Calculus III - Cognella Academic PublishingDeﬁnition 1 Assume that σis a function from an...

Calculus III

By Tunc Geveci

Included in this preview:

• Copyright Page• Table of Contents• Excerpt of Title

Section 12.1: Tangent vectors and Velocity

Section 12.7: Directional Derivatives and the Gradient

Section 13.2: Double Integrals over Non-Rectangular Regions

Section 14.2: Line Integrals

Section 14.5: Parametrized Surfaces and Tangent Planes

Section 14.6: Green’s Theorem

For additional information on adopting this book for your class, please contact us at 800.200.3908 x501 or via e-mail at

Tunc Geveci

Calculus III

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Copyright © 2011 by Tunc Geveci. All rights reserved. No part of this publication may be reprinted, reproduced, transmitted, or utilized in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information retrieval system without the written permission of University Readers, Inc.

First published in the United States of America in 2011 by Cognella, a division of University Readers, Inc.

Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe.

15 14 13 12 11 1 2 3 4 5

Printed in the United States of America

ISBN: 978-1-935551-45-4

Contents

11 Vectors 111.1 Cartesian Coordinates in 3D and Surfaces . . . . . . . . . . . . . . . . . . . . . . 111.2 Vectors in Two and Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 611.3 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1811.4 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

12 Functions of Several Variables 3512.1 Tangent Vectors and Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3512.2 Acceleration and Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4612.3 Real-Valued Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . 6112.4 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6512.5 Linear Approximations and the Differential . . . . . . . . . . . . . . . . . . . . . 7412.6 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8512.7 Directional Derivatives and the Gradient . . . . . . . . . . . . . . . . . . . . . . . 9412.8 Local Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10712.9 Absolute Extrema and Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . 121

13 Multiple Integrals 13113.1 Double Integrals over Rectangles . . . . . . . . . . . . . . . . . . . . . . . . . . . 13113.2 Double Integrals over Non-Rectangular Regions . . . . . . . . . . . . . . . . . . . 13813.3 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 14413.4 Applications of Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 15213.5 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15713.6 Triple Integrals in Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . 16713.7 Change of Variables in Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . 179

14 Vector Analysis 18714.1 Vector Fields, Divergence and Curl . . . . . . . . . . . . . . . . . . . . . . . . . . 18714.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19414.3 Line Integrals of Conservative Vector Fields . . . . . . . . . . . . . . . . . . . . . 21014.4 Parametrized Surfaces and Tangent Planes . . . . . . . . . . . . . . . . . . . . . 22014.5 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23914.6 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25914.7 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27414.8 Gauss’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

K Answers to Some Problems 285

L Basic Differentiation and Integration formulas 309

iii

Preface

This is the third volume of my calculus series, Calculus I, Calculus II and Calculus III. Thisseries is designed for the usual three semester calculus sequence that the majority of science andengineering majors in the United States are required to take. Some majors may be required totake only the first two parts of the sequence.

Calculus I covers the usual topics of the first semester: Limits, continuity, the deriv-ative, the integral and special functions such exponential functions, logarithms,and inverse trigonometric functions. Calculus II covers the material of the secondsemester: Further techniques and applications of the integral, improper integrals,linear and separable first-order differential equations, infinite series, parametrizedcurves and polar coordinates. Calculus III covers topics in multivariable calculus:Vectors, vector-valued functions, directional derivatives, local linear approxima-tions, multiple integrals, line integrals, surface integrals, and the theorems of Green,Gauss and Stokes.

An important feature of my book is its focus on the fundamental concepts, essentialfunctions and formulas of calculus. Students should not lose sight of the basic conceptsand tools of calculus by being bombarded with functions and differentiation or antidifferentia-tion formulas that are not significant. I have written the examples and designed the exercisesaccordingly. I believe that "less is more". That approach enables one to demonstrate to thestudents the beauty and utility of calculus, without cluttering it with ugly expressions. Anotherimportant feature of my book is the use of visualization as an integral part of the expo-sition. I believe that the most significant contribution of technology to the teaching of a basiccourse such as calculus has been the effortless production of graphics of good quality. Numericalexperiments are also helpful in explaining the basic ideas of calculus, and I have included suchdata.

Remarks on some icons: I have indicated the end of a proof by ¥, the end of an example by¤ and the end of a remark by ♦.

Supplements: An instructors’ solution manual that contains the solutions of all the prob-lems is available as a PDF file that can be sent to an instructor who has adopted the book. Thestudent who purchases the book can access the students’ solutions manual that contains thesolutions of odd numbered problems via www.cognella.com.

Acknowledgments: ScientificWorkPlace enabled me to type the text and the mathematicalformulas easily in a seamless manner. Adobe Acrobat Pro has enabled me to convert theLaTeX files to pdf files. Mathematica has enabled me to import high quality graphics to mydocuments. I am grateful to the producers and marketers of such software without which Iwould not have had the patience to write and rewrite the material in these volumes. I wouldalso like to acknowledge my gratitude to two wonderful mathematicians who have influencedme most by demonstrating the beauty of Mathematics and teaching me to write clearly andprecisely: Errett Bishop and Stefan Warschawski.

v

vi PREFACE

Last, but not the least, I am grateful to Simla for her encouragement and patience while I spenthours in front a computer screen.

Tunc Geveci ([email protected])San Diego, January 2011

Chapter 12

Functions of Several Variables

In this chapter we will discuss the differential calculus of functions of several variables. We willstudy functions of a single variable that take values in two or three dimensions. The imagesof such functions are parametrized curves that may model the trajectory of an object. Wewill calculate their velocity and acceleration. Geometrically, we will calculate tangentsto curves and their curvature. Then we will take up the differential calculus real-valuedfunctions of several variables. We will study their rates of change in different directionsand their maxima and minima. In the process we will generalize the ideas of local linearapproximation and the differential from functions of a single variable to functions of severalvariables.

12.1 Parametrized Curves, Tangent Vectors and Velocity

Parametrized Curves

Definition 1 Assume that σ is a function from an interval J to the plane R2. Ifσ (t) = (x (t) , y (t)), then x (t) and y (t) are the component functions of σ. The letter t isthe parameter. The image of σ, i.e., the set of points C = {(x (t) , y (t)) : t ∈ J} is said tobe the curve that is parametrized by the function σ. We say that the function σ is aparametric representation of the curve C.

We will use the notation σ : J → R2 to indicate that σ is a function from J into R2. Wecan identify the point σ (t) = (x (t) , y (t)) with its position vector, and regard σ as a vector-valued function whose values are two-dimensional vectors. We may use the notation σ (t) =x (t) i + y (t) j. It is useful to imagine that σ (t) is the position of a particle in motion at thefictitious time t. Figure 1 illustrates a parametrized curve and the "motion of the particle" isindicated by arrows. In many applications, the parameter does refer to time.

35

36 CHAPTER 12. FUNCTIONS OF SEVERAL VARIABLES

x

y

Σ�t�

Figure 1: A parametrized curve in the plane

Example 1 Let σ (t) = (cos (t) , sin (t)), where t ∈ [0, 2π].The curve that is parametrized by the function σ is the unit circle, since

cos2 (t) + sin2 (t) = 1.

The arrows in Figure 2 indicate the motion of a particle that is at σ (t) at ”time" t. ¤

1-1

1

-1

x

y

Figure 2: σ (t) = (cos (t) , sin (t))

We have to distinguish between a vector-valued function and the curve that isparametrized by that function, as in the following example.

Example 2 Let σ2 (t) = (cos (2t) , sin (2t)), where t ∈ [0, 2π].Since

cos2 (2t) + sin2 (2t) = 1,

the function σ2 parametrizes the unit circle, just as the function σ of Example 1. But thefunction σ2 6= σ. For example,

σ2 (π/4) = (cos (π/2) , sin (π/2)) = (0, 1) ,

whereas,

σ (π/4) = (cos (π/4) , sin (π/4)) =

µ1√2,1√2

¶.

If we imagine that the position of a particle is determined by σ2 (t), the particle revolves aroundthe origin twice on the time interval [0, 2π], whereas, a particle whose position is determined byσ revolves around the origin only once on the ”time" interval [0, 2π]. ¤

12.1. TANGENT VECTORS AND VELOCITY 37

Lines are basic geometric objects. Let P0 = (x0, y0) be a given point and let v = v1i+ v2j bea given vector. With reference to Figure 2, a line in the direction of v that passes through thepoint P0 can be parametrized by

σ (t) =−−−→OP 0 + tv = x0i+ y0j+ t (v1i+ v2j)

= (x0 + tv1) i+ (y0 + tv2) j.

Thus, the component functions of σ are

x (t) = x0 + tv1 and y (t) = y0 + tv2

Here, the parameter t varies on the entire number line.

x

y

P0

O

Σ�t�tv

Figure 3: A line through P0 in the direction of v

Example 3 Determine a parametric representation of the line in the direction of the vectorv = i− 2j that passes through the point P0 = (3, 4).Solution

σ (t) =−−−→OP 0 + tv = 3i+ 4j+ t (i−2j)

= (3 + t) i+ (4− 2t) j.Thus, the component functions of σ are

x (t) = 3 + t and y (t) = 4− 2t.Figure 4 shows the line.

3 6x

4

8

12

y

P0

O

tv

Σ�t�

Figure 4


We can eliminate the parameter t and obtain the relationship between x and y:

t = x− 3⇒ y = 4− 2 (x− 3) = −2x+ 10.

Note that the same line can be represented parametrically by other functions. Indeed, thedirection can be specified by any scalar multiple of v, and we can pick an arbitrary point on theline. For example, if we set t = 1 in the above representation, Q0 = (4, 2) is a point on the line,and w = 2v = 2i− 4j is a scalar multiple of v. The function

σ̃ (u) =−−−→OQ0 + uw = 4i+ 2j+ u (2i− 4j)

= (4 + 2u) i+ (2− 4u) j

is another parametric representation of the same line. ¤

Example 4 Imagine that a circle of radius r rolls along a line, and that P (θ) is a point on thecircle, as in Figure 5 ( θ = 0 when P is at the origin).

x

y

�rΘ, r�

ΘP

�rΘ, 0�x

y

r

Figure 5

We have

x (θ) = rθ − r sin (θ) = r (θ − sin (θ))

and

y (θ) = r − r cos (θ) = r (1− cos (θ)) ,

as you can confirm. The parameter θ can be any real number. Let’s set

σ (θ) = (x (θ) , y (θ)) = (r (θ − sin (θ)) , r (1− cos (θ))) ,−∞ < θ < +∞.

