_.

**Problem 76** _Let U, V, and W be three vectors in_  3. _Let M be the_ 3 × 3 _matrix whose rows are these vectors. Show that the determinant of M is equal to U_ · ( _V_ × _W_ ).

**Problem 77** _Show that the area of the parallelogram spanned by V and W is equal to_ | _V_ || _W_ | sin _θ, where θ is the angle between V and W_.

**Quiz**

**Problem 78**

_. Let f_ ( _x_ , _y_ ) _be the following function:_

_f_ ( _x_ , _y_ ) = _x_ 2 \+ 3 _xy_

_a. Find the largest slope of any tangent line to the graph of f_ ( _x_ , _y_ ) _at the point_ (1, 1).

_b. Find the critical point(s)_.

_c. Compute the value of_

_for each critical point found_.

_d. Does the graph have a max, min, or saddle at the critical point(s)?_

_. Let V_ = 〈1, 2, 3〉 _and W_ = 〈1, 1, 1〉. _Find_

_a. a unit vector which points in the same direction as V_.

_b. the cosine of the angle between V and W_.

_c. a unit vector which is perpendicular to both V and W_.
CHAPTER 8 |

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**Calculus with Parameterizations**

8.1 Differentiating Parameterizations

Suppose _ϕ_ ( _t_ ) = ( _f_ ( _t_ ), _g_ ( _t_ )) is a parameterization of some curve in  2. What should it mean to _differentiate ϕ_ at _t_ = _t_ 0? Obviously, the answer will involve two numbers, _f_ ′( _t_ 0) and _g_ ′( _t_ 0). The best way to make sense out of this pair of numbers is as a vector:

If this vector is drawn with its base at the point _ϕ_ ( _t_ 0) then it is tangent to the parameterized curve. In addition, if _t_ really represents time, and _ϕ_ ( _t_ ) the coordinates of a particle, then the _length_ of this vector tells how fast the particle is moving at time _t_ 0.

**EXAMPLE 8-1**

At time _t_ the coordinates of a car are _ϕ_ ( _t_ ) = ( _t_ 2, _t_ 3). We wish to determine how fast, and in what direction, the car is heading at time _t_ = 2. First, we differentiate the parameterization:

_ϕ_ ′( _t_ ) = 〈2 _t_ , 3 _t_ 2〉

Thus,

_ϕ_ (2) = 〈4, 12〉

The direction of the car is thus the direction that this vector is pointing. Its speed is given by the magnitude of this vector:

Nothing significant changes in three dimensions, as in our next example.

**EXAMPLE 8-2**

We find the speed of a particle moving along the helix _ϕ_ ( _t_ ) = (cos _t_ , sin _t_ , _t_ ). First, we find the tangent vector:

_ϕ_ ′( _t_ ) = 〈– sin _t_ , cos _t_ , 1〉

To find the speed we compute the magnitude of this vector:

Note that our answer does not depend on _t_ , telling us a particle moving along the spiral by the given parameterization travels with constant speed.

Differentiating parameterizations of surfaces is a bit more complicated, since now there are two variables. For example, consider the general parameterization

ψ( _u_ , _v_ ) = ( _f_ ( _u_ , _v_ ), _g_ ( _u_ , _v_ ), _h_ ( _u_ , _v_ ))

We can take the partial derivative of ψ with respect to _u_ , say, simply by taking the partial derivatives of _f_ , _g_ , and _h_ with respect to _u_. Our answer is again a vector:

Geometrically,   is a tangent vector to the surface parameterized by ψ at the point  . Another such vector is given by  .

**EXAMPLE 8-3**

We find two vectors tangent to the graph of _z_ = _x_ 2 \+ _y_ 3 at the point (2, 1, 5). This surface is parameterized by

ψ( _x_ , _y_ ) = ( _x_ , _y_ , _x_ 2 \+ _y_ 3)

The desired point is at ψ(2, 1). To find two tangent vectors we simply take the partial derivatives of ψ and evaluate at (2, 1).

and so,

Similarly,

and so,

We conclude 〈1, 0, 4〉 and 〈0, 1, 3〉 are two vectors tangent to the graph of _z_ = _x_ 2 _\+ y_ 3 at the point (2, 1, 5).

**EXAMPLE 8-4**

If ( _θ_ , _ϕ_ ) represents your longitude and latitude, then your position on the Earth's surface is given by

ψ( _θ_ , _ϕ_ ) = ( _R_ cos _ϕ_ cos _θ_ , _R_ cos _ϕ_ sin _θ_ , _R_ sin _ϕ_ )

where _R_ represents the radius of the Earth (assuming the Earth is a perfect sphere). Note the similarity to spherical coordinates. We explore the geometric meaning of the derivatives of this parameterization.

First, we differentiate with respect to _ϕ_

The magnitude of this vector is thus

Notice that the result is constant, i.e., it does not depend on your longitude or latitude. The significance of this is that if you change your latitude by roughly 1 radian then you change your position on the Earth by a constant amount. We will see in a moment that this is very different than changing your longitude. Let's try differentiating with respect to _θ_.

The magnitude of this vector is thus

This says that if you change your longitude by 1 radian, then the actual change in distance on the Earth's surface is proportional to the cosine of your latitude.

Now let's take the cross product of the vectors we have found:

The magnitude of this vector is

This says that if you stake out a parcel of land that is 1 radian of longitude by 1 radian of latitude, then its area is proportional to the cosine of your latitude.

**Problem 79** _Find a function s_ ( _t_ ) _which gives the speed of a particle at time t if its coordinates at time t are given by_ (sin _t_ 2, cos _t_ 2).

**Problem 80** _A curve in_  3 _is parameterized by c_ ( _t_ ) = ( _t_ 3, _t_ 4, _t_ 5). _Find a unit vector which is tangent to this curve at the point_ (1, 1, 1).

**Problem 81** _Let S be the paraboloid parameterized by_

ψ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _r_ 2)

0 ≤ _r_ ≤ 2, 0 ≤ _θ_ 2 _π_

_. Find two (nonparallel) vectors tangent to S at the point  _.

_. Find a vector which is perpendicular to S at the point  . (Hint: Use the cross product.)_

**Problem 82** _Let_ ψ( _t_ ) = (cos _t_ , sin _t, t_ ).

_. Find a parameterization_ Φ _of the curve parameterized by_ ψ _in which_  .

_. We may think of_   _as itself a parameterized curve. For every value of t we picture the vector_   _as being based at the origin, and pointing to some point in_  3. _We may thus differentiate this parameterization to obtain another vector_  . _Show that_   _is perpendicular to . (Hint: Compute_   _and show that the dot product with_   _is zero.)_

_. Find K so that_  , _where N is a unit vector. (The number K is called the curvature of the parameterized curve and N is called the normal.)_

_. Compute B_ =  × _N and show_ | _B_ | = 1. _(The vector B is called the binormal.)_

_. Compute_  .

_. Find τ such that_  . _(The number τ is called the torsion.)_

_. Show that_

8.2 Arc Length

Suppose _C_ is a curve, in  2 or  3, which is parameterized by the function ψ( _t_ ), where _a_ ≤ _t_ ≤ _b_. We would like to find the length of _C_. We do this by sampling _C_ at various points, looking at the length of the line segment connecting adjacent points, and adding these lengths. The result will be an approximation of the length of _C_. This approximation will get better and better as we sample more and more points along _C_. In the resulting limit, we obtain an integral which represents the length.

The steps are very straightforward.

1. Choose _n_ points in the interval [ _a_ , _b_ ]. We will call these points  .

2. Note that ψ( _t i_) is a point of _C_. For each _i_ we connect ψ( _t i_+1) to ψ( _t i_) by a line segment to approximate _C_.

3. The length of each line segment is precisely the length of the vector ψ( _t_ _i_ +1) – ψ( _t i_).

4. We now add up the lengths of all the approximating line segments to get an approximation for the length of _C_ :

A little algebraic trick will be useful in the next step:

5. Finally, we take the limit of this quantity as _n_ → ∞. Note that we can also view this as a limit as Δ _t_ → 0. But, by definition,

Hence, the desired limit is

**EXAMPLE 8-5**

Let _C_ be the curve parameterized by

ψ( _t_ ) = ( _t_ cos _t_ , _t_ sin _t_ ), 0≤ _t_ ≤ 1

Then the length of _C_ is given by

**Problem 83** _Calculate the length of the curve in_  3 _parameterized by_

ψ( _t_ ) = ( _t_ cos _t_ , _t_ sin _t_ ),

0 ≤ _t_ ≤ 1

**Problem 84** _Show that the curve in_  3 _parameterized by_

0 ≤ _t_ ≤ 1

_has the same length as the curve in_  2 _parameterized by_

( _t_ cos _t_ , _t_ sin _t_ ), 0 ≤ _t_ ≤ 1

8.3 Line Integrals

Let _C_ represent some parameterized curve in  2. Let _f_ ( _x_ , _y_ ) be a function on  2. In this section we make some sense of the integral of _f_ ( _x_ , _y_ ) over _C_. First, let's take a step backward and recall the steps used to define the integral of a function of one variable, _f_ ( _x_ ), over an interval [ _a_ , _b_ ].

1. Choose _n_ points in [ _a, b_ ], which we denote as { _x i_}.

2. Let Δ _x i_ = _x_ _i_ +1 _– x i_ .

3. For each _i_ compute _f_ ( _x i_)Δ _x i_.

4. Sum over all _i_.

5. Define   to be the limit of the sum from the previous step as _n_ → ∞.

The trick to defining the integral of _f_ ( _x_ , _y_ ) over a parameterized curve _C_ is to follow the above steps as closely as possible. As we will see presently, the most difficult step is probably the one that seems most innocent—the second.

1. Choose _n_ points along the curve _C_ , which we denote as { _p_ _i_ }. (Note that _p i_ is a label for a point of  2, and as such will have two coordinates.)

2. Let _δ_ _p i_ denote the length, along _C_ , from _p i_ to _p i+_1.

3. For each _i_ compute _f_ ( _p_ i)δ _p_ i.

4. Sum over all _i_.

5. Define   d **s** to be the limit of the sum from the previous step as _n_ → ∞.

The resulting integral is called the _line integral_ of _f_ ( _x, y_ ). Unfortunately, in practice the above steps are impossible to carry out. To make things easier we employ a parameterization of _C_ to translate the problem to an integral of a function of one variable over an interval of  1. To this end, let ψ( _t_ ), where _a_ ≤ _t_ ≤ _b_ , be a parameterization of _C_. We now repeat the above steps as closely as possible, this time using ψ.

1. Choose _n_ points in the interval [ _a_ , _b_ ]. We denote these points as { _t i_}. The set {ψ( _t_ _i_ )} is then a collection of _n_ points on _C_.

2. Let Δ _t i = ti+_1 _– t i_. Let δ _t i_ denote the length, along _C_ , from {ψ( _t i_)} to {ψ( _t_ _i_ +1)}.

3. For each _i_ compute _f_ (ψ( _t_ _i_ ))δ _t_ _i_. A little algebraic trickery yields

4. Sum over all _i_.

5. The integral   _ds_ is defined to be the limit of the sum from the previous step as _n_ → ∞. But as _n_ → ∞. it also follows that Δ _t i_ – 0. Our integrand contains the term  . As Δ _t i_ → ∞ this converges to  . Hence, our integral has become

Note that if _f_ ( _x_ , _y_ ) = 1 then the line integral of _f_ over _C_ gives precisely the length of _C_. This gives us one way to think about line integrals. Imagine a vast plane. At every point of the plane there is a number which represents how fast a point moving through that point will travel. If this number is always one, then a point moving along in the plane will always travel with unit speed. Its total travel time will then be exactly the same as the total distance it travels. But if the number at each point isn't always one then its travel time may be different than the total distance traveled. The line integral gives the total travel time.

**EXAMPLE 8-6**

We integrate the function _f_ ( _x_ , _y_ ) = _y_ over the top half of the circle of radius 1. This curve is parameterized by

ψ( _t_ ) = (cos _t_ , _t_ sin _t_ ), 0 ≤ _t_ ≤ _π_

We now compute

**Problem 85** _Let f_ ( _x_ , _y_ ) = _y_ 3. _Compute the line integral of f_ ( _x_ , _y_ ) _over the curve C parameterized by_

8.4 Surface Area

One of the most interesting applications of multiple integrals is the computation of surface area. Let _R_ be a rectangle in the _xy_ -plane. The problem is to find the area of the portion of the surface parameterized by ψ( _u, v_ ). Let _R_ denote the domain of ψ. To find the area we follow similar steps that we used to derive the formula for the volume under _f_ ( _x_ , _y_ ).

1. We begin by choosing a grid of points {( _u i, vj_)} in _R_.

2. For each grid point locate the point ψ( _u i_, _v j_) in  3. Connecting adjacent points produces a bunch of parallelograms that piece together to form an approximation of the parameterized surface.

3. We now compute the area of each parallelogram. Observe that each parallelogram is spanned by the vectors

_V_ u = ψ( _u_ _i_ +1, _v j_) – ψ( _u i_, _v j_)

and

_V_ u = ψ( _u_ j, _v_ _i_ +1) – ψ( _u i_, _v j_)

The desired area is thus the magnitude of the cross product of these vectors:

We now do some algebraic tricks:

4. The surface area is the limit, as Δ _u_ and Δ _v_ tend toward zero, of the sum of this quantity over all _i_ and _j_ :

But notice that as Δ _u_ tends toward zero the quantity   converges on  . Similarly for Δ _v_ → 0. In the end our summation becomes the integral:

**EXAMPLE 8-7**

We will use the surface area formula derived above to show that the area of a sphere of radius _R_ is 4 _πR_ 2. The most natural parameterization of the sphere comes, of course, from spherical coordinates:

ψ ( _θ_ , _ϕ_ ) = ( _R_ sin _ϕ_ cos _θ_ , _R_ sin _ϕ_ sin _θ_ , _R_ cos _ϕ_ )

0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _ϕ_ ≤ _π_

The partial derivatives of this parameterization are

The cross product of these two vectors is thus

The magnitude of this vector is

The desired surface area is thus computed as follows:

There are lots of special cases where one may want to compute surface area. In each of these cases we can derive an appropriate formula. The first case we consider is the surface area of the portion of the graph of _z_ = _f_ ( _x_ , _y_ ) that lies above some domain _R_. Such a graph can be parameterized as follows:

ψ ( _x_ , _y_ ) = ( _x_ , _y_ , _f_ ( _x_ , _y_ ))

The partial derivatives of this parameterization are

The cross product of these vectors is

The magnitude of this vector is

The desired formula for the surface area of a graph is thus

**EXAMPLE 8-8**

We will use the surface area formula derived above for the graph of a function to once again show that the area of a sphere of radius _r_ is 4 _π_ _r_ 2. An equation for the top half of the sphere is  . We will compute the area of this surface and double it to get the total area of the sphere.

The tricky part of this computation is the limits of integration. Notice that the domain of integration is the region inside a circle of radius _r_ in the _xy_ -plane. For a fixed _x_ the value of _y_ can range from  . Hence, we may set up a double integral to compute surface area as follows:

As this is the area of just the top half of the sphere we double it to obtain 4 _π_ _r 2_, the correct formula.

Another special case of a surface area formula is for a surface of revolution. This formula is typically covered in a second semester calculus class. We will show here how this formula can be derived from the one above.

Suppose we begin with the graph of a function _z_ = _f_ ( _x_ ), where 0 ≤ _a_ ≤ _x_ ≤ _b_ , and get a surface by revolving it around the _z_ -axis. The resulting surface is best parameterized with cylindrical coordinates, since it has the equation _z_ = _f_ ( _r_ ).