The curve that is parametrized by the function σ : R→ R2 is referred to as a cycloid. Figure6 shows the part of the cycloid that is parametrized by

σ (θ) = (2θ − 2 sin (θ) , 2− 2 cos (θ)) , where θ ∈ [0, 8π] .

¤


x

y

Figure 6

Similarly, we can consider parametrized curves in three dimensions:

Definition 2 Assume that σ is a function from an interval J to the three-dimensionalspace R3. If σ (t) = (x (t) , y (t) , z (t)), then x (t), y (t) and z (t) are the component func-tions of σ. The letter t is the parameter. The image of σ, i.e., the set of points C ={(x (t) , y (t) , z(t)) : t ∈ J} is said to be the curve that is parametrized by the functionσ.

We will use the notation σ : J → R3 to indicate that σ is a function from J into R3. Wecan identify the point σ (t) = (x (t) , y (t) , z (t)) with its position vector, and regard σ as avector-valued function whose values are three-dimensional vectors. We may use the notation

σ (t) = x (t) i+ y (t) j+ z (t)k.

You can imagine that σ (t) is the position at time t of a particle in motion.

Example 5 Let σ (t) = (cos (t) , sin (t) , t), where t ∈ R. The curve that is parametrized by σis a helix.

Since

cos2 (t) + sin2 (t) = 1,

the curve that is parametrized by σ lies on the cylinder x2+ y2 = 1. Figure 7 shows part of thehelix.¤

-10

1.0

-5

0

5

z

10

0.51.0

y0.0 0.5

x0.0-0.5

-0.5-1.0 -1.0

Figure 7


Tangent Vectors

Definition 3 The limit of σ : J → R2 at t0 is the vector w if

limt→t0

||σ (t)−w|| = 0

In this case we writelimt→t0

σ (t) = w.

Proposition 1 Assume that σ (t) = x (t) i+ y (t) j. We have

limt→t0

σ (t) = w = w1i+ w2j

if and only if limt→t0 x (t) = w1 and limt→t0 y(t) = w2. Thus,

limt→t0

(x (t) i+ y (t) j) =

µlimt→t0

x (t)

¶i+

µlimt→t0

y (t)

¶j.

Proof

Assume thatlimt→t0

x (t) = w1 and limt→t0

y(t) = w2.

Since||σ (t)−w||2 = (x (t)− w1)2 + (y (t)− w2)2 ,

we havelimt→t0

||σ (t)−w||2 = limt→t0

(x (t)− w1)2 + limt→t0

(y (t)− w2)2 = 0.

Thus, limt→t0 σ (t) = w = w1i+ w2j.

Conversely, assume that limt→t0 σ (t) = w = w1i+ w2. Since

|x (t)− w1| ≤ ||σ (t)−w|| and |y (t)− w2| ≤ ||σ (t)−w|| ,

and limt→t0 ||σ (t)−w|| = 0, we have

limt→t0

|x (t)− w1| = 0 and limt→t0

|y (t)− w2| = 0.

Therefore,limt→t0

x (t) = w1 and limt→t0

y(t) = w2.

¥

Definition 4 The derivative of σ : J → R2 at t is

lim∆t→0

σ (t+∆t)− σ (t)∆t

and will be denoted bydσ

dt,dσ

dt(t) ,

dσ (t)

dtor σ0 (t) .

If C is the curve that is parametrized by σ and σ0 (t) is not the zero vector, then σ0 (t) is avector that is tangent to C at σ (t).


x

y

Σ�t�

Σ�t � �t� �Σ�t�

Σ�t � �t�

Figure 8

x

y

Σ�t�

Figure 9

Proposition 2 If σ (t) = x (t) i+ y (t) j and x0 (t), y0 (t) exist, then

dσ

dt=dx

dti+

dy

dtj.

Proof

σ (t+∆t)− σ (t)∆t

=1

∆t(x (t+∆t) i+ y (t+∆t) j− x (t) i− y (t) j)

=x (t+∆t)− x (t)

∆ti+

y (t+∆t)− y (t)∆t

j.

Therefore,

dσ

dt= lim∆t→0

σ (t+∆t)− σ (t)∆t

= lim∆t→0

µx (t+∆t)− x (t)

∆t

¶i+ lim

∆t→0

µy (t+∆t)− y (t)

∆t

¶j

=dx

dti+

dy

dtj

¥

Example 6 Let σ (t) = cos (t) i+ sin(t)j. Determine σ0 (t).


Solution

σ0 (t) =µd

dtcos (t)

¶i+

µd

dtsin (t)

¶j = − sin (t) i+ cos (t) j.

Note that σ0 (t) is orthogonal to σ (t):

σ0 (t) · σ (t) = − sin (t) cos (t) + cos (t) sin (t) = 0.

¤

�1 1x

�1

1y

Σ�t�

Σ'�t�

Figure 10

Definition 5 Let C be the curve that is parametrized by the function σ : J → R2. If σ0 (t) 6= 0,the unit tangent to C at σ (t) is

T (t) =σ0 (t)||σ0 (t)||

The curve C is said to be smooth at σ (t) if σ0 (t) 6= 0.

With reference to the curve of Example 6, σ0 (t) = − sin (t) i+ cos (t) j is of unit length, since

||σ0 (t)|| =qsin2 (t) + cos2 (t) = 1.

Therefore, T (t) = σ0 (t) = − sin (t) i+ cos (t) j.

Definition 6 If C is the curve that is parametrized by σ : J → R2 and σ0 (t0) 6= 0, the linethat is in the direction of σ0 (t0) and passes through σ (t0) is the tangent line to C at σ (t0) .

Note that a parametric representation of the tangent line is the function L : R→ R2, where

L (u) = σ (t0) + uσ0 (t0) .

Another parametric representation is

l(u) = σ (t0) + uT (t0) .

Example 7 Let σ (θ) = (2θ − 2 sin (θ) , 2− 2 cos (θ)), as in Example 4, and let C be the curvethat is parametrized by σ.


a) Determine T (θ). Indicate the values of θ such that T (θ) exists.b) Determine parametric representations of the tangent line to C at σ (π/2).

Solution

a)σ0 (θ) = (2− 2 cos (θ)) i+ 2 sin (θ) j

Therefore,

||σ0 (θ)|| =q(2− 2 cos (θ))2 + 4 sin2 (θ)

= 2

q1− 2 cos (θ) + cos2 (θ) + sin2 (θ)

= 2√2p1− cos (θ).

Thus, σ0 (θ) = 0 if cos (θ) = 1, i.e., if θ = 2nπ, n = 0,±1,±2,±3, . . .. If θ is not an integermultiple of 2π, the unit tangent to C at σ (θ) is

T (θ) =σ0 (θ)||σ0 (θ)|| =

1

2√2p1− cos (θ) ((2− 2 cos (θ)) i+ 2 sin (θ) j) .

Note thatσ (2nπ) = (2θ − 2 sin (θ) , 2− 2 cos (θ))|θ=2nπ = (4nπ, 0) .

A tangent line to C does not exist at such points. Figure 6 is consistent with this observation.The curve C appears to have a cusp at (4nπ, 0). Let’s consider the case θ = 2π. If the curve isthe graph of the equation y = y (x), by the chain rule,

dy

dx=

dy

dθdx

dθ

=2 sin (θ)

2 (1− cos (θ)) .

Therefore,

limx→4π−

dy

dx= lim

θ→2π−2 sin (θ)

2 (1− cos (θ)) .

We havelimθ→2π

sin (θ) = 0 and limθ→2π

(1− cos (θ)) = 0.By L’Hôpital’s rule,

limθ→2π−

2 sin (θ)

2 (1− cos (θ)) = limθ→2π2 cos (θ)

2 sin (θ).

Since sin (θ) < 0 if θ < 2π and θ is close to 2π, and

limθ→2π

2 cos (θ) = 2 > 0, limθ→2π−

2 sin (θ) = 0,

we have

limx→4π−

dy

dx= lim

θ→2π−2 sin (θ)

2 (1− cos (θ)) = limθ→2π2 cos (θ)

2 sin (θ)= −∞.

Similarly,

limx→4π+

dy

dx= +∞.

Therefore, the curve has a cusp at (4π, 0).

b)


We haveσ (π/2) = (2θ − 2 sin (θ) , 2− 2 cos (θ))|θ=π/2 = (π − 2, 2) ,

and

T (π/2) =1

2√2p1− cos (θ) ((2− 2 cos (θ)) i+ 2 sin (θ) j)

¯̄̄̄¯π/2

=1

2√2(2i+ 2j) =

1√2(i+ j) .

A parametric representation of the line that is tangent to C at σ (π/2) is

L (u) = σ (π/2) + uσ0 (π/2)= (π − 2) i+ 2j+ u (2 (i+ j))= (π − 2 + 2u) i+ (2 + 2u) j

¤

The above definitions extend to curves in the three-dimensional space in the obvious manner:if σ : J → R3, the limit of σ is the vector w if

limt→t0

||σ (t)−w|| = 0,

and we writelimt→t0

σ (t) = w.

If σ (t) = x (t) i+ y (t) j+ z (t)k, then

limt→t0

σ (t) = w1i+ w2j+ w3k

if and only iflimt→t0

x (t) = w1, limt→t0

y (t) = w2 and limt→t0

z (t) = w3.

Thus,

limt→t0

(x (t) i+ y (t) j+ z (t)k) =

µlimt→t0

x (t)

¶i+

µlimt→t0

y (t)

¶j+

µlimt→t0

z (t)

¶k.

The derivative of σ is

dσ

dt= lim∆t→0

σ (t+∆t)− σ (t)∆t

=dx

dti+

dy

dtj+

dz

dtk,

and the unit tangent is

T (t) =1

||σ0 (t)||σ0 (t) ,

provided that ||σ0 (t)|| 6= 0.

Example 8 Let σ (t) = (cos (t) , sin (t) , t) as in Example 5, and let C be the curve that isparametrized by σ.


a) Determine T (t). Indicate the values of t such that T (t)exists.b) Determine a parametric representation of the tangent line to C at σ (π/3).Solution

a)σ0 (t) = − sin (t) i+ cos (t) j+ k.

Therefore,

||σ0 (t)|| =qsin2 (t) + cos2 (t) + 1 =

√2.

Thus, T (t) exists for each t ∈ R and

T (t) =1

||σ0 (t)||σ0 (t) =

1√2(− sin (t) i+ cos (t) j+ k) .

b) We have

σ (π/3) = cos (π/3) i+ sin (π/3) j+π

3k =

1

2i+

√3

2j+

π

3k,

and

σ0 (π/3) = − sin (π/3) i+ cos (π/3) j+ k = −√3

2i+

1

2j+ k.

A parametric representation of the line that is tangent to C at σ (π/6) is

L (u) = σ (π/3) + uσ0 (π/3)

=1

2i+

√3

2j+

π

3k+ u

Ã−√3

2i+

1

2j+ k

!

=

Ã1

2−√3

2u

!i+

Ã√3

2+1

2u

!j+

³π3+ u

´k.

¤

Velocity

Assume that σ is a function from an interval on the number line into the plane or three-dimensional space, and that a particle in motion is at the point σ (t) at time t. Let ∆t denotea time increment. The average velocity of the particle over the time interval determined by tand t+∆t is

displacementelapsed time

=σ (t+∆t)− σ (t)

∆t.

We define the instantaneous velocity at the instant t as the limit of the average velocity as ∆tapproaches 0:

Definition 7 If σ : J → R2 or σ : J → R3 and a particle is at the point σ (t) at time t, thevelocity v (t) of the particle at the instant t is

v (t) =dσ

dt= lim∆t→0

σ (t+∆t)− σ (t)∆t

.

Note that v (t)is a vector in the direction of the tangent to C at σ (t) if v (t) 6= 0.

Example 9 Let σ (t) = (2t− 2 sin (t)) i + (2− 2 cos (t)) j, as in Example 7, and let C be thecurve that is parametrized by σ. Determine v (3π/4), v (π) and v (2π).


SolutionAs in Example 7,

v(t) = σ0 (t) = (2− 2 cos (t)) i+ 2 sin (t) j.Therefore,

v (3π/4) =

µ2− 2 cos

µ3π

4

¶¶i+ 2 sin

µ3π

4

¶j =

Ã2− 2

Ã−√2

2

!!i+ 2

Ã√2

2

!j

=³2 +√2´i+√2j,

v (π) = (2− 2 cos (π)) i+ 2 sin (π) j = 4i,and

v (2π) = (2− 2 cos (2π)) i+ 2 sin (2π) j = 0.Note that the direction of the motion is not defined at t = 2π since v (2π) = 0 (and the curveC is not smooth at σ (2π)). ¤

Problems

In problems 1-4 assume that a curve C is parametrized by the function σ.a) Determine σ0 (t) , σ0 (t0) and the unit tangent T (t0) to C at σ (t0).b) Determine the vector-valued function L that parametrizes the tangent line to C at σ (t0)(specify the direction of the line by the vector σ0 (t0)).

1.σ (t) =

¡t, sin2 (4t)

¢, −∞ < t < +∞, t0 = π/3.

2.σ (t) = (2 cos (3t) , sin (t)) , 0 ≤ t ≤ 2π, t0 = π/4.

3.

σ (t) =

µ3t

t2 + 1,3t2

t2 + 1

¶, −∞ < t


a) Determine zx (x, y) and zy (x, y) directly (i.e., without relying on a formula from the text).b) Evaluate zx (1, 2) and zy (1, 2) and determine an equation for the plane that is tangent to thegraph of the given equation at (1, 2,−3).11. Assume that z (x, y) is defined implicitly by the equation

z2 + x2 − y2 − 9 = 0,and z (3, 6) = 6.

6

3x

z

6

y

a) Determine zx (x, y) and zy (x, y) directly (i.e., without relying on a formula from the text).b) Evaluate zx (3, 6) and zy (3, 6). and determine an equation for the plane that is tangent tothe graph of the given equation at (3, 6, 6).

12.7 Directional Derivatives and the Gradient

Directional Derivatives

We will begin by considering functions of two variables. Assume that f has partial derivativesin some open disk centered at (x, y). The partial derivative ∂xf (x, y) can be interpreted as therate of change of f in the direction of the standard basis vector i and the partial derivative∂yf (x, y) can be interpreted as the rate of change of f in the direction of the standard basisvector j. We would like to arrive at a reasonable definition of the rate of f at (x, y) in thedirection of an arbitrary unit vector u = u1i+ u2j.

x

y

hu

P

Ph

Figure 1

Let h 6= 0. Let’s assume that −−→OPh = −−→OP + hu so thatPh = (x+ hu1, y + hu2) .

We define the average rate of change of f corresponding to the displacement−−→PPh as

f (Ph)− f (P )h

=f (x+ hu1, y + hu2)− f (x, y)

h

12.7. DIRECTIONAL DERIVATIVES AND THE GRADIENT 95

It is reasonable to define the rate of change of f at P in the direction of u as

limh→0

f (Ph)− f (P )h

.

Another name for this limit is ”directional derivative":

Definition 1 The directional derivative of f at (x, y) in the direction of the unitvector u is

Duf (x, y) = limh→0

f (x+ hu1, y + hu2)− f (x, y)h

.

Note that Dif (x, y) is ∂xf (x, y) and Djf (x, y) is ∂yf (x, y). We can calculate the directionalderivative in an arbitrary direction in terms of ∂xf (x, y) and ∂yf (x, y):

Proposition 1 Assume that f is differentiable at (x, y) and that u = u1i+u2j is a unitvector. Then

Duf (x, y) =∂f

∂x(x, y)u1 +

∂f

∂y(x, y)u2.

Proof

Setg (t) = f (x+ tu1, y + tu2) , t ∈ R.

Then,

g0 (0) = limh→0

g (h)− g (0)h

= limh→0

f (x+ hu1, y + hu2)− f (x, y)h

= Duf (x, y)

Let’s setv (x, t) = x+ tu1 and w (x, t) = y + tu2,

so that g (t) = f (v (x, t) , w (x, t)). By the chain rule,

d

dtg (t) =

d

dtf (x+ tu1, y + tu2)

=∂f

∂v(v,w)

dv

dt+

∂f

∂w(v, w)

dw

dt

=∂f

∂v(x+ tu1, y + tu2)

d

dt(x+ tu1) +

∂f

∂w(x+ tu1, y + tu2)

d

dt(y + tu2)

=∂f

∂v(x+ tu1, y + tu2)u1 +

∂f

∂w(x+ tu1, y + tu2)u2.

Therefore,

Duf (x, y) = g0 (0) =

∂f

∂v(x+ tu1, y + tu2)u1 +

∂f

∂w(x+ tu1, y + tu2)u2

¯̄̄̄t=0

=∂f

∂v(x, y)u1 +

∂f

∂w(x, y)u2

=∂f

∂x(x, y)u1 +

∂f

∂y(x, y)u2,

as claimed. ¥

Example 1 Let f (x, y) = 36 − 4x2 − y2 and let u be the unit vector in the direction ofv = −i+ 4j. Determine Duf (1, 3).


Solution

We have

∂f

∂x(x, y) =

∂

∂x

¡36− 4x2 − y2¢ = −8x and ∂f

∂y(x, y) =

∂

∂y

¡36− 4x2 − y2¢ = −2y.

Therefore,∂f

∂x(1, 3) = −8 and ∂f

∂y(1, 3) = −6.

We have

||v|| =p1 + 42 =

√17.

Therefore,

u =v

||v|| =1√17(−i+ 4j) = − 1√

17i+

4√17j.

Thus,

Duf (1, 3) =∂f

∂x(1, 3)u1 +

∂f

∂y(1, 3)u2

= (−8)µ− 1√

17

¶+ (−6)

µ4√17

¶= − 16√

17.

¤

Remark 1 All directional derivatives may exist, but f may not be differentiable, as in thefollowing example:

Let

f (x, y) =

⎧⎨⎩ xy2

x2 + y4if (x, y) 6= (0, 0) ,

0 if (x, y) = (0, 0)

Figure 2 shows the graph of f .

2

x0

-0.52

y0

0.0z

-2-2

0.5

Figure 2

In Section 12.5 we saw that f is not continuous at (0, 0) so that it is not differentiable. Nev-ertheless, the directional derivatives of f exist in all directions at the origin. Indeed, let u be


an arbitrary unit vector, so that u = (cos (θ) , sin (θ)) for some θ. Let’s calculate Duf (0, 0),directly from the definition. We have

f (h cos (θ) , h sin (θ))

h=

h3 cos (θ) sin2 (θ)

h2 cos2 (θ) + h4 sin4 (θ)

h=

h3 cos (θ) sin2 (θ)

h3 cos2 (θ) + h5 sin4 (θ)

=cos (θ) sin2 (θ)

cos2 (θ) + h2 sin4 (θ).

Therefore,

Duf (0, 0) = limh→0

f (h cos (θ) , h sin (θ))

h=cos (θ) sin2 (θ)

cos2 (θ)=sin2 (θ)

cos (θ)

if cos(θ) 6= 0. The case cos(θ) = 0 corresponds to fy (0, 0) or −fy (0, 0). We have

fy (0, 0) = limh→0

f (0, h)− f (0, 0)h

= limh→0

0

h= 0.

Thus, the directional derivatives of f exist in all directions. ♦The gradient of a function is a useful notion within the context of directional derivatives, andin many other contexts:

Definition 2 The gradient of f at (x, y) is the vector-valued function that assigns the vector

∇f (x, y) = ∂f∂x(x, y) i+

∂f

∂y(x, y) j

to each (x, y) where the partial derivatives of f exist (read ∇f as "del f").We can express a directional derivative in terms of the gradient:

Proposition 2 The directional derivative of f at (x, y) in the direction of the unitvector u can be expressed as

Duf (x, y) =∇f (x, y) · uProof

By Proposition 1,

Duf (x, y) =∂f

∂x(x, y)u1 +

∂f

∂y(x, y)u2

By the definition of∇f (x, y), the right-hand side can be expressed as a dot product. Therefore,

Duf (x, y) =

µ∂f

∂x(x, y) i+

∂f

∂y(x, y) j

¶· (u1i+ u2j)

=∇f (x, y) · u¥Thus, the directional derivative of f at (x, y) in the direction of the unit vector u isthe component of ∇f (x, y) along u.

Example 2 Let f (x, y) = x2 + 4xy + y2.


a) Determine the gradient of f .b) Calculate the directional derivative of f at (2, 1) in the directions of the vectors i + j andi− j.Solution

a) Figure 3 displays the graph of f .