ψ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _f_ ( _r_ ))

_a_ ≤ _r_ ≤ _b_ , 0 ≤ _θ_ ≤ 2 _π_

The partial derivatives are

The cross product of these vectors is

The magnitude of this vector is

The desired surface area is now computed as an integral:

**EXAMPLE 8-9**

We compute the surface area of a sphere one final time. This time we think of the sphere as a surface of revolution and use the formula just derived. As in the previous example we will compute the area of the top half and double it.

To think of the top half of the sphere as a surface of revolution we begin with a function which gives part of a circle:

Applying the above formula for the surface area of a surface of revolution gives the following integral:

As this is just the top half of the sphere we must double it to get the correct formula, 4 _πR_ 2.

**Problem 86** _Find the area of the portion of the plane z_ = 2 _x_ \+ 3 _y that lies above the square with vertices at_ (0, 0, 0), (1, 0, 0), (0, 1, 0), _and_ (1, 1, 0).

**Problem 87** _Find the area of the portion of the graph of  that lies above the domain_

_1_. _R_ = {( _x_ , _y_ )|0 ≤ _x_ ≤ 1, 0 ≤ _y_ ≤ 1}

_2_. _D_ = {( _x_ , _y_ )| _x_ 2 \+ _y_ 2 ≤ 1}

**Problem 88** _Derive a formula for the graph of the spherical equation_ ρ = _f_ ( _ϕ_ ).

**Problem 89** _Find the area of the surface pictured below, parameterized by_

ψ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _θ_ )

–1 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ _π_

8.5 Surface Integrals

Just as we could integrate a function over a parameterized curve, we can also integrate functions over parameterized surfaces. The resulting integrals are called _surface integrals_. Here's a way to think about surface integrals. Imagine that there is a fog permeating all of  3. As the fog may not be uniform there is some function _f_ ( _x_ , _y_ , _z_ ) which gives the density of the water droplets at each point. Now imagine we have some surface in  3 cutting through the fog. We wish to find out the total amount of water droplets in the surface. If _f_ ( _x_ , _y_ , _z_ ) = 1, then this will just be the area of the surface. If not, then the answer will be the surface integral of _f_.

Recall the relationship between the arc length of a curve _C_ parameterized by a function ψ( _t_ ) and the line integral of a function _f_ ( _x_ , _y_ ) over _C_ :

The relationship between the surface area of a surface _S_ parameterized by a function ψ( _u_ , _v_ ) and the surface integral of a function _f_ ( _x_ , _y_ , _z_ ) over _S_ is the same:

The surface integral of _f_ ( _x_ , _y_ , _z_ ) over the surface S is denoted as

**EXAMPLE 8-10**

Let  . We calculate the surface integral of _f_ over the cylinder of radius 1, centered on the _z_ -axis, and lying between the plane _z_ = 0 and _z_ = 1.

First, the surface is parameterized using cylindrical coordinates as follows:

ψ ( _θ_ , _z_ ) = (cos _θ_ , sin _θ_ , _z_ )

0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _z_ ≤ 1

To compute the surface integral we will next need the partial derivatives of this parameterization:

The cross product of these vectors is

Finally, the magnitude of the cross product is then

We are now ready to compute the surface integral:

**Problem 90** _Calculate the surface integral of the function f_ ( _x_ , _y_ , _z_ ) = _x_ \+ _y_ \+ _z over the portion of the plane z_ = _x_ \+ _y that lies above the square in the xy-plane where_ 0 ≤ _x_ ≤ 1 _and_ 0 ≤ _y_ ≤ 1.

**Problem 91** _Calculate the surface integral of the function f_ ( _x_ , _y_ , _z_ ) = _z over the top half of the sphere of radius 1_.

8.6 Volume

In this section, we revisit the calculation of volumes of regions in  3. If the region is nice enough, we can just compute a triple integral of the function _f_ ( _x_ , _y_ , _z_ ) = 1 over the region. However, if the region is complicated we may want to employ a parameterization to help calculate its volume. The steps to define the volume by way of integrating a parameterization are similar to those used to define arc length and surface area. To this end, we suppose ψ( _u, v, w_ ) is a parameterization of a volume _V_ in  3, with domain _Q_. We now derive a formula for the volume of _V_ by executing the following steps:

1. Choose a lattice of points {( _u i, vj, wk_)} in _Q_. This defines a lattice {ψ( _u i, vj, wk_)} in _V_.

2. Connecting adjacent lattice points breaks up _V_ into a bunch of parallelepipeds.

3. We compute the volume of each parallelepiped. The edges are given by the vectors

The desired volume is then given by the determinant of the 3 × 3 matrix whose rows are _V u, VV_, and _V w_. We denote this as

The usual algebraic trickery transforms this as follows:

4. The desired volume is now the limit, as Δ _u_ , Δ _v_ , and Δ _w_ tend toward 0, of the sum of this quantity over all i, j, and k:

But, as in the computation of surface area,

Hence, after taking the limit the above summation becomes

**EXAMPLE 8-11**

We show that the volume bounded by a sphere of radius _R_ is  . One way to do this is to note that the top half of the sphere can be realized as the graph of the equation  . The desired volume can thus be obtained by doubling a double integral:

However, this integral is difficult to evaluate.

Instead, we employ a parameterization. The sphere is best parameterized in the usual way by using spherical coordinates:

ψ( _ρ_ , _θ, ϕ_ ) = ( _ρ_ sin _ϕ_ cos _θ_ , _ρ_ sin _ϕ_ sin _θ_ , _ρ_ cos _ϕ_ )

0 ≤ _ρ_ ≤ _R_ , 0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _ϕ_ ≤

We now compute all three partials of the parameterization:

The integrand that we will need is the determinant of the 3 × 3 matrix whose rows are the above vectors:

We now get the desired volume by integrating:

The negative sign is an unfortunate artifact of the fact that the determinant of a matrix will change sign if you swap two rows or two columns. We will see in later sections that this can be used to our advantage. For now, though, just note that a volume must always be positive. Hence, we take the absolute value of the above answer to recover the correct formula,  .

Although this method may seem very complicated at first, observe that it has a significant virtue: each step in the integration was very easy. This is the advantage of using parameterizations to compute volumes. Any volume problem can be solved by a double or triple integral. But employing a parameterization may transform difficult integrals into easy ones.

**Problem 92** _Show that the volume bounded by a cylinder of radius R and height h is π_ _R_ 2 _h by integrating with a parameterization from cylindrical coordinates_.

**Problem 93** _Calculate the volume under the cone z_ = 1 – _x_ 2 – _y_ 2, _and above the xy-plane by employing the parameterization_

ψ( _r_ , _θ_ , _z_ ) = ((1 – _z_ ) _r_ cos _θ_ , (1 – _z_ ) _r_ sin _θ_ , _z_ )

0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _z_ ≤ 1

8.7 Change of Variables

In Calculus I and II you learn how to integrate a function of one variable over a domain in  1. In Chapter 4, we learned how to integrate a function of two variables over a domain in  2, and a function of three variables over a domain in  3:

If the domain _V_ has a convenient parameterization then it may be easier to use this than to directly perform the integral. This is very similar to the relationship between line integrals and arc length, or surface integrals and surface area. As usual, we let ψ denote a parameterization of _V_ , with domain _Q_. If _V_ is two dimensional we have

And if _V_ is three dimensional

**EXAMPLE 8-12**

Recall the functions

These functions conveniently satisfy

cosh2 _t_ – sinh2 _t_ = 1

The first of these identities means that (cosh _t_ , sinh _t_ ) are the coordinates of a point on the hyperbola _x_ 2 _–_ y2 = 1. Suppose, we want to find the area of the region pictured below.

This region is parameterized by

ψ( _r_ , _t_ ) = ( _r_ cosh _t_ , _r_ sinh _t_ )

0 ≤ _r_ ≤ 1, – _T_ ≤ _t_ ≤ _T_

The area of a region _R_ of  2 is computed by integrating the function _f_ ( _x_ , _y_ ) = 1:

However, to calculate this integral directly would entail finding complicated limits of integration. Instead we use the parameterization ψ. To do this, we will need the partial derivatives of ψ:

The determinant of the matrix of partials is thus

We now compute the area:

**EXAMPLE 8-13**

We evaluate the integral

Doing this directly is difficult, so we employ a parameterization. The domain of integration is the intersection of a cylinder of height 1 and radius 1, centered on the _z_ -axis, with the positive octant. This can be parameterized with cylindrical coordinates by

The derivatives of this parameterization are

To do the integral we will need the determinant of the 3 × 3 matrix whose rows are these vectors:

We denote the function _x_ 2 \+ _y_ 2 \+ _z_ in the original integral by _f_ ( _x_ , _y_ , _z_ ). Then

**Problem 94** _Calculate the integral of the function f_ ( _x_ , _y_ ) = 2 _x_ – _y over the elliptical region in_  2 _parameterized by_

ψ( _r_ , _θ_ ) = (2 _r_ cos _θ_ , _r_ sin _θ_ )

0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ _π_

**Problem 95** _Calculate the integral of the function  over the region V of_  3 _parameterized by_

ψ( _r_ , _θ_ , _ω_ ) = ( _r_ cosh _ω_ cos _θ_ , _r_ cosh _ω_ sin _θ_ , sinh _ω_ )

1 ≤ _r_ ≤ 2, 0 ≤ _θ_ ≤ 2 _π_ , –2 ≤ _ω_ ≤ 2

( _The region V is between two hyperboloids, as depicted below_.)

**Problem 96** _Let V be the region inside the cylinder of radius 1, centered around the z-axis, and between the planes z_ = 0 _and z_ = 2. _Integrate the function f_ ( _x_ , _y_ , _z_ ) = _z over V_.

**Problem 97** _Let R be the region in the first quadrant of_  2, _below the line y_ = _x_ , _and bounded by x_ 2 \+ _y_ 2 = 4. _Integrate the function_

_over R_.

Quiz

**Problem 98**

_. a. Find a parameterization for the set of points in_  2 _that satisfies the equation x_ = sin _y_.

_b. Find a unit tangent vector to this curve at the point_ (0, 0).

_. Let Q be the region under the graph of f_ ( _x_ , _y_ ) = _x_ 2 \+ _y_ 2 _and above the square with vertices at_ (0, 0, 0), (1, 0, 0), (0, 1, 0), _and_ (1, 1, 0).

_a. Find a parameterization for Q of the form_ ψ ( _x_ , _y_ , _t_ ), _where_ 0 ≤ _x, y_ , _t_ ≤ 1 .

_b. Use the parameterization ψ to integrate over Q the function_

_. Let R be the region of_  2 _parameterized by_

_ϕ_ ( _r_ , _t_ ) = ( _r_ cosh _t_ , _r_ sinh _t_ ), 0≤ _r_ ≤ 1, –1 ≤ _t_ ≤ 1

_Integrate the function f_ ( _x_ , _y_ ) = _x_ 2 – _y_ 2 _over R_.
CHAPTER 9 |

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**Vector Fields and Derivatives**

9.1 Definition

A _vector field_ is simply a choice of vector for each point. So, for example, a vector field on  2 would have some vector at the point (1, 2), some other vector at the point (–1, 1), etc. We often draw vector fields by picking a few points and drawing the vector based at those points.

More formally, a vector field is a function from  2 to  2. What goes in to this function are the coordinates of the point where you are. What comes out are the components of the vector at that point.

**EXAMPLE 9-1**

Consider the vector field 〈 _xy_ , _x_ \+ _y_ 〉. At the point (1, 1) this vector field contains the vector 〈1, 2〉. At the point (2, 4) it contains the vector 〈8, 6〉. If we use a computer to draw it we would see something like

**EXAMPLE 9-2**

The vector field 〈sin _x_ , sin _y_ 〉 is depicted below.

**Problem 99** _Sketch the following vector fields_.

__. 〈 _x_ , _y_ 〉

__. 〈 _x_ , – _y_ 〉

__. 〈 _y_ , – _x_ 〉

9.2 Gradients, Revisited

We have already seen many vector fields, although we did not use this language. Whenever we take a function _f_ and compute its gradient ∇ _f_ at a point we get a vector. The set of all such vectors is then a vector field, which we now call "grad _f_ ."

**EXAMPLE 9-3**

Suppose _f_ ( _x_ , _y_ ) = _xy_ 2. Then the gradient of _f_ ( _x_ , _y_ ) at the point ( _x_ , _y_ ) is ∇ _f_ ( _x_ , _y_ ) = 〈 _y_ 2, 2 _xy_ 〉. If we draw this vector at various values of _x_ and _y_ , we get the picture depicted below.

**EXAMPLE 9-4**

Let _f_ ( _x_ , _y_ , _z_ ) = _xy_ 2 _z_ 3. Then ∇ _f_ = 〈 _y_ 2 _z_ 3, 2 _xyz_ 3, 3 _xy_ 2 _z_ 2〉.

**Problem 100** _For each of the following functions_ , _f_ , _compute_ ∇ _f_.

_. f_ ( _x_ , _y_ ) = _x_ \+ _y_

_. f_ ( _x_ , _y_ , _z_ ) = _x_ \+ _yz_

_. f_ ( _x_ , _y_ , _z_ ) = _xy_ \+ _xz_ \+ _yz_

9.3 Divergence

In the previous section, we saw that the gradient operator gives us a way to take a function _f_ ( _x_ , _y_ ) and get a vector field. In this section, we explore a way to take a vector field and get a function. Eventually, we will see that the value of this function at a point is a measure of how much the vector field is "spreading out" there.

Suppose 〈 _f_ ( _x_ , _y_ , _z_ ), _g_ ( _x_ , _y_ , _z_ ), _h_ ( _x_ , _y_ , _z_ )〉 is a vector field **V** on  3. Then we define the _divergence_ of **V** , "Div **V** ," to be the function

Note that the first term is associated with the first component of **V,** the second term with the second component, and the third term with the third component. This, and the fact that the terms are being added, should remind you of the dot product. This gives us a purely notational way to remember how to calculate the divergence of a vector field. We let ∇ denote the "vector"  . This, of course, is only a vector in a notational sense. But if we suspend our disbelief for a moment and allow such absurdities, we can write the formula for the divergence of a vector field in a very compact way:

Div **V** = ∇ · **V**

**EXAMPLE 9-5**

Let **V** be the vector field 〈 _x_ 2 _y_ , _x_ \+ _yz_ , _xy_ 2 _z_ 3〉. Then the divergence of **V** is calculated as follows:

**Problem 101** _Compute the divergence of the following vector fields_.

__. 〈 _y_ , _z_ , _x_ 〉

__. 〈 _x_ \+ _y_ , _x_ – _y_ , _z_ 〉

__. 〈 _x_ 2 \+ _y_ 2, _x_ 2 – _y_ 2, _z_ 2〉

**Problem 102** _Let f_ ( _x_ , _y_ , _z_ ) _be a function. What is_ ∇ · ∇ _f_ ?

**Problem 103** _Show that_

∇ · (F × **G** ) = **G** · (∇× **F** ) – **F** · (∇ × **G** )

9.4 Curl

There are many useful ways to apply partial derivatives to vector fields. We have already seen that gradients give us a way to take a function and get a vector field. Then we saw that divergence is a way to take a vector field and get a function. Here, we define a way to use partial derivatives to transform one vector field into another.

First, recall our notational absurdity, ∇ =  . In the previous section, we defined a new operation by using ∇ in a dot product. Here, we define an operation called _curl_ by using ∇ in a cross product. As before, let **V** = 〈 _f_ ( _x_ , _y_ , _z_ ), _g_ ( _x_ , _y_ , _z_ ), _h_ ( _x_ , _y_ , _z_ )〉. Then we define

**EXAMPLE 9-6**

Let **V** be the vector field 〈 _x_ 2 _y_ , _x_ \+ _yz_ , _xy_ 2 _z_ 3〉. Then the curl of **V** is calculated as follows:

Later we will see that the curl of a vector field measures how much it "twists" at each point.

**Problem 104** _Find the curl of each of the following vector fields_.

_1._ 〈 _x_ \+ _y_ , _x_ – _z_ , _y_ \+ _z_ 〉

_2._ 〈 _yz_ , – _xz_ , _xy_ 〉

**Problem 105** _Let f_ ( _x_ , _y_ , _z_ ) _be a function and_ **V** _a vector field on_  3. _Compute_

__. ∇ × (∇ _f_ )

__. ∇ · (∇ × **V** )

**Quiz**

**Problem 106**

_. Find the curl of the following vector field:_

_. Find the divergence of the following vector field:_

〈 _x_ 2 \+ _y_ 2, _y_ 2 – _x_ 2, 0〉

_. Find the gradient of_

_f_ ( _x_ , _y_ , _z_ ) = _x_ 2 sin( _y_ – _z_ )

_. Find two vector fields whose curls are_ 〈0, 0,  〉.
CHAPTER 10 |

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**Integrating Vector Fields**

In the previous chapter we introduced the concept of a vector field, and looked at various concepts of differentiation in relation to these objects. In this chapter, we look at ways to integrate a vector field. From calculus you should suspect that when we combine integration and differentiation something amazing will happen, akin to the _Fundamental Theorem of Calculus_. This is the topic of the next chapter.

**10.1 Line Integrals**

We will always integrate a vector field over a domain that can be parameterized by a line, surface, or volume. In this section, we look at the first of these situations, integrals over parameterized curves. As in Section 8.3, these will be called _line integrals_. However, the definition of the integral of a vector field over a parameterized curve is a bit different than the definition of a function over the same curve, so there is potential for confusion here.

Before giving the definition, it may help to give a physical motivation for line integrals. Drop a leaf into a flowing stream and follow it. Eventually, you may see it encounter a stick caught on some rocks. If the stick is perpendicular to the flow of water, the leaf will get stuck. If the stick is parallel to the flow, then the leaf rushes by as the force of the flowing water pushes it along. But if the stick is at some angle to the flow the leaf may hit it, and then slowly work its way past.

One can explain this by looking at the force the water can exert on the leaf. The leaf can only travel in a direction parallel to the stick, which we can represent as a vector, _L_. The water flowing under the stick exerts some force on the leaf, which can also be represented as a vector, _W_. If these vectors are perpendicular then the leaf does not move, so the net force _F_ on the leaf must be zero. On the other hand, the leaf moves the fastest when _L_ and _W_ are parallel, so _F_ must be greatest in this situation. It should be no surprise, then, that _F_ is proportional to the dot product, _L_ · _W_.

Before proceeding further, we must dispense with an important technicality. In the above figure, there were two choices for the vector _L_. One of these is depicted in the figure, and the other points in the exact opposite direction (but still along the stick). If the other choice was made, then the value of _L_ · _W_ would have exactly the opposite sign. Which is the correct choice? There is no right answer to this. We just have to declare, before we do any problem, which is correct. Such a declaration is called an _orientation_. We will say more about this idea shortly. For now, just note that if the wrong choice was made then we can correct things by changing the sign of our final answer.

Now we suppose that somehow the leaf is confined to some parameterized curve, _C_ , as the water rushes past. Sometimes the water may be perpendicular to the curve, and sometimes it may be parallel. The problem is to evaluate the total force that the water exerts on the leaf as it travels along the curve. (If the leaf always moves with unit speed then this is called the _work_ done by the water.)

Let ψ( _t_ ) be a parameterization of _C_. Let **W** be a vector field that gives the direction and speed of the water at each point of the stream. The direction that the leaf is moving at the point ψ( _t_ ) is the vector   Let W(ψ( _t_ )) be the vector of the vector field **W** that is based at the point ψ( _t_ ). Then the quantity we want to evaluate is

This integral is called the _line integral_ of **W** over _C_ , and is often denoted as

Our answer, unfortunately, is still incomplete. Recall that at some point we must make a choice, called an orientation. This amounts to deciding if the vector   points the "right way" or the "wrong way." An orientation can be denoted pictorially just by an arrow along _C_ , showing the correct choice. But this does not mean that to do the integral, we must choose a parameterization whose derivative points the right way. If it does not then we just need to change the sign of our final answer.

**EXAMPLE 10-1**

Let **W** = 〈 _xy_ , _x_ \+ _y_ 〉. Let _C_ be the oriented curve depicted below.

The orientation is given by the direction of the arrow in the figure. The curve is parameterized by

ψ( _t_ ) = ( _t_ 2, 3 _t_ ), 0 ≤ _t_ ≤ 2

At the point ψ(1) one can easily check that   This vector points in the direction opposite to the orientation. This just means that after we integrate we will have to flip the sign of our final answer.

We now calculate the line integral of **W** along _C_ using ψ. First, note that

**W** (ψ( _t_ )) = 〈( _t_ 2)(3 _t_ ), _t_ 2 \+ 3 _t_ ) = (3 _t_ 3, _t_ 2 \+ 3 _t_ 〉

and

As the orientation disagrees with ψ we must now flip the sign, obtaining our final answer,

An easy way to give an orientation on a curve parameterized by a function ψ( _t_ ) is to simply declare that   points in the right direction. In this case, we may go ahead and use ψ to evaluate the integral of **W** without any further considerations. In this case, we say the orientation of _C_ is the one _induced_ by ψ.

**EXAMPLE 10-2**

Let _C_ be the curve parameterized by

with the orientation induced by ψ.

Let  . We integrate **W** over _C_.

First, note that

Next, observe

We now integrate

As the orientation of _C_ is chosen to agree with ψ we do not need to worry about changing the sign of our answer.

One may wonder about the connection between line integrals of vector fields as defined here and line integrals of functions defined previously. Let _f_ denote a function on  _n_ and _C_ a curve in  _n_ parameterized by ψ( _t_ ). Let **W** be a vector field chosen so that at the point ψ( _t_ ) of _C_ the vector **W** (ψ( _t_ )) is tangent to _C_ (i.e., it points in the same direction as   and has length _f_ (ψ( _t_ )). Then we have

The definition of the line integral of **W** over _C_ then gives

This last equality is precisely the definition of the line integral of _f_ over _C_.

**Problem 107** _Calculate the line integral of_ 〈− _y, x_ 〉 _over the circle of radius 1, oriented counterclockwise_.

**Problem 108** _Let_ **W** = 〈 _xy_ , _xz_ 2, _y_ \+ _z_ 〉. _Let C be the curve parameterized by_

ψ( _t_ ) = ( _t_ 2, _t_ , 1 − _t_ ), 0 ≤ _t_ ≤ 1

_with the induced orientation. Calculate the line integral of_ **W** _over C_.

**Problem 109** _Let f_ ( _x_ , _y_ ) = _xy_ 2. _Let C be the portion of the parabola y_ = _x_ 2 _parameterized by_

ψ( _t_ ) **=** ( _t_ , _t_ 2), −1 ≤ _t_ ≤ 2

_with the induced orientation_.

_. Calculate the integral of the vector field_ ∇ _f over C_.

_. Calculate f_ (ψ (2)) − _f_ (ψ (−1)).

**Problem 110** _Let C be the subset of the graph of y_ = _x_ 2 _where_ 0 ≤ _x_ ≤ 1 ( _oriented away from the origin_ ) _. Let_ **W** _be the vector field_ 〈− _x_ 4, _xy_ 〉. _Integrate_ **W** _over C_.

**10.2 Surface Integrals**

In Section 8.3 we defined line integrals of functions, and in the previous section of this chapter we defined line integrals of vector fields. By analogy we will now define surface integrals of vector fields, just as we had defined surface integrals of functions.

The physical motivation for surface integrals of vector fields can again be seen by looking at a river of flowing water. Imagine a net placed vertically in the river. Suppose you wanted to calculate the amount of water flowing through the net at some particular moment in time. Let _W_ denote a vector representing the flow of water at some point of the net. Let _N_ be a vector which is perpendicular to the net. If the net is parallel to the direction of flow, then no water passes through it. In this situation, _N_ is perpendicular to _W_. The most water passes through the net when the net is perpendicular to the flow. In this case, _N_ and _W_ are parallel. It stands to reason, then, that the amount of water flowing through the net is proportional to _W_ · _N_.

Once again we encounter here a technical difficulty. Why did we draw the vector _N_ in the figure the way we did? If the only condition on _N_ is that it is perpendicular to the net, then why couldn't _N_ point in the exact opposite direction? The answer is that it can! Which choice is "correct" will again depend on a choice, called an _orientation_. We will discuss this more shortly.

Now let _S_ be a surface (representing the net), parameterized by ψ( _u_ , _v_ ). The vectors   are both tangent to S. Hence, the vector   is perpendicular, or _normal, to S_. Let **W** be a vector field representing the water flow, and **W** (ψ( _u_ , _v_ )) the vector of **W** at the point ψ( _u_ , _v_ ). Then the total amount of water flowing through _S_ is given by the integral

We call this the _surface integral_ of **W** over _S_ , and denote it by

But this formula isn't the whole story. We still must deal with the issue of orientations. Otherwise, two people evaluating   may get different answers, depending on which parameterization of _S_ they use to evaluate the integral. One way to give an orientation is to say which way is "up" at each point of _S_. This can be done by giving a vector _O_ which is perpendicular to _S_ at some point. Our parameterization ψ "agrees" with the choice of orientation if the vector   points in the same direction as _O_. If our parameterization does not agree with the specified orientation then we can remedy things simply by negating our final answer.

**EXAMPLE 10-3**

Let **W** = 〈 _y, x, z_ 〉. Let _S_ be the portion of a cylinder parameterized by

The vector 〈−1, 0, 0〉 defines an orientation on _S_ at the point (1, 0, 0).We now compute the integral of **W** over the surface _S_ with this orientation.

First, we compute the partials,

And so,

Now notice that the point (1, 0, 0) = ψ(0, 0). The vector   equal to 〈1,0, 0〉 at this point. This is exactly opposite to the specified orientation of S. Hence, if we use the parameterzation ψ to compute the integral of **W** over _S_ we will have to remember to negate our final answer.

We now integrate

As the parameterization ψ disagreed with the specified orientation the correct answer is

**Problem 111** _Let_ **W =** 〈 _x, y, z_ 〉 _. Integrate_ **W** _over the unit sphere, with orientation given by the vector_ 〈1, 0, 0〉 _at the point_ (1, 0, 0).

**Problem 112** _Let_ **W =** 〈 _xz, yz_ , 0〉. _Integrate_ **W** _over the surface S paramaterized by_

ψ( _u_ , _v_ ) = ( _u_ , _v_ , _u_ 2 \+ _v_ 2)

0 ≤ _u_ ≤ 1, 0 ≤ _v_ ≤ 1

_Use the induced orientation_.

**Problem 113** _Let S be the surface given by the following parameterization:_

ψ ( _θ_ , _ϕ_ ) = (cos _ϕ_ cos _θ_ , cos _ϕ_ sin _θ_ , sin _ϕ)_

( _Note that this is not quite spherical coordinates_.)

_Integrate over S the vector field_

**Problem 114** _Let_ **W =** 〈 _xz, yz_ , 0〉. _Let S denote the intersection of the unit sphere with the positive octant. An orientation on S is given by the vector_ 〈0, − 1, 0〉 _at the point_ (0, 1, 0). _Compute_

**Quiz**

**Problem 115**

_. Let C be the portion of the graph of x_ = _y_ 2 _where_ 0 ≤ _x_ ≤ 1, _as depicted in the figure. Integrate the vector field_ 〈1, 1〉 _over C with the indicated orientation_.

_. Let S be the frustum parameterized by_

ψ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _r_ ), 1 ≤ _r_ ≤2, 0 ≤ _θ_ ≤2 _π_

_Integrate the vector field  over S with the induced orientation_.
CHAPTER 11 |

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**Integration Theorems**

A centerpiece of the calculus is the "Fundamental Theorem." This important result says that when you take a function, differentiate it, and then integrate the result something special happens. In this chapter, we explore what happens when we do the same with vector fields. As there are many different ways to both differentiate and integrate a vector field this gives us many different theorems. It should be pointed out that all of these results (including the Fundamental Theorem of Calculus) are special cases of the _generalized Stokes' Theorem_ , which we will not cover here. This approach uses the theory of differential forms, rather than vector calculus, and can be found in the book _A Geometric Approach to Differential Forms_ , published by Birkhäuser, 2006.

**11.1 Path Independence**

The first way we saw vector fields related to derivatives was through the gradient. In this section we see what happens when you start with a function _f_ , take its gradient to get a vector field, and then perform a line integral along a curve _C_ on this field.Formally, we study integrals of the form

To really understand what is going on we need to go back to the definition of a line integral. Suppose _C_ is parameterized by ψ( _t_ ), and the domain of ψ is the interval [ _a_ , _b_ ]. The steps to define the integral of ∇ _f_ over _C_ are then

1. Choose _n_ evenly spaced points in the interval [a, b]. Call these points { _t_ _i_ }. This gives us _n_ point {ψ( _t i_} on the curve _C_.

2. Let Δ _t_ = _t_ _i_ +1 − _t_ _i_.

3. For each _i_ compute   at the point ψ( _t i_).

4. Sum over all _i_.

5. The integral   is defined to be the limit of the sum from the previous step, as _n_ → ∞.

Let us now focus on the quantity computed in Step 3 above. To estimate ∇ _f_ · _V_ at a point _p_ we may substitute

This approximation gets better and better as Δ _t_ approaches zero. Furthermore, if Δ _t_ is small one can also make the approximation

Putting these together we have

Hence, the term   computed in Step 3 above can be approximated by _f_ (ψ( _t_ _i_ +1)) − _f_ (ψ( _t_ _i_ )).

We now shift our attention to Step 4:

But this last expression telescopes:

We conclude with the following theorem:

Suppose ψ is a parameterization of _C_ with domain [ _a_ , _b_ ] which agrees with the orientation on _C_. Then for any function _f_ we have

This should look a lot to you like the Fundamental Theorem of Calculus. There is no such convenient name for it. Often it is referred to as the _path independence of line integrals of gradient fields_. This is because it says that the result of a line integral of a gradient field only depend on the endpoints of the curve, and not the path used to get from one endpoint to the other.

**EXAMPLE 11-1**

Suppose _f_ ( _x_ , _y_ ) = _x_ 3 \+ _y_ 2. Let _C_ be the top half of the unit circle, oriented counterclockwise. We compute

All we really have to know is the endpoints of the curve _C_. The first is (1, 0) and the second is (−1, 0). (Which one is which is determined by the orientation on _C_.) Hence,

There is an important application of the independence of path of line integrals of gradient fields that may be familiar. Recall that a line integral of a vector field says something about how much _work_ you have to do to move an object along a curve _C_ in the presence of a force **W**. Let's say you want to know how much work you have to do against the force of gravity to get a heavy package up a mountain. Suppose the mountain is represented by the graph of a function _f_ ( _x_ , _y_ ). That is, at the map cooridnates ( _x_ , _y_ ) the function _f_ ( _x_ , _y_ ) gives you your altitude. Then the force of gravity at the point ( _x_ , _y_ ) is proportional to the vector ∇ _f_ ( _x_ , _y_ ) (i.e., the steeper the mountain, the harder you have to work to get up it). Let's suppose you have identified your route on a map. You start at the point ( _x_ 0, _y_ 0), follow some curve _C_ , and end up at ( _x_ 1, _y_ 1). Then the work you have to do to overcome gravity is

Notice that the result is just the difference in elevation between your beginning and ending point, and doesn't matter what path you take to get from one to the other!