0

4

50

z

100

24

y0 2

x0-2

-2-4 -4

Figure 3

We have∂f

∂x(x, y) =

∂

∂x

¡x2 + 4xy + y2

¢= 2x+ 4y

and∂f

∂y(x, y) =

∂

∂y

¡x2 + 4xy + y2

¢= 4x+ 2y.

Therefore,

∇f (x, y) = ∂f∂x(x, y) i+

∂f

∂y(x, y) j =(2x+ 4y) i+ (4x+ 2y) j.

Figure 4 displays ∇f (x, y) at certain points (x, y) as arrows located at these points.

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

x

y

Figure 4

b) A unit vector in the direction of i+ j is

u1 =1√2i+

1√2j,

and a unit vector in the direction of i− j is

u2 =1√2i− 1√

2j.

We have∇f (2, 1) = (2x+ 4y) i+ (4x+ 2y) j|x=2,y=1 = 8i+ 10j.


Therefore,

Du1f (2, 1) =∇f (2, 1) · u1= (8i+ 10j) ·

µ1√2i+

1√2j

¶=

8√2+10√2=18√2= 9√2,

and

Du2f (2, 1) =∇f (2, 1) · u2= (8i+ 10j) ·

µ1√2i− 1√

2j

¶=

8√2− 10√

2= − 2√

2= −√2.

Thus, at (2, 1) the function increases in the direction of u1 and decreases in the direction ofu2. Figure 5 shows some level curves of f , the vectors u1 and u2, and a unit vector along∇f (2, 1) . ¤

�13

�13

�5

�5

0

0

0

0

55

13

13

�4 �2 2 4x

�4

�2

1

2

4

y

�2,1�fu1

u2

Figure 5

Proposition 3 The maximum value of the directional derivatives of a function f ata point (x, y) is ||∇f (x, y)||, the length of the gradient of f at (x, y), and it is attainedin the direction of ∇f (x, y). The minimum value of the directional derivatives of fat (x, y) is − ||∇f (x, y)|| and it is attained the direction of −∇f (x, y) .Proof

Since Duf (x, y) =∇f (x, y) · u, and ||u|| = 1, we haveDuf (x, y) = ||∇f (x, y)|| ||u|| cos (θ) = ||∇f (x, y)|| cos (θ) ,

where θ is the angle between ∇f (x, y) and the unit vector u. Since cos (θ) ≤ 1,Duf (x, y) ≤ ||∇f (x, y)|| .

The maximum value is attained when cos (θ) = 1, i.e., θ = 0. This means that u is the unitvector in the direction of ∇f (x, y) .


Similarly,Duf (x, y) = ||∇f (x, y)|| cos (θ)

attains its minimum value when cos (θ) = −1, i.e., θ = π. This means that u = −∇f (x, y).The corresponding directional derivative is − ||∇f (x, y)|| .¥

Example 3 Let f (x, y) = x2 + 4xy + y2, as in Example 2. Determine the unit vector alongwhich the rate of increase of f at (2, 1) is maximized and and the unit vector along which therate of decrease of f at (2, 1) is maximized. Evaluate the corresponding rates of change of f .

Solution

In Example 2 we showed that ∇f (2, 1) = 8i+ 10j. Therefore,

||∇f (2, 1)|| = ||(8i+ 10j)|| = √64 + 100 = √164 = 2√41.

Thus, the value of f increases at the maximum rate of 2√41 in the direction of the unit vector

1

||∇f (2, 1)||∇f (2, 1) =1

2√41(8i+ 10j) =

4√41i+

5

41j.

The value of the function decreases at the maximum rate of 2√41 (i.e., the rate of change is −

2√41) in the direction of the unit vector

− 1||∇f (2, 1)||∇f (2, 1) = −4√41i− 5

41j.

¤

The Chain Rule and the Gradient

A special case of the chain rule says that

d

dtf (x (t) , y (t)) =

∂f

∂x(x (t) , y (t))

dx

dt+

∂f

∂y(x (t) , y (t))

dy

dt,

as In Proposition 1 of Section 12.6. The notion of the gradient enables us to provide a usefulgeometric interpretation of the above expression. Let’s begin by expressing the chain rule asfollows:

Proposition 4 Assume that the curve C in the plane is parametrized by the functionσ where

σ (t) = (x(t), y(t)) .

Thend

dtf (σ (t)) =∇f (σ (t)) · dσ

dt

Proof

We have

d

dtf (σ (t)) =

d

dtf (x (t) , y (t)) ,

∇f (σ (t)) = ∂f∂x(x (t) , y (t)) i+

∂f

∂y(x (t) , y (t)) ,


anddσ

dt=dx

dti+

dy

dtj.

Therefore,

∇f (σ (t)) · dσdt=

µ∂f

∂x(x (t) , y (t)) i+

∂f

∂y(x (t) , y (t))

¶·µdx

dti+

dy

dtj

¶=

∂f

∂x(x (t) , y (t))

dx

dt+

∂f

∂y(x (t) , y (t))

dy

dt.

By the chain rule,

∂f

∂x(x (t) , y (t))

dx

dt+

∂f

∂y(x (t) , y (t))

dy

dt=d

dtf (x (t) , y (t)) =

d

dtf (σ (t)) .

Thus,d

dtf (σ (t)) =∇f (σ (t)) · dσ

dt,

as claimed. ¥

Remark 2 With the notation of Proposition 4, assuming that σ0 (t) 6= 0,

d

dtf (σ (t)) =∇f (σ (t)) · dσ

dt=∇f (σ (t)) ·

dσ

dt(t)¯̄̄̄¯̄̄̄

dσ

dt(t)

¯̄̄̄¯̄̄̄ ¯̄̄̄¯̄̄̄dσdt(t)

¯̄̄̄¯̄̄̄

=

¯̄̄̄¯̄̄̄dσ

dt(t)

¯̄̄̄¯̄̄̄∇f (σ (t)) ·T (t) ,

where T (t) is the unit tangent to C at σ (t). The quantity∇f (σ (t))·T (t) is DT(t)f (σ (t)), thedirectional derivative of f at σ (t) in the direction of the unit tangent T (t), i.e., the componentof ∇f (σ (t)) along T (t). Thus,

d

dtf (σ (t)) =

¯̄̄̄¯̄̄̄dσ

dt(t)

¯̄̄̄¯̄̄̄DT(t)f (σ (t)) .

If σ (t) is the position of a particle at time t, ||σ0 (t)|| is its speed at t. In this case,d

dtf (σ (t)) = the speed of the object at t× component of ∇f (σ (t)) along T (t) .

♦

Proposition 5 Let C be the level curve of f that passes through the point (x0, y0).The gradient of f at (x0, y0) is perpendicular to C at (x0, y0), in the sense that ∇fis orthogonal to the tangent line to C at (x0, y0).

A Plausibility Argument for Proposition 5

We will assume that a segment of C near (x0, y0) can be parametrized by the function σ : J → R2such that σ (t0) = (x0, y0) and σ0 (t0) 6= 0. By Proposition 4,

d

dtf (σ (t)) =∇f (σ (t)) · dσ

dt(t) .


Since C is a level curve of f , the value of f on C is a constant. Therefore,

d

dtf (σ (t)) = 0

for each t ∈ J . In particular,

0 =d

dtf (σ (t0)) =∇f (σ (t0)) · dσ

dt(t0) =∇f (x0, y0) · dσ

dt(t0) .

This implies that ∇f (x0, y0) is orthogonal to the vector σ0 (t0) that is tangent to C at σ (t0) =(x0, y0). ¥

Example 4

a) The point (1, 3) is on the curve C that is the graph of the equation x2 + 4xy + y2 = 22.Determine a vector that is orthogonal to C at (1, 3).b) Determine the line that is tangent to the C at (1, 3) .

Solution

a( We set f (x, y) = x2 + 4xy + y2, as in Example 2, so that C is a level curve of f . We have

∇f (1, 3) =∇f (1, 3) = (2x+ 4y) i+ (4x+ 2y) j|x=1,y=3 = 14i+ 10j.

Therefore, 14i+10j. is orthogonal to C at (1, 3). Figure 6 shows C and a vector along ∇f (1, 3).The picture is consistent with our claim.

1x

3

y

�1,3�

Figure 6

b) If P0 = (1, 3) and P = (x, y) is an arbitrary point on the line that is tangent to C at P0, wehave

∇f (1, 3) ·−−→P0P = 0.Therefore,

(14i+ 10j) · ((x− 1) i+ (y − 3)) = 0⇔

14 (x− 1) + 10 (y − 3) = 0.Figure 7 shows the tangent line.


1x

3

y

�1,3�

Figure 7

Functions of three or more variables

Our discussion of directional derivatives and the gradient extend to functions of more than twovariables in a straightforward fashion. Let’s consider functions of three variables to be specific.Thus, assume that f is a scalar function of the variables x, y and z. If u = u1i+ u2j + u3kis a three-dimensional unit vector, the directional derivative of f at (x0, y0, z0) in thedirection of u is

Duf (x0, y0, z0) = limh→0

f (x0 + hu1, y0 + hu2, z0 + hu3)− f (x0, y0, z0)h

The gradient of f at (x, y, z) is

∇f (x, y, z) = ∂f∂x(x, y, z) i+

∂f

∂y(x, y, z) j+

∂f

∂z(x, y, z)k.

If f is differentiable at (x0, y0, z0),

Duf (x0, y0, z0) =∂f

∂x(x0, y0, z0)u1 +

∂f

∂y(x0, y0, z0)u2 +

∂f

∂z(x0, y0, z0)u3

=∇f (x0, y0; z0) · uWe have

− ||∇f (x0, y0, z0)|| ≤ Duf (x0, y0, z0) ≤ ||∇f (x0, y0, z0)|| ,Thus, the rate of change of f at (x0, y0, z0) has the maximum value ||∇f (x0, y0, z0)|| in the direc-tion of∇f (x0, y0, z0). The function decreases at the maximum rate of ||∇f (x0, y0, z0)|| (i.e., therate of change is − ||∇f (x0, y0, z0)||) at the point (x0, y0, z0) in the direction of −∇f (x0, y0, z0).If σ : J → R3, and σ (t) = (x (t) , y (t) , z (t)), by the chain rule,d

dtf (x (t) , y (t) , z (t))

=∂f

∂x(x (t) , y (t) , z (t))

dx

dt(t) +

∂f

∂y(x (t) , y (t) , z (t))

dy

dt(t) +

∂f

∂z(x (t) , y (t) , z (t))

dz

dt(t) .

so that.d

dtf (σ (t)) =∇f (σ (t)) · dσ

dt.