**EXAMPLE 11-2**

Let **W** be the vector field 〈 _y_ 2 _z_ 3, 2 _xyz_ 3, 3 _xy_ 2 _z_ 2〉. We would like to integrate **W** over the curve _C_ parameterized by

ψ( _t_ ) = ( _t_ 3, _t_ 2, _t_ ), 1 ≤ _t_ ≤2

with the induced orientation.

This integral can be done directly, but the astute reader will notice that

**W** = ∇( _xy_ 2 _z_ 3)

Hence, we may use the independence of path of line integrals:

**Problem 116** _Let_ **W** = 〈1, 1, 1〉. _Suppose C is some curve that goes from_ (1, 0, 1) _to_ (1, 1, 1). _Calculate the integral of_ **W** _over C_.

**Problem 117** _Let C be the curve pictured below. Let_ **W** = 〈sin _y_ , _x_ cos _y_ 〉. _Calculate_

**Problem 118** _Suppose f_ ( _x_ , _y_ ) = _x_ 2 − 4 _x_ \+ 4 + _y_ 2 \+ 2 _y_ \+ 1. _Let C be a curve that starts at_ (2, − 1) _and ends at some other point. Show that_   _is (strictly) larger than zero_.

**Problem 119** _Suppose C is a closed curve, i.e., one whose beginning and ending points are the same. Show that the integral of any gradient field over C is zero_.

**Problem 120** _Deduce the Fundamental Theorem of Calculus from the independence of path of line integrals of gradient fields. (Hint: Begin by letting C be a curve in_  1 _from a to b.)_

**Problem 121** _In this section we saw that the work you must do to carry a package halfway up a mountain is just proportional to the difference in your starting and ending altitudes, and does not depend on the path you take. But it certainly seems like if you started at the bottom of a mountain, went to the top, and then came down to the halfway point, you'd be doing more work than if you just went straight up to the halfway point. Explain_.

**11.2 Green's Theorem on Rectangular Domains**

The next theorem that relates derivatives, integrals, and vector fields is special to  2. Recall that the curl of a vector field

**V** = 〈 _f_ ( _x_ , _y_ , _z_ ), _g_ ( _x_ , _y_ , _z_ ), _h_ ( _x_ , _y_ , _z_ )〉

in  3 is defined to be the quantity

Although the curl is only defined for vector fields in  3, we can use it to define a natural (and important!) operation on vector fields in  2. First, given a vector field **W** = 〈 _f_ ( _x_ , _y_ ), _g_ ( _x_ , _y_ )〉 on  2 we can easily create a vector field on  3: 〈 _f_ ( _x_ , _y_ ), _g_ ( _x_ , _y_ ), 0〉. We will abuse the notation and continue to call this new vector field **W.** We now compute its curl:

Hence,

So, if **W** is a vector field on  2 then the result of |∇ × **W** | is a function on  2. This function acts much like the derivative of **W.** We now ask, what happens when we integrate the result of such a derivative? That is, if _Q_ is some domain in  2, then what can we say about

For simplicity, we will assume here that the domain of integration _Q_ is a rectangle. In the next section, we will explore more general domains.

To understand what is going on here, we will have to go all the way back to the definition of a multiple integral. Recall the steps to define the value of

1. Choose a lattice of evenly spaced points ( _x i_, _y j_) in _Q_.

2. Define

Δ _x_ = _x_ _i_ +1 − _x i_, Δ _y_ = _y_ _j_ +1 − _y_ _j_

3. For each _i_ and _j_ compute _h_ ( _x i_, _y j_)Δ _x_ Δ _y_.

4. Sum over all _i_ and _j_.

5. Take the limits of the resulting number as Δ _x_ and Δ _y_ approach zero.

To define the value of   we follow the above steps, with _h_ ( _x_ , _y_ ) =   Hence, in Step 3 above we compute

If Δ _x_ and Δ _y_ are small, we may make the following approximations:

and

Hence, the quantity computed in Step 3 can be approximated by

We now assume _Q_ is the rectangle pictured in Figure 11-1. As in the figure, let _L_ , _R_ , _B_ , and _T_ denote the Left, Right, Bottom, and Top sides of this rectangle, with the indicated orientations.

**Figure 11-1** The rectangle, _Q_

We now move on to Step 4. In this step, we are instructed to compute the sum of the values in Step 3. Using our approximation of each term, we can approximate

this sum by

Note that each term in the final expression is a sum involving values of _f_ or _g_ at points on the sides of _Q_. For example, in the first sum we see the expression _g_ ( _x n_, _y j_). The number _n_ is a constant, and represents the largest index of the _x_ -values. Hence, a point of the form ( _x n_, _y j_) must lie on _R_ , the Right side of _Q_.

We are now prepared to consider Step 5, where we take limits as Δ _x_ and Δ _y_ go to zero. We do this for each term in the above sum. For example, consider the limit of the first term:

But this integral is the same as the integral of **W** over _R_ with the orientation indicated in Figure 11-1. To see this, first parameterize _R_ by

ψ( _y_ ) = ( _x n_, _y_ ), _y 0_ ≤ _y_ ≤ _y m_

Then,

and so

The other three terms in the sum from Step 4 give similar integrals over the remaining sides of _Q_. We are now ready for the conclusion:

The sides _R_ , _L_ , _T_ , and _B_ of _Q_ , with the orientations as indicated in Figure 11-1, when taken together are referred to as its _boundary_. The usual notation for this is ∂ _Q_. Using this notation we can write our conclusion much more succinctly:

This final equation is known as "Green's Theorem."

**EXAMPLE 11-3**

Let _Q_ be the rectangle in the plane with corners at (0, 0), (2, 0), (0, 3), and (2, 3). Let **W** = 〈 _xy_ , _x_ 2 \+ _y_ 2〉. We will use Green's Theorem to evaluate the integral of **W** over ∂ _Q_.

Green's Theorem enables us to see the geometric significance of the value of   at a point ( _x_ 0, _y_ 0). Let _Q_ denote a _very_ small rectangle around this point. Then at each point of _Q_ the value of   is roughly constant, and equal to its value at ( _x_ 0, _y_ 0). Hence,

But Green's Theorem says

Putting these together gives us

or,

So, the function   is a measure of the "circulation" of **W** around each point. There is a way to experience such a function for yourself. Many people enjoy going "tubing"—floating down a river in an inner tube. Suppose you are tubing and decide you want to stop somewhere to enjoy the scenery, so you drop an anchor. Then you notice that the water on your left is rushing past you faster than the water on your right. What happens? Your inner tube starts to turn. This turning is a measure of the strength of  .

**Problem 122** _Let Q be the rectangle_ {( _x_ , _y_ )|0 ≤ _x_ ≤ 1, 0 ≤ _y_ ≤ 1}. _Use Green's Theorem to evaluate the integral of_ 〈− _y_ 2, _x_ 2〉 _over_ ∂ _Q_.

**Problem 123** _Let C be the curve pictured below. Show that the integral of_ **W** = 〈 _y_ , _x_ 〉 _over C does not depend on b_.

**11.3 Green's Theorem over More General Domains**

Although we have only demonstrated the validity of Green's Theorem for rectangular regions, it holds in much more generality. For other shaped regions _Q_ the expression _∂Q_ should generally be interpreted as the "edge" of _Q_ , with a counterclockwise orientation. But there is an important technical restriction on the types of regions _Q_ for which this is valid.

We begin by examining what happens when we look at Green's Theorem applied to two neighboring rectangles.

We apply Green's Theorem to the rectangle _Q_ 1 ∪ _Q_ 2:

But the integral of **W** along _R_ 1 will cancel with the integral of **W** along _L_ 2, so the last integral equals

as Green's Theorem would predict.

The cancellation of the integrals over _R_ 1 and _L_ 2 illustrates an important phenomenon. For example, suppose _Q_ is the following region.

We can use Green's Theorem to analyze the integral of   over _Q_ by breaking it up into rectangles. The boundary of each rectangle then gets an orientation. When we add up the integral of **W** over the boundary of every rectangle, there is much cancellation, as in the following figure.

We end up with the integral of   being equal to the integral of **W** over the squares depicted in the following figure, with the indicated orientations. Hence, we say that this is the appropriate boundary of _Q_.

In general, we may use the following rule-of-thumb to figure out the orientation on each loop of its boundary:

If _Q_ is a connected region in  2 then the "outermost" loop of its boundary is oriented counterclockwise and all other loops of the boundary are oriented clockwise.

**EXAMPLE 11-4**

We use Green's Theorem to integrate the function _x_ 2 \+ _y_ 2 over the inside of the unit circle. If we denote this region as _Q_ , then the boundary of _Q_ is the unit circle itself, with a counterclockwise orientation. This can be parameterized in the usual way by

ψ( _t_ ) = (cos _t_ , sin _t_ ), 0 ≤ _t_ ≤ 2 _π_

To use Green's Theorem we must find functions _f_ and _g_ so that

A suitable choice for _g_ ( _x_ , _y_ ) can be found by integrating _y_ 2 with respect to _x_ , yielding the function _xy_ 2. Similarly, we may find _f_ ( _x_ , _y_ ) by integrating − _x_ 2 with respect to _y_ , yielding − _x_ 2 _y_.So

〈 _f_ ( _x_ , _y_ ), _g_ ( _x_ , _y_ )〉 = 〈− _x_ 2 _y_ , _xy_ 2〉

Finally, we integrate

**Problem 124** _Let_ **W** = 〈− _y_ 2, _x_ 2〉. _Let σ be the region in_  2 _parameterized by the following_ :

_ϕ_ ( _u_ , _v_ ) = (2 _u_ − _v_ , _u_ \+ _v_ )

_where_ 1 ≤ _u_ ≤ 2 _and_ 0 ≤ _v_ ≤ 1. _Calculate_

**Problem 125** _Calculate the area enclosed by the unit circle by integrating some vector field around its boundary_.

**Problem 126**

_. Suppose_ **W** = 〈 _f_ ( _x_ ), _g_ ( _x_ )〉 _is a vector field which is defined everywhere except at_ (0, 0). _If_   _then show that the integral of_ **W** _along every circle centered on the origin, oriented counterclockwise, is the same_.

_. If_ **W** =   _then show that the integral of_ **W** _along every circle centered on the origin, oriented counterclockwise, is the same_.

_. Calculate the integral of_   _over the unit circle_.

**Problem 127** _Let σ be the region parameterized by_

_ϕ_ ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ ), 0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ _π_

_Suppose_ **W** = 〈 _x_ 2, _e_ _y_ 〉.

_. Use Green's Theorem to show that_

_. Let C be the horizontal segment connecting_ (−1, 0) _to_ (1, 0). _Calculate_

_. Use your previous answers to determine the integral of_ **W** _over the top half of the unit circle (oriented counterclockwise)_.

**11.4 Stokes' Theorem**

In the previous section, we saw that if **W** is a vector field in  2, then we can view |∇ × **W** | as a kind of derivative. When we integrated this "derivative" we saw something special happen, namely Green's Theorem:

We now move our attention to  3, and explore a similar phenomenon. Suppose now **W** is a vector field in  3, and _S_ is a surface. Then we wish to explore   A reasonable guess, based on our experience from the previous section, would be

This turns out to be the case, and is called _Stokes' Theorem_. We will not prove it here, as the proof is extremely similar to that of Green's Theorem. The strategy, once again, is to choose a lattice of points in S. This breaks up _S_ into a bunch of parallelograms, and we can chase through the definition of   on each. At every parallelogram, we see that we get the same as the integral of **W** over the boundary of the parallelogram. But, because of orientation considerations, the integrals over the boundaries of neighboring parallelograms cancel. The result is the integral of **W** over the boundary of _S_.

One potential complication in using Stokes' Theorem is determining the boundary of the surface _S_ in question. To get the proper orientation on ∂ _S_ you need to know the orientation of _S_. Recall that this is often given by an outward-pointing normal vector, _v_. To get the orientation on the boundary, we use the "right-hand rule." To do this point the thumb of your right hand in the direction of _v_. Your fingers will then curl in the sense that determines the orientation on the boundary.

**EXAMPLE 11-5**

Let _S_ denote the top half of the unit sphere, with orientation given by the normal vector 〈1, 0, 0〉 at the point (1, 0, 0). We use Stokes' Theorem to integrate the curl of the vector field 〈− _y_ , _x_ , 0〉 over _S_.

First, note that Stokes' Theorem says that the answer will be the same as the integral of 〈− _y_ , _x_ , 0〉 around ∂ _S_. The boundary of _S_ (with proper orientation) is parameterized by

ψ( _t_ ) = (cos _t_ , sin _t_ , 0), 0 ≤ _t_ ≤ 2 _π_

Thus, we may integrate

**EXAMPLE 11-6**

Let _S_ denote the portion of the paraboloid _z_ = 2 − _x_ 2 − _y_ 2 that lies above the plane _z_ = 1, with an orientation determined by an upward pointing normal. Let **W** = 〈cos _z_ , sin _z_ , 0〉. We will use Stokes' Theorem indirectly to find

First, let _D_ be the disk in the plane _z_ = 1 bounded by the unit circle, with orientation given by an upward pointing normal. Then ∂ _S_ = ∂ _D_. Stokes' Theorem says that the integral of ∇ × **W** over both _D_ and _S_ is equal to the integral of **W** over ∂ _S_. So, to get an answer to the original problem we may evaluate the integral of ∇ × **W** over _D_ instead of _S_.

To do the integral, note that

∇ × **W** = 〈− cos _z_ , sin _z_ , 0〉

So, on the plane _z_ = 4 we have ∇ × **W** = 〈−cos 1, sin 1, 0〉. A parameterization for _D_ is given by

ψ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , 1), 0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ 2 _π_

The reader may check that

We now integrate

**Problem 128** _Let_ **W** = 〈 _xy_ , _xz_ , _y_ 〉. _Let S be the surface parameterized by_

ψ( _r_ , _θ_ ) = (0, _r_ cos _θ_ , _r_ sin _θ_ ), 0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ 2 _π_

_with the induced orientation. Calculate_

**Problem 129** _Suppose_ **W** = 〈 _f_ ( _x_ , _y_ ), _g_ ( _x_ , _y_ ),0〉 _and S is a region of_  3 _that lies in the xy-plane. Show that Stokes' Theorem applied to_ **W** _and S is equivalent to Green's Theorem_.

**Problem 130** _Let S be the portion of the cylinder x_ 2 \+ _y_ 2 = 1 _that lies between the planes z_ = 0 _and z_ = 1, _with orientation given by the normal vector_ 〈1, 0, 0〉 _at the point_ (1, 0, 0). _Let_ **W** = 〈− _yz_ , _xz_ , 0〉. _Calculate the integral of_ ∇ × **W** _over S_.

**Problem 131** _If_ **W** _is a vector field defined on all of_  3, _then show that the integral of_ ∇ × **W** _over the unit sphere is zero_.

**11.5 Geometric Interpretation of Curl**

Just as Green's Theorem gave us a way to interpret the function   geometrically, we can use Stokes' Theorem to give a geometric interpretation of the curl of a vector field. Let **W** = 〈 _f_ , _g_ , _h_ 〉 be a vector field on  3 and _p_ a point of  3. Our goal is to understand the meaning of the vector ∇ × **W** ( _p_ ).

Let _D_ be a small, flat disk centered on _p_. If _D_ is small enough, then ∇ × **W** is roughly constant at every point of D. Let ψ( _u_ , _v_ ) denote a parameterization of _D_ , with domain _R_ , in which

In particular, this implies

Since _D_ is part of a plane the vectors normal to _D_ are all parallel. Hence, our assumption that the magnitude of   is constant implies

for some fixed unit vector _N_ which is normal to _D_.

We now examine the integral of the curl of **W** over _D_ :

Recall that the dot product of two vectors is the product of their magnitudes times the cosine of the angle between them. Hence, if we choose _D_ so that _N_ points in the same direction as ∇ × **W** ( _p_ ) we get

(∇ × **W** ( _p_ )) · _N_ = |∇ × **W** ( _p_ )|

and hence

To go further we must appeal to Stokes' Theorem:

Putting these together then gives

or,

Our conclusion is that the magnitude of ∇ × **W** at the point _p_ is a measure of how much **W** circulates around it. The direction of ∇ × **W** is the same as the direction of _N_ , which was chosen to be perpendicular to the plane in which this circulation is greatest. In this sense, the curl of a vector field is the three-dimensional version of the function   which appears in Green's Theorem.

**EXAMPLE 11-7**

Let **W** = 〈− _y_ , _x_ , 0〉. Then

This is a vector that points up. Notice that a plane perpendicular to an upward pointing vector is the plane that contains the "circulation" of **W.** In contrast, the circulation in a plane which contains ∇ × **W** would be zero.

**Problem 132** _Among all circles C in_  3 _centered at the origin with radius 1, suppose the circulation_   _is greatest when C lies in the xz-plane. Suppose, furthermore, that when C is such a loop_ ,

_Estimate_ ∇ × **W** _at the origin_.

**Problem 133** _Suppose some vector field has the property that the direction of every vector is up, and in any vertical plane parallel to the xz-plane the vector field is constant. Then what direction would the curl of this vector field point?_

**11.6 Gauss' Theorem**

We know that one way to "differentiate" a vector field is to take its curl. It is then not surprising that the integral of the curl of a vector field should be special. Another way to "differentiate" a vector field in  3 is to take its divergence. In this section, we explore what happens when we integrate the divergence of a vector field. To this end, suppose

**W** = 〈 _f_ ( _x_ , _y_ , _z_ ), _g_ ( _x_ , _y_ , _z_ ), _h_ ( _x_ , _y_ , _z_ )〉

Recall the definition of divergence:

The result is a function on  3. We may thus integrate this function over volumes _V_ :

As in the previous sections, we might guess that there is a relationship between this and the integral of **W** over the boundary of _V:_

This equality is in fact true, and is known as _Gauss' Divergence Theorem_. The proof is again similar to the proof of Green's Theorem. We choose a three-dimensional lattice of points in _V_ , and approximate   as a sum over the points of this lattice. The lattice breaks up _V_ into little cubes, and we find that the integral of ∇ · **W** over each cube is approximately equal to the integral of **W** over the boundary of each cube, with suitable orientations. But faces of cubes inherit opposite orientations from neighboring cubes, so in the sum all that is left are the faces of the cubes on the boundary of _V_.

To properly orient ∂ _V_ , we simply choose a normal vector that points "out" of _V_.

**EXAMPLE 11-8**

Let **W** = 〈 _x_ , _y_ , _z_ 〉. We would like to find the value of the integral of ∇ · **W** over the volume _V_ bounded by the unit sphere. According to Gauss' Theorem, this is equal to the integral of **W** over the unit sphere _S_.

To evaluate this, we first parameterize the unit sphere in the usual way with spherical coordinates:

ψ( _θ_ , _ϕ_ ) = (sin _ϕ_ cos _θ_ , sin _ϕ_ sin _θ_ , cos _ϕ_ )

0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _ϕ_ ≤ _π_

Now we compute

To check orientations, note that at   this vector is equal to 〈1, 0, 0〉, which does indeed point out of _V_. Hence, we do not need to worry about negating the value of an integral that is computed using ψ.

Finally, we integrate:

**EXAMPLE 11-9**

Let **W** = 〈0, 0, _z_ 2〉. We integrate ∇ · **W** over the volume _V_ which is inside the cylinder _x_ 2 \+ _y_ 2 = 1, above the plane _z_ = 0, and below the plane _z_ = 1.

To use Gauss' Theorem we will have to parameterize each "side" of _V_. First, the cylinder, _C_ :

ψ( _θ_ , _z_ ) = (cos _θ_ , sin _θ_ , _z_ ), 0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _z_ ≤ 1

Then the bottom, _B_ :

ψ−( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , 0), 0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ 2 _π_

And finally the top, _T_ :

ψ+( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , 1), 0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ 2 _π_

Now we must check orientations by computing normal vectors:

This vector points out of _V_.

This vector points "up," which is into _V_. We will have to remember to negate the value of any integral that is computed using ψ−.

This vector again points "up," but at the top this is pointing out of _V_.

Gauss' Theorem says

We compute each of these integrals individually:

Hence,

**Problem 134** _Integrate_ **W** = 〈 _x_ 2 _yz_ , _xy_ 2 _z_ , _xyz_ 2〉 _over the boundary of the unit cube in_  3.

**Problem 135** _Let_ **W** = 〈0, 0, _e z_〉. _Calculate the integral of_ ∇ · **W** _over the ball bounded by the unit sphere_.

**Problem 136** _Let_ **W** = 〈 _x_ 3, _y_ 3, 0〉. _Let V be the region between the cylinders of radii 1 and 2 (centered on the z-axis), in the positive octant, and below the plane z_ = 2. _Calculate_

**Problem 137** _The surfaces C and D are defined by_

_1. C is the graph of the cylindrical equation r_ = cos _θ in_  3, _where_ 0 ≤ _θ_ ≤ _π_ _and_ 0 ≤ _z_ ≤ 1.

_2. D is the set of points in the plane z_ = 1, _which are within_   _of a unit away from the point_ ( , 0, 1).

_Let_ **W** _be the vector field_ 〈0, _xyz_ , 0〉. _Calculate_

**Problem 138** _Suppose_ ∇ · **W** = 0, _and S_ 1 _and S_ 2 _are oriented surfaces with the same oriented boundary. Show that_

_(For simplicity you may assume that S_ 1 _and S_ 2 _only meet in their boundary.)_

**11.7 Geometric Interpretation of Divergence**

Recall that Green's Theorem gave us a geometric interpretation of   and Stokes' Theorem gave us a geometric interpretation for ∇ × **W**. We now use Gauss' Theorem to give a geometric interpretation of ∇ . **W**. This will be completely analogous.

Let **W** denote a vector field on  3. Let _p_ be a point in  3 and let _B_ be a small ball centered on _p_. If _B_ is chosen small enough then the function ∇ · **W** is approximately equal to ∇ · **W** ( _p_ ) on all of _B_. Hence,

Now, Gauss' Theorem says

Combining these we get

or,

The quantity   is a measure of the net amount of **W** which leaves the ball _B_. So, if the same amount of **W** enters and leaves _B_ then this integral will be zero. One way to think about this is that   and hence ∇ · **W** ( _p_ ), is a measure of how much **W** "spreads out" at _p_.

**EXAMPLE 11-10**

The vector fields pictured below have more "leaving" each point than "entering," and so have positive divergence. For example, the second one pictured might be something like 〈 _x_ , _y_ , _z_ 〉, whose divergence is 3. The third one might be 〈0, 0, _z_ 〉, whose divergence is 1. Can you think of a vector field like the first one?

**EXAMPLE 11-11**

The vector fields pictured below have zero divergence. The first one is a constant vector field. The second might be something like 〈 _y_ , − _x_ , 0〉.

**Problem 139** _Suppose B is a ball of radius_ 1 _centered around the origin. Let_ **W** _be a vector field and suppose_

_Estimate the value of_ ∇ · **W** _at the origin_.

**Quiz**

**Problem 140**

_. Let C be any curve in_  3 _from_ (0, 0, 0) _to_ (1, 1, 1). _Let_ **W** _be the vector field_ 〈 _y_ 2 _z_ 2, 2 _xyz_ 2, 2 _xy_ 2 _z_ 〉. _Calculate_

_. Let σ be the region parameterized by the following:_

_ϕ_ ( _u_ , _v_ ) = ( _uv_ 2, _u_ 3 _v_ ), 1 ≤ _u_ ≤ 2, 1 ≤ _v_ ≤ 2

_Calculate_

_. Let C1 and C2 be curves given by the following parameterizations (with the induced orientations):_

_Show that for any vector field_ **W** _such that_ ∇ × **W** = 〈0, 0, 0〉 _the following is true:_

**Final Exam**

**Problem 141**

_. Let f_ ( _x_ , _y_ ) _be the following function:_

_f_ ( _x_ , _y_ ) = _xy_ \+ _x_ – 2 _y_ \+ 4

_a. Sketch the intersections of f_ ( _x_ , _y_ ) _with the xz- and yz-planes_.

_b. Find the critical point(s) and compute the value of_

_Can you say if the graph has a max, min, or saddle at the critical point_ ( _s_ ) _?_

_c. Find the slope of the tangent line to the graph of f_ ( _x_ , _y_ ), _in the direction of_ 〈1, 2〉, _at the point_ (0, 1).

_d. Find the volume under the graph of f_ ( _x_ , _y_ ), _and above the rectangle in the xy-plane with vertices at_ (0, 0), (1, 0), (0, 2), _and_ (1, 2).

_. Parameterize the portion of the graph of z_ = 4 – _x_ 2 – _y_ 2 _that lies above the xy-plane_.

_. Parameterize the curve that lies on a sphere of radius 1 such that θ_ = _ϕ_

_. Let_ **W** = 〈 _xz_ 2, 0, _xz_ 2〉. _Calculate_ ∇ · **W** _and_ ∇ × **W**.

_. Let V be the volume in the first octant, inside the cylinder of radius 1, and below the plane z_ = 1. _Integrate the function_

_over V_.

_. Let C be the curve parameterized by the following:_

_ϕ_ ( _t_ ) = (2 cos _t_ , 2 sin _t_ , _t_ 2), 0 ≤ _t_ ≤ 2

_Integrate the vector field_ 〈0, 0, _x_ 2 \+ _y_ 2〉 _over C_.

_. Let_ **F** _be the vector field_ 〈0, – _z_ , 0〉. _Let P be the portion of a paraboloid parameterized by_

_ϕ_ ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _r_ 2)

0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤

_Calculate _

_. Let_ **W** = 〈 _yz_ , _xz_ , _xy_ 〉. _Let C be the curve parameterized by_

_with the induced orientation. Evaluate_   _without integrating!_

_. Let S be the can-shaped surface in_ ℝ3 _whose side is the cylinder of radius 1 (centered on the z-axis), and whose top and bottom are in the planes z_ = 1 _and z_ = 0. _Let_ **W** = 〈0, 0, _z_ 2〉. _Use Gauss' Theorem to calculate_

**Answers to Problems**

**Chapter 1: Functions of Multiple Variables**

**Problem 1**

. _f_ (3, 2) = 32 \+ 23 = 9 + 8 = 17

. _g_ (0,  ) = sin(0) + cos( ) = 0 + 0 = 0

. _h_ (2,  ) = 22 sin( ) = (4)(1) = 4

**Problem 2**

(b), (d)

**Problem 3**

**Problem 4**

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**Problem 5**

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**Problem 6**

. The circles get larger as _z_ increases, but they begin to "bunch up."

. The circles get larger as _z_ gets larger, and their spacing does not change.

. The circles get smaller as _z_ gets larger, eventually becoming very close to the origin.

. At _z_ = 0 the level set consists of infinitely many circles. For | _z_ | > 1 there are no level curves.

**Problem 7**

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**Problem 8**

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**Problem 9**

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**Problem 10**

The function (cos _t_ 2, sin _t_ 2) describes a point moving clockwise around the circle, and slowing down until _t =_ 0. (It then reverses direction when _t_ becomes positive.) In contrast, the function (cos _t_ , sin _t)_ describes a point moving counterclockwise around the circle, at uniform speed, no matter what _t_ is.

**Problem 11**

_c_ ( _t_ ) = ( _t, f_ ( _t_ ))

**Problem 12**

. The curve spirals around the _z_ -axis, rising faster and faster, so that there is more and more space between successive coils.

. The curve spirals around the _z_ -axis, descending, becoming closer and closer to the _xy_ -plane.

. The curve spirals up around the cone

**Chapter 1 Quiz**

**Problem 13**

. Left handed

.(a) The intersection with the _xy_ -plane looks like the _x_ -axis. The intersection with the _xz_ -plane also looks like the _x_ -axis. The intersection with the _yz_ -plane is the line _z_ = _y_ , pictured here.

(b) |
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(c) |

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**Chapter 2: Fundamentals of Advanced Calculus**

**Problem 14**

. Along the _y_ -axis we have

Along the _x_ -axis

. Along the _y_ -axis

Along the line _y_ = _x_

. Along the _y_ -axis

Along the line _y_ = _x_

. Along the _y_ -axis

Along the curve _y_ = _x 2_

**Problem 15**

. _y_ ≠ _x_

. _y_ ≥ _x_ 2

. The domain is empty.

**Problem 16**

Yes. For small values of _t_ the function sin( _t_ ) is approximately equal to _t_. The value of _x_ 2 \+ _y_ 2 is the square of the distance from ( _x_ , _y_ ) to (0, 0), which is small for points near the origin.

**Chapter 2 Quiz**

**Problem 17**

. On the _x_ -axis we know _y =_ 0, and so the function _f_ ( _x, y_ ) = 0 (as long as _x_ is not also 0). Similarly, on the _y_ -axis we know _x_ = 0, and so again the function _f_ ( _x, y_ ) = 0. But when _x = y_ we have

To evaluate the limit of this as _x_ → 0 we use L'Hopital's rule twice

Hence, the limit does not exist.

. As long as both _x_ and _y_ are not zero, the function   Hence, the function _f_ ( _x, y_ ) = 1 for _every_ value of _x_ and _y_. The function _f_ ( _x, y_ ) is thus a constant function, which is continuous everywhere.

. The first problem we may encounter is a zero in the denominator, in which case _x_ − _y_ 2 = 0. The second problem is taking the log of a nonpositive number, in which case   But this second situation implies _x_ − _y_ 2 ≤ 0, and hence includes the first.

If _x_ − _y_ 2 ≤ 0, then _x_ ≤ _y_ 2. Hence, the domain of _f_ ( _x, y_ ) is all the points ( _x_ , _y_ ) such that _x_ > _y_ 2.

**Chapter 3: Derivatives**

**Problem 18**

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**Problem 19**

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**Problem 20**

  Setting these both equal to zero gives the system of equations

The first equation tells us _y_ 2 = 1, or _y_ = ±1. If _y_ = 1 then the second equation becomes 0 = 2 _x_ − 2, so _x_ = 1. If _y_ = −1 then the second equation becomes 0 = −2 _x_ \+ 2, so _x_ = 1. Hence, the solutions are (1, 1) and (1, −1).

**Problem 21**

First, note that _ϕ_ (1) = (2, 1). So

Now, note that _x_ ( _t_ ) = 2 _t_ , so   Similarly, _y_ ( _t_ ) = _t_ 2, so   and   Finally, we compute

**Problem 22**

First, note that   and   Now we compute:

**Problem 23**

. First, note that _x_ (1, 2) = 2 and _y_ (1, 2) = 5. Hence, if ( _u, v_ ) = (1, 2) then ( _x, y_ ) = (2, 5). Next, note that   and  , and so   and   We now compute:

. _x_ (2, 1) = 2 and y(2, 1) = 3. Hence, if ( _u, v_ ) = (2, 1) then ( _x, y_ ) = (2, 3). Next, note that   and   and so   We now compute:

**Problem 24**

. First, note that   and   Now,

. First, note that   Similarly,   Finally, note that   Hence,

. Note that   so

**Problem 25**

.

|
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.

|

.

|

**Chapter 3 Quiz**

**Problem 26**

.