If S is a level surface of f = f (x, y, z) so that f has a constant value on S, and (x0, y0, z0)lies on S, then ∇f (x0, y0, z0) is orthogonal to S at (x0,y0, z0), in the sense that ∇f (x0, y0, z0)is orthogonal at (x0, y0, z0) to any curve on S that passes through (x0, y0, z0). Indeed, if such acurve C is parametrized by σ : J → R3 andσ (t0) = (x0, y0, z0) we have

d

dtf (σ (t)) = 0 for t ∈ J,

since f (σ (t)) has a constant value. Therefore,

∇f (σ (t)) · dσdt(t) =

d

dtf (σ (t)) = 0.

In particular,

∇f (x0, y0, z0) · dσdt(t0) =∇f (σ (t0)) · dσ

dt(t0) = 0,

so that ∇f (x0, y0, z0) is orthogonal to the vector σ0 (t0) that is tangential to the curve C. Itis reasonable to declare that the plane that is spanned by all such tangent vectors and passesthrough (x0, y0, z0) is the tangent plane to the level surface S of f . Since ∇f (x0, y0z0)is normal to that plane, the equation of the tangent plane is

∇f (x0, y0z0) · ((x− x0) i+ (y − y0) j+ (z − z0)k) = 0,

i.e.,∂f

∂x(x0, y0, z0) (x− x0) + ∂f

∂y(x0, y0, z0) (y − y0) + ∂f

∂z(x0, y0, z0) (z − z0) = 0.

Example 5 Let f be the function such that

f(x, y, z) =x2

9+y2

4+ z2

a) Determine∇f (x, y, z) and the directional derivative of f at (2, 1, 1) in the direction of (1, 1, 1).

b) Determine the tangent plane to the surface

x2

9+y2

4+ z2 =

61

36

at (2, 1, 1).

Solutiona) We have

∇f (x, y, z) = ∂f∂x(x, y, z) i+

∂f

∂y(x, y, z) j+

∂f

∂z(x, y, z) =

2x

9i+

y

2j+ 2zk.

Therefore,

∇f (2, 1, 1) = 49i+

1

2j+ 2k.

The unit vector u along (1, 1, 1) is

1√3i+

1√3j+

1√3k.


Therefore,

Duf (2, 1, 1) =∇f (2, 1, 1) · u=

µ4

9i+

1

2j+ 2k

¶·µ1√3i+

1√3j+

1√3k

¶=

4

9√3+

1

2√3+

2√3=53

54

√3.

b) We can express the equation of the required plane as

∇f (2, 1, 1) · ((x− 2) i+ (y − 1) j+ (z − 1)k) = 0,

i.e., µ4

9i+

1

2j+ 2k

¶· ((x− 2) i+ (y − 1) j+ (z − 1)k) = 0.

Thus,4

9(x− 2) + 1

2(y − 1) + 2 (z − 1) = 0.

Figure 8 illustrates the surface and the above tangent plane. ¤

Figure 8

Problems

In problems 1-6,a) Compute the gradient of fb) Compute the directional derivative of f at P in the direction of the vector v.

1.f (x, y) = 4x2 + 9y2, P = (3, 4) , v = (−2, 1)

2.f (x, y) = x2 − 4x− 3y2 + 6y + 1, P = (0, 0) , v = (1,−1)

3.f (x, y) = ex

2−y2 , P = (2, 1) , v = (−1, 3)4.

f (x, y) = sin (x) cos (y) , P = (π/3,π/4) , v = (2,−3)


5.f (x, y, z) = x2 − y2 + 2z2, P = (1,−1, 2) , v = (1,−1, 1)

6.f (x, y, z) =

1px2 + y2 + z2

, P = (2,−2, 1) , v = (3, 4, 1)

In problems 7 and 8,a) Determine a vector v such that the rate at which f increases at P has its maximum value inthe direction of v, and the corresponding rate of increase of f,b) Determine a vector w such that the rate at which f decreases at P has its maximum valuein the direction of w, and the corresponding rate of decrease of f.

7.f (x, y) =

1px2 + y2

, P = (2, 3)

8.f (x, y) = ln

³px2 + y2

´, P = (3, 4)

In problems 9 and 10,a) Compute

d

dtf (σ (t))

¯̄̄̄t=t0

by making use of the chain rule,b) Compute the directional derivative of f in the tangential direction to the curve that isparametrized by σ at the point σ (t0) .

9.f (x, y) = ex

2−y2 , σ (t) = (2 cos (t) , 2 sin (t)) , t0 = π/6

10

f (x, y) = arctan

Ãyp

x2 + y2

!, σ (t) =

µ1− t21 + t2

,2t

1 + t2

¶, t0 = 2.

In problems 11-13,a) Find a vector that is orthogonal at the point P to the curve that is the graph of the givenequation,b) Determine an equation of the line that is tangent to that curve at P :

11.2x2 + 3y2 = 35, (2, 3)

12.x2 − y2 = 4,

³3,√5´

13.e25−x

2−y2 = 1, (3, 4)

In problems 14-17,a) Find a vector that is orthogonal at the point P to the surface that is the graph of the givenequationb) Determine an equation of the plane that is tangent to the given surface at P.

14.z − x2 + y2 = 0, P = (4, 3, 7)

15.x2 − y2 + z2 = 1, P = (2, 2, 1)

12.8. LOCAL MAXIMA AND MINIMA 107

16.

x2 − y2 − z2 = 1, P = (3, 2, 2)

17.

x− sin (y) cos (z) = 0, P = (1,π/2, 0)

12.8 Local Maxima and Minima

The Definitions and Preliminary Examples

Definition 1 A function f has a local maximum at the point (x0, y0) if there exists an opendisk D containing (x0, y0) such that f (x, y) ≤ f (x0, y0) for each (x, y) ∈ D. A function f hasa local minimum at the point (x0, y0) if there exists an open disk D containing (x0, y0) suchthat f (x, y) ≤ f (x0, y0) for each (x, y) ∈ D.

Definition 2 The absolute maximum of a function f on the set D ⊂ R2 is f (x0, y0) iff (x, y) ≤ f (x0, y0) for each (x, y) ∈ D. The absolute minimum of a function f on the setD ⊂ R2 is f (x0, y0) if f (x, y) ≥ f (x0, y0) for each (x, y) ∈ D.

Example 1 Let Q (x, y) = x2 + y2. Since Q(x, y) ≥ 0 for each (x, y) ∈ R2 and Q (0, 0) = 0, Qattains its absolute minimum value 0 on R2 at (0, 0). Of course, the only point at which Q hasa local minimum is also (0, 0). The function does not attain an absolute maximum on /R2 sinceit attains values of arbitrarily large magnitude. For example,

limx→+∞Q (x, x) = limx→+∞ 2x

2 = +∞.

¤

0

5

10z

15

2

y20

x0

-2 -2

Figure 1

Example 2 Let Q (x, y) = −x2 − y2. Q attains its maximum value 0 at (0, 0) .The functiondoes not attain an absolute minimum on R2 since it attains negative values of arbitrarily largemagnitude. For example,

limx→+∞Q (x, x) = limx→+∞

¡−2x2¢ = −∞.¤

138 CHAPTER 13. MULTIPLE INTEGRALS

1. Z y=2y=1

µZ x=3x=0

¡x2 + 3xy2

¢dx

¶dy

2. Z x=1x=0

ÃZ y=π/3y=π/4

x sin (y) dy

!dx

3. Z y=4y=2

µZ x=3x==1

exyydx

¶dy

4. Z x=2x=0

µZ y=3y=1

xyp1 + y2dy

¶dx

In problems 5-8, calculate the double integral:

5. Z ZD

xy2

x2 + 1dA where D = {(x, y) : 0 ≤ x ≤ 1, −2 ≤ y ≤ 2}

6. Z ZD

xyex2ydA where D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2}

7. Z ZD

x2

1 + y2dA where D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}

8. Z ZD

y cos (x+ y) dA, D = {(x, y) : 0 ≤ x ≤ π3, 0 ≤ y ≤ π

2}

13.2 Double Integrals over Non-Rectangular RegionsIn this section we will discuss double integrals on regions that are not rectangular. Let D bea bounded region in the Cartesian coordinate plane that has a boundary consisting of smoothsegments. As in the case of a rectangle, let’s begin with a function f that is continuous andnonnegative on D. In order to approximate the volume of the region G in the xyz-space that isbetween the graph of z = f (x, y) and D we form a mesh in the xy-plane that consists of verticaland horizontal lines. In this case D is not necessarily the union of rectangles. Let’s consideronly those rectangles that have nonempty intersection with D and label them as

Rjk = [xj−1, xj ]× [yk−1, yk] , where j = 0, 1, 2, . . . ,m and k = 0, 1, 2, . . . , n.

We set∆xj = xj − xj−1 and ∆yk = yk − yk−1.

We can approximate the volume of G by a sum

nXk=1

mXj=1

f¡x∗j , y

∗k

¢∆Ajk =

nXk=1

mXj=1

f¡x∗j , y

∗k

¢∆xj ∆yk,

where¡x∗j , y

∗k

¢ ∈ Rjk..The double integral of f on D is the numberZ ZD

f (x, y) dA =

Z ZD

f (x, y) dxdy

13.2. DOUBLE INTEGRALS OVER NON-RECTANGULAR REGIONS 139

that can be approximated by such sums as accurately as desired provided that the maximumof ∆Ajk is small enough. Such a number exists for any continuous function f on D even if thesign of f may vary.

There are counterparts of the Fubini Theorem that allowed us to calculated a double integralon a rectangle as an iterated integral. Assume that the region D in the xy-plane is bounded bythe lines x = a, x = b and the graphs of functions f (x) and g (x), as in Figure 1. We will referto such a region as an x-simple region, or as a region of type 1.

Figure 1

Assume that F is continuous on D. We can evaluate the double integral as an iterated integral:Z ZD

F (x, y) dA =

Z x=bx=a

ÃZ y=g(x)y=f(x)

F (x, y) dy

!dx.

We will leave the proof to a course in advanced calculus. The statement should be plausible, asin the case of rectangular regions: As x varies from a to b,.the region is swept by rectangles of"infinitesimal" thickness dx that extend from y = g (x) to y = f (x) .The roles of x and y may be reversed: A region D is a y-simple region, or a region of type2, if it is bounded by the lines y = a, y = b and the graphs of x = f (y) and x = g (y), as inFigure 2. We have Z Z

D

F (x, y) dA =

Z y=by=a

ÃZ x=g(y)x=f(y)

F (x, y) dx

!dxy.

x

y

y � b

y � a

x � f�y� x � g�y�

Figure 2

Example 1 Evaluate Z ZD

p1 + x2dxdy,

where D is the triangle that is bounded by y = x, x = 1 and the x-axis.