.

|
---|---

. First note from Question 1 that   and   Now,

. Since _x_ ( _u, v_ ) = _u_ 2 \+ _v_ , it follows that _x_ (1, 1) = 2. Hence, when ( _u, v_ ) = (1, 1), ( _x, y_ ) = (2, 1). Now, note

Now, we have

And so,

from which it follows that

**Chapter 4: Integration**

**Problem 27**

.

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.

|

.

|

**Problem 28**

**Problem 29**

**Problem 30**

To find the answer just set _y_ = 1 and integrate with respect to _x_

**Problem 31**

.

|
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.

|

**Problem 32**

The graph of _y_ = _x_ 2 − _x_ − 2 is depicted below.

The _x_ -intercepts of the graph are _x_ = − 1 and _x_ = 2. Fixing a value of _x_ in between these two numbers, _y_ will range from _x_ 2 − _x −_ 2 to 0. This tells us how to set up the integral

**Problem 33**

The region _R_ is depicted below

• The range of possible _x_ -values is 0 to 1. If we fix a value of _x_ between these numbers, then _y_ can vary from _x_ 3 to _x_. So the integral can be set up as

• The range of possible _y_ -values is also 0 to 1. If we fix a value of _y_ between these numbers, then _x_ can vary from _y_ to   Hence, the integral can be set up as

**Problem 34**

The original limits of integration tell us the shape of the region over which we are integrating the function sin( _y_ 3). The outside limits tell us that the domain of integration lies between the vertical lines _x_ = 0 and _x_ = _π_ 2. Fixing a value of _x_ between these numbers, _y_ can vary from   to _π_ (the inside limits of integration). This tells us the shape of the domain of integration is

Now notice from the figure that the possible range of _y_ values is from 0 to _π_. Fixing a value of _y_ between these numbers, _x_ can range from 0 to _y_ 2. Hence, we may rewrite the integral and evaluate as follows:

**Problem 35**

The solid is pictured above. To find its volume, note that if we fix _x_ and _y_ then _z_ varies from 0 to 1 − _y_ 2. If we fix _y_ then _x_ varies from 0 to 1. Finally, the range of values that _y_ can take on goes from −1 to 1. Hence, we have the triple integral:

**Problem 36**

The volume can be computed with a triple integral as follows:

**Problem 37**

**Problem 38**

**Chapter 4 Quiz**

**Problem 39**

.(a)

|
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(b) The trick with this integral is to reverse the order of integration:

. The domain of integration is a disk of radius 1 in the _xy_ -plane. Hence, if we fix _x_ then _y_ can vary from   The biggest and smallest possible values of _x_ are −1 and 1. Hence, the desired volume can be computed by the integral

**Chapter 5: Cylindrical and Spherical Coordinates**

**Problem 40**

. ( ,  , −1)

. (0, 0, 3)

. (2,  , 0)

**Problem 41**

. The _xz_ -plane.

. The _yz_ -plane.

. The _xy_ -plane.

. A right circular cone centered on the z-axis.

. A sphere of radius 1.

**Problem 42**

.

.

.

**Problem 43**

. (0, 0, 0)

. (− 3, 0, 0)

. ( , , 2)

**Problem 44**

. A single point at the origin.

. The _xy_ -plane.

. The negative z-axis.

. A cylinder of radius 2, centered on the _z_ -axis. (Recall that _ρ_ sin _ϕ_ = _r_ , so this is just the graph of the cylindrical equation _r_ = 2.)

. A horizontal plane at height 2. (Recall that _ρ_ cos _ϕ_ = _z_ , so this is just the graph of the rectangular equation _z_ = 2.)

**Problem 45**

.

.

**Chapter 5 Quiz**

**Problem 46**

. a)

|
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b) |

. (a) In polar coordinates the equation _r_ = cos _θ_ describes a circle of radius  , centered on the point ( , 0). In cylindrical coordinates, we see the same picture no matter what _z_ is. Hence, the graph is the cylinder pictured below.

(b) Consider the sphere of radius 1. On this sphere the points where _θ_ = _ϕ_ forms a curve which starts at the top (where _z_ = 1), ends at the bottom (where _z_ = −1), and passes through the point where _y_ = 1. Now we see the same picture no matter what _ϕ_ is, so we see this curve on a sphere of every radius, as in the figure below.

**Chapter 6: Parameterizations**

**Problem 47**

. _ψ_ ( _x_ , _y_ ) = ( _x_ , _y_ , _x_ 2)

. _ψ_ ( _θ_ , _z_ ) = ( _θ_ 2 cos _θ_ , θ2sin _θ_ , _z_ )

. _ψ_ ( _θ, ϕ_ )= ( _θ_ 2 sin _ϕ_ cos _θ_ , _θ_ 2 sin _ϕ_ sin _θ_ , _θ_ 2 cos _ϕ_ )

**Problem 48**

. _ϕ_ = _θ_

. _y_ = _x_ \+ _xz_

. _z_ = sin _r_

**Problem 49**

.

.

**Problem 50**

. A portion of a cone that lies above a square.

. A portion of a cone that lies above a circle.

. A portion of a cone that lies above half of a circle.

**Problem 51**

. Start with the graph of _x = y_ 2 \+ _z_. Now stretch this graph in the _y_ -direction by a factor of 2. Finally, translate the entire picture −1 unit in the _z_ -direction.

. Start with the graph of _z = x_ \+ sin _y_. Now reflect in the plane _y_ = _x_ , switching the roles of _x_ and _y_.

. Start with the cylinder _r_ = 1. Stretch in the _x_ -direction by 2 and the _y_ -direction by 3, so that cross sections are now ellipses. Now move the central axis 1 unit in the _y_ -direction and −1 unit in the _z_ -direction. Note that this last translation does nothing to the shape.

**Problem 52**

If the curves shown were circles, with the one corresponding to _z_ = 1 having radius 1, then the uniform spacing of the level curves would tell us that the shape is a cone. Such a cone would be the graph of the cylindrical equation _z_ = _r_ , and would thus be parameterized by

ψ( _r_ , _θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _r_ )

0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ 2 _π_

The actual level curves we are given are ellipses. These can be obtained from circular level curves by multiplying the _x-_ and _y_ -coordinates by a factor which will stretch them the appropriate amount:

ψ( _r, θ_ ) = (2 _r_ cos _θ_ , 3 _r_ sin _θ, r_ )

0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ 2 _π_

**Problem 53**

.

.

**Problem 54**

. The region between spheres of radii 1 and 2, and above the _xy_ -plane.

. The region inside a cylinder of height 2 and radius 1, between the planes _z_ = 1 and _z_ = 3.

**Problem 55**

ψ( _r_ , _θ_ , _t_ ) = ( _r_ cos _θ, r_ sin _θ_ , _tr_ )

0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _t_ ≤ 1

**Chapter 6 Quiz**

**Problem 56**

. For each point on a cylinder of radius 2, we know _r_ = 2. If such points are also on the graph of _z_ = 2 _r_ , then _z_ = 2(2) = 4. Hence, the problem is asking for the points of a cylinder of radius 2 for which 0 ≤ _z_ ≤ 4. This is most easily done by utilizing cylindrical coordinates:

ψ( _θ_ , _z_ ) = (2 cos _θ_ , 2 sin _θ_ , _z_ )

0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _z ≤_ 2

. If the 2 were not at the _x_ -coordinate then this parameterization would look just like spherical coordinates. The restrictions on _ρ_ , _θ_ , and _ϕ_ would then specify one-quarter of a ball. The factor of 2 then stretches this shape in the _x_ -direction. The result is the figure below.

**Chapter 7: Vectors and Gradients**

**Problem 57**

. _V_ \+ _W_ = 〈7, 7〉, − _V_ = 〈−1, −6〉, _V_ − _W_ = 〈−5, 5〉

. _V_ \+ _W_ = 〈1, 2〉, − _V_ = 〈0, 0〉, _V_ − _W_ = 〈−1, −2〉

. _V_ \+ _W_ = 〈0, −1〉, − _V_ = 〈1, −1〉, _V_ − _W_ = 〈−2, 3〉

**Problem 58**

.

.

**Problem 59**

The magnitude of 〈2, 1〉 is   To get a unit vector that points in the same direction we just divide by the magnitude:

**Problem 60**

The vector 〈5,12〉 lies on a line of slope   A perpendicular line will have a slope which is the negative reciprocal of this,   A vector that lies in this line is 〈−12, 5〉. To get a unit vector that points in the same direction, we divide by the magnitude:

**Problem 61**

. (2)(−3) + (4)(1) = −2

. (0)(5) + (7)(2) = 14

. (−2)(6) + (−1)(−3) = −9

**Problem 62**

.

.

**Problem 63**

_V_ · _W_ = | _V_ || _W_ | cos _θ_ = 9 cos _θ_. So _V_ · _W_ is largest when the value of cos _θ_ is largest. Since the largest possible value of cos _θ_ is one (when _θ_ = 0), the largest possible value of _V_ · _W_ is 9. Similarly, the smallest possible value for cos _θ_ is −1, so the smallest possible value for _V_ · _W_ is −9. If _V_ and _W_ are perpendicular then cos _θ_ = 0, so the smallest possibility for | _V_ · _W_ | is zero.

**Problem 64**

. No, _V_ · _W_ = 4 ≠ 0.

. Yes, _V_ · _W_ = 0.

. Yes, _V_ · _W_ = 0.

. No, _V_ · _W_ = −9 ≠ 0.

**Problem 65**

. • ∇ _f_ ( _x_ , _y_ ) = (ln _y_ ,  )

• ∇ _f_ (1, 1) ·   = 〈0, 1〉 · ( ) =

• The direction of the maximum rate of change at (1, 1) is ∇ _f_ (1, 1) = 〈0, 1〉. This is already a unit vector.

• |∇ _f_ (1, 1)| = 1

• Such a line is perpendicular to ∇ _f_ (1, 1) = 〈0, 1〉. Such a vector is 〈1, 0〉.

. • ∇ _f_ ( _x_ , _y_ ) = 〈2, 3〉

• ∇ _f_ (1, 1) · ( ) = 〈2, 3〉 · ( ) =

• The direction of maximum rate of change is ∇ _f_ (1, 1) = 〈2, 3〉. A unit vector that points in this direction is found by dividing by  , the magnitude of this vector. The result is

• |∇ _f_ (1, 1)| =

• A line perpendicular to the one containing ∇ _f_ (1, 1) = 〈2, 3〉 would have slope   A vector in this line is 〈−3, 2〉. A unit vector pointing in the same direction is

. • ∇ _f_ ( _x_ , _y_ ) = 〈2 _xy_ \+ _y_ 3, _x_ 2 \+ 3 _xy_ 2〉

• ∇ _f_ (1, 1) · ( ) = 〈3,4〉 · 〈 〉 = 5

• The direction of maximum rate of change is ∇ _f_ (1, 1) = 〈3, 4〉. A unit vector that points in the same direction is 〈 〉.

• |∇ _f_ (1, 1)| = 5

• A line perpendicular to the one containing ∇ _f_ (1, 1) = 〈3, 4〉 would have slope   A vector in this line is 〈−4, 3〉. A unit vector pointing in the same direction is

**Problem 66**

. First, we find the critical points by looking for places where ∇ _f_ ( _x_ , _y_ ) = 〈0, 0〉. Setting the vector ∇ _f_ ( _x_ , _y_ ) = 〈 _y_ \+ 2, _x_ −3〉 equal to 〈0, 0〉, we see that _x_ must be 3 and _y_ must be −2. So the only critical point is at (3, −2). Now note that _D_ ( _x_ , _y_ ) = −1 for all ( _x_ , _y_ ). In particular, _D_ (3, −2) = −1 < 0, so the critical point corresponds to a saddle.

. Setting ∇ _f_ ( _x_ , _y_ ) = 〈3 _x_ 2 − _y_ , − _x_ \+ 2 _y_ 〉 equal to 〈0, 0〉, we see that the coordinates of any critical point must satisfy the system

The second equation tells us _x_ = 2 _y_. Plugging this into the first equation gives 3(2 _y_ )2 − _y_ = 0. Solving for _y_ then gives solutions at _y_ = 0 and   Plugging these numbers back into _x_ = 2 _y_ gives us _x_ values of 0 and   So the coordinates of the critical points are (0, 0) and

We now compute

_D_ ( _x_ , _y_ ) = (6 _x_ )(2) − (−1)2 = 12 _x_ −1

We conclude _D_ (0, 0) = −1 < 0, so (0, 0) is a saddle. On the other hand,   so there is either as maximum or a minimum at   it must be a minimum.

**Problem 67**

All second partials are equal to − sin(x + _y_ ), so

Inspection of the graph, pictured below, shows that there are infinitely many maxima and minima, but no saddles.

**Problem 68**

Recall that   Since   is always bigger than or equal to zero,   can only be positive if   is positive. But if the product of two numbers is positive, and you know the first of those two is also positive, then the second must also be positive.

**Problem 69**

. First we look for critical points in the interior of _D_ by setting the partials equal to zero:

From the first equation _x_ = 0, and from the second _y_ = −1. So (0, −1) is a critical point. (To find out what type we look at the matrix of second partials:

Since   is also greater than 0, the point (0, −1) represents a minimum.) We now look for minima on the boundary of _D_ using the method of LaGrange multipliers. The boundary of _D_ satisfies the equation   Hence, we must look for points ( _x_ 0, _y_ 0) that satisfy this and

∇ _f_ ( _x_ 0, _y_ 0) = λ∇ _g_ ( _x_ 0, _y_ 0)

Note that ∇ _f_ = (2x, 2 _y_ \+ 2) and ∇ _g_ =   Hence, we get two equations:

The first equation tells us either λ = 1 or _x_ = 0. If λ = 1 then the second equation becomes

Solving for _y_ then gives   However, if we try to find corresponding values for _x_ by plugging this into   (the equation for the boundary of D) we get imaginary answers.

If _x_ = 0, then the equation   gives us y = ±2. So on the boundary of D the only potential minima are at (0, 2) and (0,−2).

. To check which critical point represents the minimum of f we just plug them in:

_f_ (0,−1) = 1 −2 −1 = −2

_f_ (0, 2) = 4 + 4 −1 = 7

_f_ (0,−2) = 4 −4 −1 = −1

So the smallest value attained by _f_ ( _x_ , _y_ ) on _D_ is−2, and this happens at (0,−1).

**Problem 70**

.

.

.

**Problem 71**

Since _V_ and _W_ are parallel there is a number _k_ such that _V_ = _kW_. So if V = 〈 _a_ , _b_ 〉 then _W_ = 〈 _ka_ , _kb_ 〉. We now compute

**Problem 72**

.

.

**Problem 73**

.

.

.

**Problem 74**

. First we compute the cross product:

The magnitude of this vector is   So a unit vector that points in the right direction is

. The cross product is

The magnitude of this is   So a unit vector that points in the right direction is

**Problem 75**

Let _V_ = 〈 _a, b, c_ 〉 and _W_ = 〈 _c, d, e_ 〉. We now compute _V_ × _W_ and _W_ × _V_ :

**Problem 76**

Let _U_ = 〈 _a, b, c_ 〉, _V_ = 〈 _d, e, f_ 〉 and _W_ = 〈 _g, h, i_ 〉. Now compute:

**Problem 77**

Inspection of the following figure reveals that | _W_ | sin _θ_ is the height of the parallelogram, and | _V_ | is the length of the base. The area is just the product of the base and the height.

**Chapter 7 Quiz**

**Problem 78**

.(a) The gradient of _f_ ( _x_ , _y_ ) at ( _x_ , _y_ ) is

∇ _f_ ( _x_ , _y_ ) = 〈2 _x_ \+ 3 _y_ , 3 _x_ 〉

The largest slope at (1, 1) is the magnitude of the gradient

(b) To find the critical points we set each partial derivative equal to 0:

2 _x_ \+ 3 _y_ = 0, 3 _x_ = 0

From the second equation _x_ = 0. The first then implies _y_ = 0. So the only critical point is at (0, 0).

(c) We compute

(d) Since the determinant of the matrix of second partials is negative, the critical point (0, 0) is a saddle.

. (a) The magnitude of _V_ is

Hence, the desired unit vector is

(b) Since _V_ · _W_ = | _V_ || _W_ | cos _θ_ , it follows that

(c) First we find the cross product:

The magnitude of this vector is

Hence, a unit vector that points in the desired direction is

**Chapter 8: Calculus with Parameterizations**

**Problem 79**

The speed is the magnitude of the derivative of the parameterization:

**Problem 80**

A tangent vector is given by the derivative of the parameterization: _c_ ′( _t_ ) = 〈3 _t_ 2, 4 _t_ 3, 5 _t_ 4〉. The point (1, 1, 1) = _c_ (1). A tangent vector at this point is then _c_ ′(1) = 〈3, 4, 5〉. The magnitude of this vector is   A unit tangent vector is thus

**Problem 81**

. The derivatives of the parameterization are

If _r_ = 1 and   these vectors are

.

**Problem 82**

. First we compute

Hence, if we multiply ψ by   we get the required parameterization:

. First, note that

It now follows that

We now compute the dot product

. We compute the magnitude of

Hence, if we let   then _N_ is a unit vector and

So, the required _K_ is

. We compute

We now compute the magnitude of _B_ :

.

. Since   it must be that

. First, notice

We now do some calculation:

**Problem 83**

The length of ψ( _t_ ) is computed as follows:

**Problem 84**

The length of ψ( _t_ ) is calculated as follows:

This is precisely the integral we obtained for the length of ( _t_ cos _t, t_ sin _t_ ) in Example 8-5.

**Problem 85**

**Problem 86**

**Problem 87**

.

|
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. Note that the surface over the domain _D_ is a surface of revolution. In cylindrical coordinates it has the equation _z_ = _r_. So the area is given by

**Problem 88**

A parameterization for the surface is given by

ψ( _θ_ , _ϕ_ ) = ( _f_ ( _ϕ_ ) sin _ϕ_ cos _θ_ , _f_ ( _ϕ_ ) sin _ϕ_ sin _θ_ , _f_ ( _ϕ_ ) cos _θ)_

The partials of this are

The cross product is

With some work, the magnitude of this vector is

And so,

**Problem 89**

**Problem 90**

The portion of the plane specified is parameterized by

ψ( _x_ , _y_ ) = ( _x_ , _y_ , _x_ \+ _y_ )

0 ≤ _x_ ≤ 1, 0 ≤ _y_ ≤ 1

We now compute

**Problem 91**

A parameterization is given by

ψ( _θ_ , _ϕ_ ) = (sin _ϕ_ cos _θ_ , sin _ϕ_ sin _θ_ , cos _ϕ_ )

0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _ϕ_ ≤

Now calculate:

**Problem 92**

A parameterization is given by

ψ( _r_ , _θ_ , _z_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _z_ )

0 ≤ _r_ ≤ _R_ , 0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _z_ ≤ _h_

The derivatives of this are

We now compute volume by integrating:

**Problem 93**

First, we calculate the derivatives of the parameterization:

Now we integrate:

**Problem 94**

Letting _R_ denote the elliptical region given by the parameterization ψ, and _f_ ( _x_ , _y_ ) = 2 _x_ − _y_ , we have

**Problem 95**

**Problem 96**

The region _V_ is easily parameterized using cylindrical coordinates:

_Psi_ ( _r_ , _θ_ , _z_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _z_ )

0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _z_ ≤ 2

To use this parameterization to evaluate the integral, we will need the determinant of the matrix of partial derivatives of Psi:

We now integrate

**Problem 97**

The region _R_ is easily parameterized with polar coordinates:

ψ( _r, θ_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ )

0 ≤ _r_ ≤ 2, 0 ≤ _θ_ ≤

The determinant of the matrix of partials of ψ is _r_. Hence,

**Chapter 8 Quiz**

**Problem 98**

. (a) ψ( _y_ ) = (sin _y_ , _y_ )

(b) The point (0, 0) is ψ(0). So a tangent vector is given by the derivative of the parameterization,  , at y = 0. This is the vector (1, 1). A unit tangent vector is thus

. (a) |

---|---

(b) We will need the partials of ψ :

Next we must compute the determinant of the matrix of partials:

Finally, we integrate

. First, we compute the partials of _ϕ_ :

Next we will need the determinant of the matrix of partials:

Now we integrate

**Chapter 9: Vector Fields and Derivatives**

**Problem 99**

.

.

.

**Problem 100**

. 〈 1, 1〉

. 〈 1, _z, y_ 〉

. 〈 _y_ \+ _z_ , _x_ \+ _z_ , _x_ \+ _y_ 〉

**Problem 101**

.

.

.

**Problem 102**

**Problem 103**

We let **F** = 〈 _f x_, _f y_, _f z_〉 and **G** = 〈 _g x_, _g y_, _g z_〉. Then

**F** × **G** = 〈 _f ygz_ − _f zgy_, _f zgx_ − _f xgz_, _f xgy_ − _f ygx_〉

and so

**Problem 104**

.

|
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. 〈 2 _x_ , 0, −2 _z_ 〉

**Problem 105**

.

|
---|---

.

|

**Chapter 9 Quiz**

**Problem 106**

.

.

. 〈2 _x_ sin( _y_ − _z_ ), _x_ 2 cos( _y_ − _z_ ), − _x_ 2 cos( _y_ − _z_ )〉

. Suppose **W** = 〈 _f_ , _g_ , _h_ 〉. Then the last component of   This can equal   if either   In the former case _g_ ( _x_ , _y_ , _z_ ) = _y_ ln _x_. In the latter case   So two vector fields that have the required curl are 〈0, _y_ ln _x_ , 0〉 and

**Chapter 10: Integrating Vector Fields**

**Problem 107**

The circle is parameterized by

ψ( _t_ ) = (cos _t_ , sin _t_ ), 0 ≤ _t_ ≤ 2 _π_

Note that this parameterization agrees with the specified orientation. Now we may integrate

**Problem 108**

**Problem 109**

. First, note that ∇ _f_ = 〈 _y_ 2, 2 _xy_ 〉. Now we integrate

. _f_ (ψ(2)) − _f_ ((−1))| = _f_ (2, 4) − _f_ (−1, 1)
---|---
| = 32 − (−1)
| = 33

**Problem 110**

The curve _C_ is parameterized by

ψ( _t_ ) = ( _t, t_ 2), 0 ≤ _t_ ≤ 1

The derivative of this parameterization is 〈1, 2 _t_ 〉.

**Problem 111**

The unit sphere is parameterized with spherical coordinates by

ψ( _θ_ , _ϕ_ ) = (sin _ϕ_ cos _θ_ , sin ϕ sin _θ_ , cos _ϕ_ )

0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _ϕ_ ≤ _π_

The derivatives of this are

And so,

Notice that at the point   the vector   is (1, 0, 0), which agrees with the specified orientation. We may thus integrate to find the correct answer:

**Problem 112**

First, we calculate the partials of the parameterization:

Now, the cross product is given by

Finally, we integrate

**Problem 113**

First, we calculate the partials of the parameterization:

Now, the cross product of these vectors is

We are now ready to integrate

**Problem 114**

First, we compute ∇ × **W:**

A parameterization for _S_ is given (via spherical coordinates) by

The partials of this are

And so,

Now notice that at the point   the vector   is equal to 〈0, 1, 0〉, which disagrees with the specified orientation. We must therefore remember to switch the sign of our final answer. Finally, we can integrate

The correct answer is therefore

**Chapter 10 Quiz**

**Problem 115**

. A parameterization for _C_ is given by

ψ( _t_ ) = ( _t_ 2, _t_ ), 0 ≤ _t_ ≤ 1

The derivative of this is

Note that at the point ψ(0) = (0, 0) this vector is 〈0, 1〉, which disagrees with the orientation indicated in the figure. Thus, we will have to negate our final answer. We now integrate

. The partials of ψ are given by

The cross product of these vectors is

We are now prepared to integrate

**Chapter 11: Integration Theorems**

**Problem 116**

Notice that **W** = 〈1, 1,1〉 = ∇ ( _x_ \+ _y_ \+ _z_ ). Hence, if we let _f_ ( _x_ , _y_ , _z_ ) = _x_ \+ _y_ \+ _z_ then

**Problem 117**

The trick is to notice that **W** = ∇ f , where _f_ ( _x_ , _y_ ) = _x_ sin _y_. Then

**Problem 118**

First, notice that _f_ ( _x_ , _y_ ) = _x_ 2 −4 _x_ \+ 4 + _y_ 2 \+ 2 _y_ \+ 1 = ( _x_ −2)2 \+ ( _y_ \+ 1)2. Thus the graph of _f_ ( _x_ , _y_ ) can be obtained from the graph of _z_ = _x_ 2 \+ _y_ 2 (a paraboloid) by shifting 2 units in the positive _x_ -direction and 1 unit in the negative _y_ -direction. This puts the "bottom" of the paraboloid at the point (2, −1). Hence, if ( _a, b_ ) is any point other than (2, −1) then _f_ ( _a, b_ ) > 0. The result now follows easily:

**Problem 119**

Suppose the beginning and ending point of _C_ is _p_. Then

**Problem 120**

Suppose _f_ ( _x_ ) is a function. Let ψ be a parameterization of [ _a, b_ ] given by

ψ( _x_ ) = _x_ , _a_ ≤ _x_ ≤ _b_

By the theorem,

We now follow the definitions:

Putting this all together gives

**Problem 121**

When you are coming down the mountain the force of gravity is working _with_ you, not _against_ you. So the work done against gravity during this portion of the trip is negative. This cancels out the extra positive work you did to get to the top of the mountain.

**Problem 122**

**Problem 123**

Let _D_ be the horizontal line segment that connects (0, 0) to ( _a_ , 0). Then _C_ ∪ _D_ bounds a rectangular region _Q_. Green's Theorem says

We conclude

and so the answer only depends on _a_.

**Problem 124**

By Green's Theorem,

To evaluate this integral we will need the determinant of the matrix of partial derivatives of _ϕ:_

We now integrate

**Problem 125**

The area of any region _Q_ in  2 can be calculated by the integral

Green's Theorem says that this will be equal to the integral of **W** = 〈 _f_ ( _x_ ), _g_ ( _x_ )〉 around _∂Q_ if

One suitable choice for **W** would thus be 〈0, _x_ 〉. We now integrate this vector field over the unit circle, by using the usual parameterization

**Problem 126**

. Let _C_ 1 and _C_ 2 be two circles centered on the origin, oriented counterclockwise. Let _Q_ be the region between these circles. Then _∂Q_ has two pieces. One is _C_ 1, and the other is _C_ 2 with the opposite orientation. As a shorthand only we write this as _C_ 1 − _C_ 2. Green's Theorem now says

And hence,

. Using the previous part, we just have to show

. Note that on the unit circle _x_ 2 \+ _y_ 2 = 1, so, on this circle the integral of W is the same as the integral of 〈− _y, x_ 〉.

**Problem 127**

. By Green's Theorem,

. The curve _C_ is parameterized by

ψ( _t_ ) = ( _t_ , 0), −1 ≤ _t_ ≤ 1

The derivative of this parameterization is 〈1, 0〉, so

. Let _D_ be the top half of the unit circle (oriented counterclockwise). Then _C_ ∪ _D_ = _∂σ_. So we have

Hence,

**Problem 128**

The surface _S_ is a disk in the _yz_ -plane. Its boundary is thus the unit circle in the _yz_ -plane. The tricky part is determining the proper orientation on the boundary circle. A normal vector giving the orientation of _S_ is determined by

For _r_ ≠ 0 this is a vector which is parallel to the positive _x_ -axis. The orientation on _∂S_ is then given by the right-hand rule, as in the following figure.

The usual parameterization for the unit circle in the _yz_ -plane would be

ϕ( _t_ ) = (0, cos _t_ , sin _t),_ 0 ≤ _t_ ≤ 2 _π_

Notice that   〈0, 0, 1〉. This agrees with the orientation given in the figure. We are now prepared to integrate, using Stokes' Theorem:

As the chosen parameterization _ϕ_ agrees with the orientation on _∂S_ there is no need to change sign.

**Problem 129**

Let _Q_ be the region in  2 such that ( _x_ , _y_ ) is in _Q_ if ( _x_ , _y_ , 0) is in S. Then we may parameterize _S_ as follows:

ψ( _x_ , _y_ ) = ( _x_ , _y_ , 0), ( _x_ , _y_ ) ∈ _Q_

It follows that   and   Thus

By definition

Also by definition

But Stokes' Theorem says

So we get

which is Green's Theorem.

**Problem 130**

The boundary of _S_ is parallel unit circles _C_ 0 and _C_ 1 in the planes _z_ = 0 and _z_ = 1, respectively. However, the orientations on these circles are opposite, as can be seen in the following figure.

A parameterization for _C_ 0 is given by

ψ0( _t_ ) = (cos _t_ , sin _t_ , 0), 0 ≤ _t_ ≤ 2 _π_

At the point ψ0(0) = (1, 0, 0) the derivative   is the vector 〈0, 1, 0〉. This agrees with the arrow in the picture.

Similarly, a parameterization for _C_ 1 is given by

ψ1( _t_ ) = (cos _t_ , sin _t_ , 1), 0 ≤ _t_ ≤ 2 _π_

At the point ψ0(0) = (1, 0, 1) the derivative   is again the vector 〈0, 1 , 0〉. This is opposite to the arrow in the picture, so we will have to remember to negate the integral over this curve.

Stokes' Theorem says

We do each of these integrals separately

But, for orientation reasons, we have to negate our answer, yielding −2 _π_. Our final answer is thus

**Problem 131**

We assume the sphere is oriented with an outward normal (just reverse the picture if not). Dividing the sphere _S_ into a northern hemisphere _S_ + and a southern hemisphere _S_ −, we see in the following figure that the equator gets opposite orientation as the boundary of each.

As the induced orientations on the equator are opposite, it follows that

We now employ Stokes' Theorem

**Problem 132**

The direction of ∇ × **W** will be perpendicular to the plane containing the loop with greatest circulation. Since this is the xz-plane, the direction of ∇ × **W** will be parallel to the _y_ -axis. The magnitude of ∇ × **W** can be estimated by

where _D_ is the disk whose boundary is the loop _C_. Since the radius of _C_ is .1, the area of _D_ is _π_ (.1)2. This gives us

We conclude that at the origin

∇ × **W** ≈ 〈0, 15.92, 0〉

**Problem 133**

The circulation of the vector field around any loop in a plane parallel to the xz-plane would be zero, since the vector field is constant on such a plane. A vector _V_ that is perpendicular to such a loop points in the _y_ -direction. The circulation around any horizontal loop would also have to be zero, since the vector field is perpendicular to such a loop. A vector _W_ that is perpendicular to such a loop points in the _z_ -direction. The curl must then be perpendicular to both _V_ and _W_ , and so must point in the _x_ -direction.

One can also see this algebraically. Such a vector field must look like 〈0, 0, _f_ ( _y_ )〉. The curl of this is

**Problem 134**

Let _V_ be the unit cube. Then Gauss' Theorem tells us

**Problem 135**

By Gauss' Theorem the integral of ∇ · **W** over the ball _B_ is equal to the integral of **W** over the unit sphere _S_ , with outward-pointing normal vector. The unit sphere is parameterized in the usual way by

ψ( _θ_ , _ϕ_ ) = (sin _ϕ_ cos _θ_ , sin _ϕ_ sin _θ_ , cos _ϕ_ )

0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _ϕ_ ≤ _π_

As in Example 11-8,

which agrees with the orientation on _S_.