Solution

Figure 3 shows the triangle D. The integral is the volume of the region between the graph ofz =√1 + x2 and the triangle D in the xy-plane, as shown in Figure 4.

0 x1

0

1

02

2y1

z

2

Figure 4

The region D is both x-simple and y-simple. We will treat it as an x-simple region. Thus,

Z ZD

p1 + x2dxdy =

Z 1x=0

µZ xy=0

p1 + x2dy

¶dx =

Z 1x=0

p1 + x2

µZ xy=0

1dy

¶dx

=

Z 1x=0

p1 + x2 (y|x0) dx

=

Z x=1x=0

p1 + x2xdx.

We set u = 1 + x2, so that du = 2xdx. Therefore,Z x=1x=0

p1 + x2xdx =

1

2

Z 2u=1

√udu =

1

2

Ãu3/2

3/2

¯̄̄̄2u=1

!=23/2 − 13

.

¤


x sin (y) dA,

where D is the region bounded by y = x, y = x2, x = 0 and x = 1.

Solution

Figure 5 shows the region D.

1x

1

y

Figure 5


This is a region of type 1 and of type 2. We will treat it as a region of type I, and integratewith respect to y first:

Z ZD

x sin (y) dA =

Z x=1x=0

µZ y=xy=x2

x sin (y) dy

¶dx =

Z x=1x=0

x

µZ y=xy=x2

sin (y) dy

¶dx

=

Z x=1x=0

x³− cos (y)|y=xy=x2

´dx

=

Z x=1x=0

x¡− cos (x) + cos ¡x2¢¢ dx

= −Z 10

x cos (x) dx+

Z 10

x cos¡x2¢dx.

The first integral requires integration by parts: If we set u = x and dv = cos (x), then

du = dx and v =Zcos (x) dx = sin (x) .

Thus, Zx cos (x) dx =

Zudv = uv −

Zvdu

= x sin (x)−Zsin (x) dx

= x sin (x) + cos (x) .

Therefore, Z 10

x cos (x) dx = sin(1) + cos (1)− cos (0) = sin(1) + cos (1)− 1.

As for the second integral, we set u = x2 so that du = 2xdx. Thus,Zx cos

¡x2¢dx =

1

2

Zcos (u) du =

1

2sin¡x2¢.

Therefore, Z 10

x cos¡x2¢dx =

1

2sin (1) .

Thus, Z ZD

x sin (y) dA = −Z 10

x cos (x) dx+

Z 10

x cos¡x2¢dx

= − sin (1)− cos (1) + 1 + 12sin (1)

= 1− cos (1)− 12sin (1) .

¤


yexdA

where D is the region bounded by x = y2/4, x = 1, y = 1 and y = 2.


Solution

Figure 6 shows the region D.

14 1 2

x

1

2

y

Figure 6

This is a region of type 1 and of type 2. We will treat it as a region of type 2, and integratewith respect to x first:

Z ZD

yex2

dA =

Z y=2y=1

ÃZ x=1x=y2/4

yexdx

!dy =

Z y=2y=1

y

ÃZ x=1x=y2/4

exdx

!dy

=

Z y=2y=1

y³ex|1y2/4

´dy

=

Z y=2y=1

y³e− ey2/4

´dy

=e

2y2 − 2ey2/4

¯̄̄21

= 2e− 2e− e2+ 2e1/4

= −e2+ 2e1/4.

¤

In some cases, changing the order of integration enables us to evaluate an integral that isintractable in its original form, as in the following example.

Example 4 Evaluate the iterated integralZ 1x=0

µZ y=2y=2x

e−y2

dy

¶dx.

Solution

We cannot evaluate the given iterated integral in terms of familiar antiderivatives. The firstthing to do is to sketch the region of integration:


1 2x

1

2

y

y = 2x

Figure 7

We will reverse the order of integration:Z 1x=0

µZ 2y=2x

e−y2

dy

¶dx =

Z y=2y=0

ÃZ x=y/2x=0

e−y2

dx

!dy =

Z y=2y=0

e−y2

ÃZ x=y/2x=0

dx

!dy

=

Z y=2y=0

e−y2³x|y/20

´dy

=

Z y=2y=0

e−y2 y

2dy

= −14e−y

2

¯̄̄̄20

= −14e−4 +

1

4.

¤

Problems

In problems 1 and 2 evaluate the given iterated integral:

1. Z y=4y=0

Z x=√yx=0

xy2dxdy

2. Z θ=π/2θ=0

Z r=cos(θ)r=0

esin(θ)drdθ

In problems 3-6, sketch the region D and evaluate the given double integral:3. Z Z

D

4x3ydA,

where D is the region bounded by the graphs of y = x2 and y = 2x.

4. Z ZD

y√xdA,

where D is the region bounded by the graphs of y = x2, y = 0, x = 0 and x = 4.

5. Z ZD

y2exydA,

where D is the region bounded by the graphs of y = x, y = 1 and x = 0.

6. Z ZD

y2 sin¡x2¢dA,


where D is the region bounded by the graphs of y = −x1/3, y = x1/3, x = 0 and x = √π.In problems 7-9 determine the volume of the region D in R3:

7. D is bounded by the paraboloid z = x2+y2+1, the xy-plane, the surfaces y = x2 and y = 1.

8. D is bounded by the graphs of z = ex, z = −ey, x = 0, x = 1, y = 0 and y = 1.9. D is the tetrahedron bounded by the plane 3x+ 2y + z = 6 and the coordinate planes.

In problems 10-12,a) Sketch the region of integration,b) Evaluate the integral by reversing the order of integration:

10. Z y=1y=0

Z 3x=3y

ex2

dxdy

11. Z y=√π/2y=0

Z x=√π/2x=y

cos¡x2¢dxdy.

12. Z y=1y=0

Z π/2x=arcsin(y)

cos (x)p1 + cos2 (x)dxdy.

13.3 Double Integrals in Polar Coordinates

In some cases it is convenient to evaluate a double integral by converting it to an integral inpolar coordinates.Assume that D is the set of points in the plane with polar coordinates (r, θ) such that α ≤ θ ≤ βand a ≤ r ≤ b,and we would like to approximateZ Z

D

f (x, y) dA

by a sum that will be constructed as follows: Let

α = θ0 < θ1 < θ2 < · · · < θj−1 < θj < · · · < θm = β,

be a partition of the interval [α,β], and let

a = r0 < r1 < r2 < · · · < rk−1 < rk < · · · < rn = b

be a partition of [a, b]. We set

∆θj = θj − θj−1 and ∆rk = rk − rk−1.

The rays θ = θj and the arcs r = rk form a grid that covers D, as illustrated in Figure 1.

194 CHAPTER 14. VECTOR ANALYSIS

14.2 Line Integrals

The Integral of a Scalar Function with respect to arc length

Let C be a curve in the plane that is parametrized by the path σ : [a, b] → R2. Assumethat f is a real-valued function of two variables. You may imagine that C is a wire, andf (σ (t)) = f (x (t) , y (t)) is the lineal density of the wire. How can we calculate the mass of thewire? It is reasonable to approximate mass by sums of the form

nXk=1

f (σ (tk))∆sk =nXk=1

f (x (tk) , y (tk))∆sk,

where {t0, t1, . . . , tn−1, tn} is a partition of [a, b] and ∆sk is the length of the part of C corre-sponding to the interval [tk−1, tk]. Thus

∆sk =

Z tktk−1

||σ0 (t)|| dt.

By the Mean Value Theorem for integrals,Z tktk−1

||σ0 (t)|| dt = σ0 (t∗k) (tk − tk−1) = σ0 (t∗k)∆tk,

where t∗k is some point between tk−1 and tk. Therefore,

nXk=1

f (σ (tk))∆sk =nXk=1

f (σ (tk)) ||σ0 (t∗k)||∆tk

This is almost a Riemann sum for the integralZ ba

f (σ (t)) ||σ0 (t)|| dt

and approximates the integral with desired accuracy provided that f and σ0 (t) are continuous.This leads to the following definition:

Definition 1 The integral of the scalar function f with respect to arc length on thecurve C that is parametrized by the function σ : [a, b]→ R2 is defined asZ b

a

f (σ (t)) ||σ0 (t)|| dt

Symbolically,

ds =ds

dtdt =

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dt,

so that we can express the integral as ZC

fds.

In terms of the coordinate functions x (t) and y (t) of σ (t),

ZC

fds =

Z ba

f (x (t) , y (t))

sµdx

dt

¶2+

µdy

dt

¶2dt.

14.2. LINE INTEGRALS 195

Example 1 Evaluate ZC

x2y2ds,

where C is the semicircle x2 + y2 = 4 that is parametrized so that C is traversed from (2, 0) to(−2, 0).Solution

We can setσ (t) = (2 cos (t) , 2 sin (t)) , 0 ≤ t ≤ π.

Then σ parametrizes C as required.

�2 �1 1 2x

1

2y

Figure 1

Therefore,

σ0 (t) = −2 sin (t) i+ 2 cos (t) j⇒ ||σ0 (t)|| =q4 sin2 (t) + 4 cos2 (t) = 2.

Thus, ZC

x2y2ds =

Z π0

¡4 cos2 (t)

¢ ¡4 sin2 (t)

¢ ||σ0 (t)|| dt = 16Z π0

cos2 (t) sin2 (t) (2) dt

= 32

Z π0

cos2 (t) sin2 (t) dt.

As we saw in Section 7.3,Zcos2 (t) sin2 (t) dt =

1

8x− 1

8sin (x) cos3 (x) +

1

8sin3 (x) cos (x) .

Therefore,

32

Z π0

cos2 (t) sin2 (t) dt = 4x− 4 sin (x) cos3 (x) + 4 sin3 (x) cos (x) .¯̄π0= 4π

¤A curve C can be parametrized by different functions and may be traversed from point A topoint B or vice versa. Thus, there is a notion of "the orientation" of a curve. Assume thatC is a curve in the plane that is parametrized initially by σ : [a, b] → R2 and σ (t) traces Cfrom "the initial point" P1 to "the terminal point" P2 as t increases from a to b. Wewill say that σ̃ : [α,β] → R2 is an orientation preserving parametrization of C if σ̃ (τ)traces C from the initial point P1 to the terminal point P2 as the new parameter τ increasesfrom α to β. We will say that σ̃ : [α,β]→ R2 is an orientation reversing parametrizationof C if σ̃ (τ) traces C from the terminal point P2 to the iniital point P1 as the new parameterτ increases from α to β. The positive orientation of C is the orientation provided by the


initial parametrization σ and the negative orientation of C is the orientation provided byany orientation reversing parametrization.You can find the precise defintions at the end of thissection. The line integral of a scalar function is does not change if the curve is parametrized byany orientation preserving or orientation reversing parametrization:

Proposition 1 If C is parametrized by σ : [a, b] → R2 and σ̃ : [α,β] → R2 is an orientationpreserving or orientation reversing parametrization of C, thenZ

C

fds =

Z t=bt=a

f (σ (t))

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dt =

Z τ=βτ=α

f (σ̃ (τ))

¯̄̄̄¯̄̄̄dσ̃

dτ

¯̄̄̄¯̄̄̄dτ

You can find the proof of Proposition 1 at the end of this section.

Example 2 Let P1 = (1, 2) and P2 = (2, 4), and let C be the directed line segment P1P2. Wecan parametrize C by

σ (t) = OP1 + tP1P2 = (1, 2) + t (1, 2) = (1 + t, 2 + 2t) , 0 ≤ t ≤ 1.

If we set

σ̃ (τ) = OP1 + 2τP1P2 = (1, 2) + 2τ (1, 2) = (1 + 2τ , 2 + 4τ) , 0 ≤ τ ≤ 12,

then σ is an orientation preserving parametrization of C.If we set

σ∗ (μ) = OP2 + μτP2P1 = (2, 4) + μ (−1,−2) = (2− μ, 4− 2μ) , 0 ≤ μ ≤ 1,

then σ∗ is an orientation reversing parametrization of C. Confirm that the statements of Propo-sition 1 are valid in the special case Z

C

sin³π2(x− y)

´ds

Solution

We havedσ

dt= (1, 2) so that

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄=√1 + 4 =

√5.

Therefore ZC

sin³π2(x− y)

´ds =

Z 10

sin³π2((1 + t)− (2 + 2t))

´ ¯̄̄̄¯̄̄̄dσdt

¯̄̄̄¯̄̄̄dt

=

Z 10

sin³π2(−t− 1)

´√5dt

=√5

Z 10

− sin³π2(t+ 1)

´dt.

We setu =

π

2(t+ 1)⇒ du = π

2dt.

Thus,

√5

Z 10

− sin³π2(t+ 1)

´dt =

2√5

π

Z ππ/2

− sin (u) du = 2√5

π

³cos (u)|u=πu=π/2

´= −2

√5

π.


Now let’s use the orientation preserving parametrization σ. We have

dσ

dτ= (2, 4) so that

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄=√4 + 16 =

√20 = 2

√5.

Therefore ZC

sin³π2(x− y)

´ds =

Z 1/20

sin³π2((1 + 2τ)− (2 + 4τ))

´ ¯̄̄̄¯̄̄̄dσdτ

¯̄̄̄¯̄̄̄dτ

=

Z 1/20

sin³π2(−2τ − 1)

´2√5dτ

= −2√5

Z 1/20

sin³π2(2τ + 1)

´dτ .

We set

u =π

2(2τ + 1)⇒ du = πdτ .

Thus

2√5

Z 1/20

− sin³π2(2τ + 1)

´dτ =

2√5

π

Z ππ/2

− sin (u) du = −2√5

π,

as in the case of the parametrization by σ.

σ∗ (μ) = OP2 + μτP2P1 = (2, 4) + μ (−1,−2) = (2− μ, 4− 2μ) , 0 ≤ μ ≤ 1,

Now let’s use the orientation reversing parametrization σ∗. We have

dσ∗

dμ= (−1,−2) so that

¯̄̄̄¯̄̄̄dσ∗

dμ

¯̄̄̄¯̄̄̄=√5.

ThereforeZ 10

sin³π2((2− μ)− (4− 2μ))

´ ¯̄̄̄¯̄̄̄dσ∗dμ

¯̄̄̄¯̄̄̄dμ =

Z 10

sin³π2(μ− 2)

´√5dμ

=√5

Z 10

− sin³π2(2− μ)

´dμ.

We set

u =π

2(2− μ)⇒ du = −π

2dμ.

Thus√5

Z 10

− sin³π2(2− μ)

´dμ =

2√5

π

Z π/2π

sin (u) du =2√5

π.

Therefore Z 10

sin (σ∗ (μ))¯̄̄̄¯̄̄̄dσ∗

dμ

¯̄̄̄¯̄̄̄dμ = −

ZC

fds,

as predicted by Proposition 1


The Line Integral of a Vector Field in the Plane

Assume that F (x, y) = M (x, y) i + N (x, y) j is a two-dimensional force field, σ : [a, b] → R2,and a particle is at σ (t) = (x (t) , y (t)) at time t. Let C be the curve that is parametrized by σ.How should we calculate the work done by the force F as the particle moves from (x (a) , y (a))to (x (b) , y (b)) along C? Our starting point is the definition of the work done by a constantforce F on a particle that moves along a line. If the displacement of the particle is representedby the vector w, the work done is

F ·w = ||F|| ||w|| cos (θ) ,

where θ is the angle between F and w. Let us subdivide the interval [a, b] into subintervals bya partition

P = {a = t0 < t1 < · · · < tk−1 < tk < · · · < tn−1 < tn = b} .As usual we set ∆tk = tk − tk−1.Let ∆σk = σ (tk)−σ (tk−1). If the norm of the partition P issmall, we can approximate the work done by F as the particle moves from σ (tk−1) to σ (tk) by

F (σ (tk)) ·∆σk = F (σ (tk)) · (σ (tk)− σ (tk−1)) ∼= F (σ (tk)) · σ (tk)− σ (tk−1)∆tk

∆tk

∼= F (σ (tk)) · dσdt(tk)∆tk

Therefore, the total work done is approximately

nXk=1

F (σ (tk)) · dσdt(tk)∆tk

This is a Riemann sum for the integralZ ba

F (σ (t)) · dσdtdt.

We will define the work done by the force F on the particle as it moves along thecurve C that is parametrized by σ by this integral. In more general terms, this is a line integral:

Definition 2 Assume that the curve C in the plane is parametrized by σ : [a, b] → R2 and Fis a continuous vector field in the plane. The line integral of F on C isZ

C

F·dσ =Z ba


The notationRCF·dσ is appropriate since the line integral is approximated by sums of the form

nXk=1

F (σ (tk)) ·∆σk.

Symbolically,

dσ =dσ

dtdt.


Σ�t�Σ'�t�

F�Σ�t��

Figure 2

Example 3 LetF (x, y) = 2xi+yj and assume that C is parametrized by σ (t) = (cosh (t) , sinh (t)) ,where 0 ≤ t ≤ 3. Calculate Z

C

F·dσ

Solution

Figure 3 shows the curve C.

1 3 6 9x

2

4

6

8

y

Figure 3

We havedσ

dt= sinh (t) i+ cosh (t) j.

Therefore, ZC

F·dσ =Z 30

(2 cosh (t) i+ sinh (t) j) · (sinh (t) i+ cosh (t) j) dt

=

Z 30

(2 cosh (t) sinh (t) + sinh (t) cosh (t)) dt

=

Z 30

3 cosh (t) sinh (t) dt.

If we set u = cosh (t) then du = sinh (t) dt, so thatZ 30

3 cosh (t) sinh (t) dt. = 3

Z cosh(3)cosh(0)

udu = 3

Ãu2

2

¯̄̄̄cosh(3)1

!=3

2

¡cosh2 (3)− 1¢

¤


We stated that the integral of a scalar function with respect to arc length does not depend onan orientation preserving or orientation reversing parametrization of C. In the case of the lineintegral of a vector field, an orientation reversing reparametrization introduces a minus sign:

Proposition 2 Assume that the curve C is parametrized by σ : [a, b]→ R2. If σ̃ : [α,β]→ R2is an orientation preserving parametrization of C, thenZ β

α

F (σ̃ (τ)) · dσ̃ (τ)dτ

dτ =

Z ba


If σ̃ : [α,β]→ R2 is an orientation reversing parametrization of C, then

Z βα

F (σ̃ (τ)) · dσ̃ (τ)dτ

dτ = −Z ba

F (σ (t)) · dσdtdt

You can find the proof of Proposition 2 at the end of this section.

Remark Assume that the curve C is parametrized by σ : [a, b]→ R2. and that σ̃ : [α,β]→ R2is an orientation reversing parametrization of C. We will setZ

−CF · d σ =

Z βα

F (σ̃ (τ)) · dσ̃ (τ)dτ

dτ .

By Proposition 2, Z−CF · dσ = −

Z ba

F (σ (t)) · dσdtdt = −

ZC

F · dσ.

If C is a piecewise smooth curve that can be expressed as a sequence of smooth curves C1, C2, . . . , Cm,we will write

C = C1 + C2 + · · ·+ Cm.With this understanding, we setZ

C

F·dσ =ZC1+C2+···+Cm

F·dσ =ZC1

F·dσ +ZC2

F·dσ + · · ·+ZCm

F·dσ

♦

Example 4 LetF (x, y) = −yi+ xj.

Evaluate ZC1+C2

F · dσ,

where C1 is the line segment from (−1, 0) to (1, 0) on the x-axis, and C2 is the semicircle ofradius 1 centered at the origin traversed in the counterclockwise direction

Solution

�1 1x

1y

C1

C2

Figure 4


We can parametrize the line segment by σ1 (t) = (t, 0), where t ∈ [−1, 1]. The semicircle can beparametrized by σ2 (t) = (cos (t) , sin (t)), where t ∈ [0,π]. Therefore,Z

C1

F · d σ =Z 1−1F (σ (t)) · dσ

dtdt =

Z 1−1tj · idt = 0,

andZC2

F · dσ =Z π0

F (σ (t)) · dσdtdt =

Z π0

(− sin (t) i+ cos (t) j) · (− sin (t) i+ cos (t) j) dt

=

Z π0

¡sin2 (t) + cos2 (t)

¢dt =

Z π0

1dt = π.

Thus, ZC1+C2

F · d σ =ZC1

F ·dσ +ZC2

F·dσ = π.