We now compute

**Problem 136**

The region _V_ is parameterized by

ψ( _r_ , _θ_ , _z_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _z_ )

We use Gauss' Theorem to transform the integral:

We now use the parameterization to change variables:

**Problem 137**

First, we observe that the _polar_ equation _r_ = cos _θ_ is a circle of radius  , centered on the point   Hence, the surface _C_ is a cylinder of height 2. The surface _D_ is a disk, which caps off the top of the cylinder _C_ , like an upside-down can. Let _E_ denote the "bottom" of the can. That is, _E_ is the set of points in the _xy_ -plane which are within   of a unit away from the point ( , 0, 0). We assume an orientation is given on _E_ so that _C, D_ , and _E_ together form the (oriented) boundary of _V_ , the points inside the "can." Applying Gauss' Theorem gives us

Now notice that on _E_ the vector field **W =** 〈0, 0, 0〉, so

Putting these results together gives us

We may thus obtain the desired answer by evaluating the integral on the left-hand side of this last equation. The first thing we will need to evaluate this is the divergence of **W:**

Next, we will need to parameterize _V_. Notice that _V_ is a solid cylinder of radius  , whose central axis has been translated   of a unit away from the z-axis, in the positive _x_ -direction. A parameterization is thus given by

The partials of this parameterization are

The determinant of the matrix which consists of these vectors is

We now use ψ to evaluate the integral:

**Problem 138**

_S_ 1, and _S_ 2 _with the opposite orientation_ , together bound a volume _V_ of  3.

By Gauss' Theorem

But ∇ · **W** = 0 implies   We conclude

**Problem 139**

**Chapter 11 Quiz**

**Problem 140**

. The key to this problem is to notice that **W** = ∇ ( _x_ , _y_ , _z_ ), where _f_ ( _x_ , _y_ , _z_ ) = _xy_ 2 _z_ 2. So,

. Using Green's Theorem

To evaluate this integral we will need the partials of the parameterization:

The determinant of the matrix of partials is thus

We now integrate:

. The curves _C_ 1 and _C_ 2 lie in the _xy_ -plane, and are pictured below.

Let   denote the curve _C_ 1 with the opposite orientation. Then the curves   and _C_ 2 together form the (oriented) boundary of a region _R_ of the _xy_ -plane. Stokes' Theorem then says

But the statement of the problem specifies ∇ × **W** = 〈0, 0, 0〉, so the integral on the right (and hence the integral on the left) is 0. Now notice

Since this is zero, the desired result follows.

**Final Exam**

**Problem 141**

1.(a)|
---|---

(b) To find the critical points we set the partial derivatives equal to zero:

The first equation tells us _y_ = −1 and the second tells us _x_ = 2. So (2, −1) is the only critical point. Now we compute

Since this is negative the critical point (2, −1) corresponds to a saddle.

(c) First, we find a unit vector that points in the desired direction:

The desired slope is the directional derivative in this direction:

(d) |
---|---

. Since _r_ 2 = _x_ 2 \+ _y_ 2, the desired surface has the cylindrical equation _z_ = 4 − _r_ 2. Where the graph hits the _xy_ -plane we know _z_ = 0, and hence 0 = 4 − _r_ 2, or _r_ = 2. A parameterization is thus given utilizing cylindrical coordinates by

. There are several ways to do this. We will utilize spherical coordinates, since that is how the problem was stated. Being on a sphere of radius 1 says _ρ_ = 1. We now plug this, and the information _θ_ = _ϕ_ , into the usual spherical coordinates:

ψ( _θ_ ) = (sin _θ_ cos _θ_ , sin _θ_ sin _θ_ , cos _θ)_

To get the whole circle the domain should be 0 ≤ _θ_ ≤ 2 _π_.

.

|
---|---

. Since the region _V_ is cylindrical, this is best done by utilizing a parameterization. We parameterize _V_ in the usual way with cylindrical coordinates:

ψ( _r_ , _θ_ , _z_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _z_ )

To do the integral we will need the determinant of the matrix of partial derivatives, which simplifies to _r_. We can thus integrate as follows:

. The derivative of the parameterization is

We now integrate:

. First, we compute the partials of the parameterization:

The cross product of these vectors is

We now integrate:

. The key to this problem is to notice that **W** = ∇ _f_ , where _f_ ( _x_ , _y_ , _z_ ) = _xyz_. Then, using the independence of path of line integrals

. The surface _S_ bounds a volume _V_ which is parameterized by

ψ( _r_ , _θ_ , _z_ ) = ( _r_ cos _θ_ , _r_ sin _θ_ , _z_ )

0 ≤ _r_ ≤ 1, 0 ≤ _θ_ ≤ 2 _π_ , 0 ≤ _z_ ≤ 1

To evaluate an integral over this region we will need the determinant of the matrix of partials. We have done this calculation several times. The reader may check that the answer is _r_. We now integrate

INDEX |

---|---

A

algebra, using with partial derivatives

arc length, relationship to line integral

area. _See also_ surface area

computing

computing for parallelogram

computing for region _of xy-plane_

determining relative to change of variables

finding for parallelograms

parameterizing

axes

labeling as _r_ and z

labeling when plotting points with multiple coordinates

B

boundary, determining for surface _S_

C

chain rule

computing derivatives with

using

circle of radius one, parameterizing

components, relationship to vectors

composition

of functions of multiple variables

relationship to parameterized curves

visualizing

constants, relationship to derivatives

continuous function, definition of

coordinate planes, plotting intersection of graphs with

coordinate system

definition of

identifying type of

importance of

coordinates, plotting points with

cross product

determining

versus dot product

magnitude of

rewriting definition of

of vectors for surface of revolution

curl, geometric interpretation of

curl of vector fields

computing

formula for

curl operation, defining relative to vector fields

curves

finding lengths of

parameterizing

cylindrical coordinates

converting to rectangular coordinates from

expressing restriction on domains in

parameterizing domain of integration with

parameterizing surfaces with

translating to rectangular coordinates

cylindrical equations, graphing

D

derivatives. _See also_ partial derivatives; second derivatives

computing with chain rule

determinant

computing

for volume calculation

of a 3 × 3 matrix

relationship to parallelograms

writing for matrix

direction, finding for level curve

directional derivative, explanation of

divergence

calculating for vector field **V**

defining for **V,** "Div **V** "

definition of

geometric interpretation of

positive divergence

taking for vector fields

zero divergence

domain of integration

definition of

parameterizing with cylindrical coordinates

restricting

dot product

versus cross product

definition of

finding cosine of angle between vectors with

purpose of

relationship to geometric interpretation of curl

source of

taking

usefulness of

writing directional derivative as

E

earth's surface, determining position on

equations

for cross product

for curl of vector field

for directional derivative

for divergence of vector field

for parabolas

of spheres in spherical coordinates

for surface area of graph

examples

continuity of functions

cosine of angle between vectors

cross product

curl of vector field **V,**

direction for level curve

divergence of vector field **V,**

dot product

finding area of _R_

finding gradient for saddle

functions

Gauss' Divergence Theorem

geometric interpretation of curl

geometric interpretation of divergence

gradient of function

gradient of _f_ ( _x_ , _y_ ) at point ( _x_ , _y_ )

gradient vector

graphing cylindrical equations

graphing functions

Green's Theorem

integral evaluation

locating points in rectangular coordinates

optimization

oriented curve relative to line integral

overcoming gravity

parallelogram area computation

parameterization

parameterization for graph of spherical equation

parameterizations

parameterized surfaces

particle speed moving along helix

rate of change for functions

rectangular coordinates

signed volume of parallelepiped

Stokes' Theorem

surface area of sphere

surface integrals

translation from polar to rectangular coordinates

unit vector

vector addition

vector fields

vector subtraction

volume bounded by sphere

volume computations for integration

F

Figures

contours of _f_ ( _x, y_ ) and domain _D_

level curves

level curves forming saddle

local and absolute maxima

paraboloid _z_ = _x_ 2 \+ _y_ 2

perpendicular axes drawn in perspective

plotting points

point and vector

rectangle, Q

topographic map related to level curves

weather map showing level curves

formulas. _See_ equations; examples; problems

Fubini's Theorem, relationship to integration

function of one variable, integral of

functions

change of variables related to

continuity of

defining integrals of

differentiating for variables

evaluating at indicated points

evaluating limits of

examining matrices for

examples of

graphing

integrating

integrating over parameterized surfaces

integrating via Green's Theorem

limits of

mental model of

picturing as graphs

plotting for variables

producing from vector fields

of three variables

functions of multiple variables. _See also_ variables

composition of

limit for

Fundamental Theorem of Calculus

deducing

significance of

G

Gauss' Divergence Theorem

applying

example of

Generalized Stokes' Theorem

geometric interpretation

of curl

of divergence

grad _f_ vector field, producing

gradient, finding for saddles

gradient of function, example of

graphing

functions

level curves

graphs

formula for surface area of

picturing functions as

sketching

gravity, overcoming

Green's Theorem

versus Gauss' Divergence Theorem

geometric interpretation produced by

on Rectangular Domains

relationship to Stokes' Theorem

versus Stokes' Theorem

I

**i** vector, defining

integrals

cancellation relative to Green's Theorem

computing relative to Gauss' Theorem

computing surface area of sphere with

computing surface of revolution with

computing volume with triple integrals

of curl of **W** over _D_

defining for functions

determining limits of integration for

evaluating

over nonrectangular domains

setting up

for Stokes' Theorem

integrand, determining for volume

integration

act of

over nonrectangular domains

performing

relationship to rectangles

integration order, determining for triple integral

J

**j** vector, defining

K

**k** vector, defining

L

Lagrange multipliers method, using

Law of Cosines, explanation of

level curves

finding direction of

graphing

level sets, plotting

limit properties, relationship to integration and rectangles

limits

for curve

calculation for continuous functions

definition of

evaluating for functions

of functions

proving non-existence of

significance of

surface area as

volume as

limits of integration

determining for outer integral

finding

significance of

line integrals

defining

determining

interpreting

physical motivation for

recalculating

relationship to arc length

relationship to vector fields

representing

significance of

local maximum and minimum, determining

M

matrix

tracking information in

writing determinant of

maxima

detecting via derivative test

distinguishing from saddles

determining local maximum

finding

minima

detecting via derivative test

distinguishing from saddles

seeking for _f_

determining local minimum

multiple integral, definition of

O

orientation

calculating for rectangles relative to Green's Theorem

checking relative to Gauss' Theorem

determining for surface _S_

example of

producing

relevance to surface integrals

significance of

P

parabola, equation for

paraboloid

definition of

example of

parallelepiped, explanation of

parallelograms

computing area of

computing signed area of

parameterizations

of areas and volumes

computing partials of

computing volumes with

determining shapes from

differentiating

finding for graphs of equations

restricting domains of

using with volume

parameterized curves

composition with

examples of

parameterized surfaces, definition of

partial derivatives. _See also_ derivatives

applying to vector fields

computing for parameterization

computing relative to chain rule

example of

for parameterizations

representing with respect to _x_

for surface area of sphere

for surface of revolution

particle speed, determining along helix

path independence of line integrals of gradient fields

perpendicular vectors, detecting

points

locating in rectangular coordinates

locating with spherical coordinates

plotting for functions

plotting with three coordinates

plugging into functions

versus vectors

polar coordinates

adapting to three dimensions

translating to rectangular coordinates

problems

area of parallelogram spanned by vectors

area of parameterized plane

calculating curve length

circles for level curves

computing integral of _f_ ( _x, y_ ) over parameterized curve

continuous function

converting to rectangular coordinates

coordinate system

cosine of angle between vectors

curl of vector fields

curve parameterization

derivatives

determinant

divergence of vector fields

domain of functions

dot product for perpendicular vectors

dot product of vectors

function evaluation at point

Gauss' Divergence Theorem

geometric interpretation of curl

graphs of equations

graphs of functions

graphs of spherical equations

Green's Theorem

Green's Theorem over more general domains

integrals of functions

integration

integration and integrals

intersections of graphs of functions with coordinate planes

level curves

line integrals

local maxima, minima, and saddles

parameterization for volume below cone

parameterizations

parameterizations for graphs of equations

parameterized sphere

parameterizing surface with level curves

partial derivatives

plotting points on axes

right handed coordinate systems

second partial derivatives

shape of cylindrical equations

shapes with parameterizations

showing parameterized curve

signed area of parallelepiped

signed area of parallelogram

for Stokes' Theorem

surface integral

triple integrals

unit vector perpendicular to vectors

vector fields

vector fields with curls

vectors

volume calculation

volumes of parameterizations

writing rectangular coordinates of points

problems, optimizing

Pythagorean Theorem, applying to magnitude of vectors

R

_r_ and _z_ , labeling axes as

rectangles

applying Green's Theorem to

relationship to integration

rectangular coordinates

converting from cylindrical coordinates to

converting spherical coordinates to

explanation of

locating points in

translating cylindrical coordinates to

translating polar coordinates to

region

computing area of

cutting into slabs

finding volume of

parameterizing relative to change of variables

revolution, determining surface of

right hand rule, relationship to labeling axes

right handed coordinate system, using

river analogy

applying to line integrals

applying to surface integrals

S

saddles

definition of

determining

distinguishing from minima and maxima

example of

finding gradients for

scalar multiplication, relationship to vectors

second derivatives, shorthand notations for. _See also_ derivatives

shapes

defining with rectangular coordinates

describing with spherical coordinates

determining from parameterizations

with parameterizations

sphere of radius 1, equation of

spheres

bounding volumes by

computing surface area of

determining surface area of

parameterizing

spherical coordinates

converting to rectangular coordinates

describing shapes with

equation of sphere in

locating points with

parameterizing unit sphere by means of

spherical equations, graphing

spiral, parameterizing

Stokes' Theorem

applying

geometric interpretation produced by

versus Green's Theorem

producing geometric interpretation of curl with

stream analogy

applying to line integrals

applying to surface integrals

summation properties, relationship to integration and rectangles

surface area. _See also_ area

computing

computing for sphere

computing with double integral

determining for sphere

as limit

relationship to surface integral

surface integrals

calculating

defining for vector fields

definition of

relationship to surface area

surface of revolution, determining

surfaces

determining boundary of

differentiating parameterizations of

parameterizing

parameterizing with cylindrical coordinates

T

tangent plane

theorems

Fubuni's Theorem

Gauss' Divergence Theorem

Green's Theorem

Green's Theorem on Rectangular Domains

Green's Theorem over more general domains

path independence of line integrals of gradient fields

Stokes' Theorem

triangle, applying Law of Cosines to

trigonometry, applying to dot product

triple integrals, computing volume with

U

unit sphere, parameterizing relative to Gauss' Theorem

unit vector

definition of

perpendicular example of

pointing in same direction

V

vector fields

applying partial derivatives to

defining surface integrals for

definition of

differentiating

divergence of

integrating divergence of

producing functions from

relationship to line integrals

vectors

adding

computing cross product for

computing magnitude of

defining **i, j,** and **k**

detecting perpendicular vectors

defining product of

finding cosine of angles between

magnitude for surface of revolution

magnitude of

multiplying by numbers

versus points

producing

producing from cross product

product of

subtracting

volumes

bounding by spheres

computing for integration

computing with parameterizations

computing with triple integrals

determining

above nonrectangular areas

under graph and above rectangle

parameterizing

positive state of

of thin slab

volumes of regions, finding

X

_xy_ -plane

example of

finding area of region _R_ of

relationship to integration

_xz_ -plane, finding intersection with

label: mathematics

title: Advanced Calculus Demystified

authors: David Bachman