¤

The Line Integral as an Integral with respect to Arc Length

We can express the line integral of F on the curve C as the integral of the tangentialcomponent of F with respect to arc length (Recall that T (t) denotes the unit tangentto C at σ (t)):

ZC

F · dσ =Z ba


Z ba

F (σ (t)) ·dσ

dt¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄ ¯̄̄̄¯̄̄̄dσdt

¯̄̄̄¯̄̄̄dt =

Z ba

F (σ (t)) ·T (t)¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dt

=

Z ba

F (σ (t)) ·T (t) dsdtdt

=

ZC

F ·Tds.

Thus, we have the following fact:

Proposition 3 ZC

F · d σ =ZC

F ·Tds.

Example 5 LetF (x, y) = −yi+ xj

and let C be the circle of radius 2 that is centered at the origin and is traversed counterclockwise.Evaluate the line integral of F on C as an integral with respect to arc length.

Solution

We can parametrize C via the function σ (t) = (2 cos (t) , 2 sin (t)), where 0 ≤ t ≤ 2π. We haveσ0 (t) = −2 sin (t) i+ 2 cos (t) j,

so thatF (σ (t)) = F (2 cos (t) , 2 sin (t)) = −2 sin (t) i+ 2 cos (t) j,


ds = ||σ0 (t)|| dt =q4 sin2 (t) + 4 cos2 (t)dt = 2dt,

and

T (t) =σ0 (t)||σ0 (t)|| = − sin (t) i+ cos (t) j.

Therefore,ZC

F · d σ =ZC

F ·Tds =Z 2π0

F (σ (t)) ·T (t) ds

=

Z 2π0

(−2 sin (t) i+ 2 cos (t) j) · (− sin (t) i+ cos (t) j) 2dt

=

Z 2π0

¡4 sin2 (t) + 4 cos2 (t)

¢dt

=

Z 2π0

4dt = 8π.

¤

The Differential Form Notation

There is still another useful way to express a line integral. If F (x, y) = M (x, y) i + N (x, y) jand the curve C is parametrized by σ (t) = (x (t) , y (t)) , where a ≤ t ≤ b,Z

C

F · dσ =Z ba

(M (x (t) , y (t)) i+N (x (t) , y (t)) j) ·µdx

dti+

dy

dtj

¶dt

=

Z ba

µM (x (t) , y (t))

dx

dt+N (x (t) , y (t))

dy

dt

¶dt

=

Z ba

M (x (t) , y (t))dx

dtdt+

Z ba

N (x (t) , y (t))dy

dtdt.

Using the formalism,

dx =dx

dtdt and dy =

dy

dtdt,

we can express the integrals asZ ba

M (x (t) , y (t)) dx+

Z ba

N (x (t) , y (t)) dy.

This motivates the notation ZC

Mdx+Ndy

for the line integral of F =M i+N j on the curve C.

Definition 3 If M and N are functions of x and y, and C is a smooth curve in the plane thatis parametrized by σ : [a, b]→ R2 we setZ

C

Mdx+Ndy =

Z ba

M (x (t) , y (t))dx

dtdt+

Z ba

N (x (t) , y (t))dy

dtdt.


Thus, ZC

F · d σ =ZC

Mdx+Ndy

if F =M i+N j. The expression Mdx+Ndy is referred to as a differential form, and we willrefer to the above expression for a line integral as the integral of a differential form on thecurve C.

Example 6 Evaluate the line integral of F (x, y) = y3i+x2j on the ellipse that is parametrizedby

σ (t) = (4 cos (t) , sin (t)) , 0 ≤ t ≤ 2π,by expressing the line integral as the integral of a differential form.

4-4

-1

1

x

y

Figure 5

Solution

ZC

F·dσ =ZC

y3dx+ x2dy =

Z 2π0

µsin3 (t)

dx

dt+ 16 cos2 (t)

dy

dt

¶dt

=

Z 2π0

¡sin3 (t) (−4 sin (x)) + 16 cos2 (t) (cos (t))¢ dt

=

Z 2π0

¡−4 sin4 (t) + 16 cos3 (t)¢ dt = −3π(Check: You will have to refresh your memory with respect to the integrals of powers of sinesand cosines as we discussed in Section 7.3). ¤

Example 7 Let C be the boundary of the square [−1, 1] × [−1, 1] that is traversed in thecounterclockwise direction. Calculate Z

C

−ydx+ xdy.

1-1

1

-1

x

y

Figure 6


Solution

We can express C as C1 + C2 + C3 + C4,as indicated in Figure 4.We can parametrize C1 by setting x (t) = t and y (t) = −1, where −1 ≤ t ≤ 1. Then,Z

C1

−ydx+ xdy =Z 1−1

µdx

dt+ tdy

dt

¶dt =

Z 1−1dt = 2

We can express C3 as −C3, where −C3 is parametrized by setting x (t) = t and y (t) = 1, where−1 ≤ t ≤ 1. Then,Z

C3

−ydx+ xdy = −Z−C3−ydx+ xdy = −

Z 1−1

µ−dxdt+ tdy

dt

¶dt =

Z 1−1dt = 2

We can parametrize C2 by setting x (t) = 1 and y (t) = t, where −1 ≤ t ≤ 1. Then,ZC2

−ydx+ xdy =Z 1−1

µ−tdxdt+dy

dt

¶dt =

Z 1−1dt = 2

Similarly, C4 = −C2, where −C2 is parametrized by x (t) = −1 and y (t) = t, −1 ≤ t ≤ 1.Then, Z

C4

−ydx+ xdy = −Z−C2−ydx+ xdy = −

Z 1−1

µ−dydt

¶dt =

Z 1−1dt = 2

Therefore, ZC

−ydx+ xdy. =4Xk=1

ZCk

−ydx+ xdy = 2 + 2 + 2 + 2 = 8.

¤

Curves in R3

Our discussion of integrals with respect to arc length and line integrals extends to curves inthree dimensions. Thus, assume that σ : [a, b]→ R3 is a smooth function and parametrizes thecurve C. If f is a continuous scalar function of three variables, we define the integral of f withrespect to arc length on C as Z

C

fds =

Z ba

f (σ (t)) ||σ0 (t)|| dt.

If σ (t) = (x (t) , y (t) , z (t)),ZC

fds =

Z ba

f (x (t) , y (t) , z (t))

sµdx

dt

¶2+

µdy

dt

¶2+

µdz

dt

¶2dt.

As in the case of curves in the plane, the above integral is the same for orientation preservingand orientation reversing parametrizations of C.If F is a vector field inR3 so that

F(x, y, z) =M (x, y, z) i+N (x, y, z) j+ P (x, y, z)k,

the line integral of F on the curve C isZC

F·dσ =Z ba



In the differential form notation,ZC

F·dσ =ZC

Mdx+Ndy + Pdz

=

Z ba

µM (x (t)) , y (t) , z (t)

dx

dt+N (x (t)) , y (t) , z (t)

dy

dt+ P (x (t)) , y (t) , z (t)

dz

dt

¶dt.

The line integral can be expressed as an integral with respect to arc length:ZC

F·d σ =ZC

F ·Tds,

where T denotes the unit tangent to the curve C. As in the case of two-dimensional vectorfields, Z

C

F·dσ

is unchanged by an orientation preserving parametrization of C and multiplied by −1 if theparametrization reverses the orientation.

Example 8 Let C be the helix that is parametrized by

σ (t) = (cos (t) , sin (t) , t) , 0 ≤ t ≤ 4π,

and let F (x, y, z) = −yi+ xj+zk. Evaluatea) Z

C

zds,

b) ZC

F · dσ

Solution

The picture shows the helix.

z

yx

Figure 7

a) We haveσ0 (t) = − sin (t) i+ cos (t) j+ k

Therefore,

ds = ||σ0 (t)|| =qsin2 (t) + cos2 (t) + 1 =

√1 + 1 =

√2.


Thus, ZC

zds =

Z 4π0

t ||σ0 (t)|| dt =Z 4π0

t√2dt =

√2

Ãt2

2

¯̄̄̄4π0

!=√2

µ16π2

2

¶= 8√2π2

b) ZC

F · dσ =Z 4π0

(− sin (t) i+ cos (t) j+ tk) · (− sin (t) i+ cos (t) j+ k) dt

=

Z 4π0

¡sin2 (t) + cos2 (t) + t

¢dt

=

Z 4π0

(1 + t) dt = t+t2

2

¯̄̄̄4π0

= 4π +16π2

2= 4π + 8π2.

¤

Precise Definitions and Proofs

Let’s begin by stating the precise definitions of orientation preserving and orientation reversingparametrizations of a curve:

Definition 4 Assume that C is a curve in the plane that is parametrized by σ : [a, b] → R2.We say that σ̃ : [α,β] → R2 is an orientation preserving parametrization of C if thereexists a differentiable increasing function h : [α,β] → [a, b] such that σ̃ (τ) = σ (h (τ)) foreach τ ∈ [α,β]. The function σ̃ is an orientation reversing parametrization of C if thereexists a differentiable decreasing function h : [α,β]→ [a, b] such that σ̃ (τ) = σ (h (τ)) for eachτ ∈ [α,β].Now let’s prove Proposition 1:

Assume that C is a curve in the plane that is parametrized by σ : [a, b] → R2 and thatσ̃ : [α,β]→ R2 is an orientation preserving or reversing parametrization of C. ThenZ b

a

f (σ (t))

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dt =

Z βα

f (σ̃ (τ))

¯̄̄̄¯̄̄̄dσ̃

dt

¯̄̄̄¯̄̄̄dτ .

Proof

Assume that σ̃ (τ) = σ (h (τ)), where h is an increasing or decreasing and differentiable functionfrom [α;β] onto [a, b]. By the chain rule,

dσ

dτ=dh

dτ

dσ

dt(h (τ)) .

Therefore, ¯̄̄̄¯̄̄̄dσ

dτ

¯̄̄̄¯̄̄̄=

¯̄̄̄dh

dτ

¯̄̄̄ ¯̄̄̄¯̄̄̄dσ

dt(h (τ))

¯̄̄̄¯̄̄̄Thus, Z β

α

f (σ (h (τ)))

¯̄̄̄¯̄̄̄dσ

dτ

¯̄̄̄¯̄̄̄dτ =

Z βα

f (σ (h (τ)))

¯̄̄̄dh

dτ

¯̄̄̄ ¯̄̄̄¯̄̄̄dσ

dt(h (t))

¯̄̄̄¯̄̄̄dτ

If dh/dτ > 0, Z βα

f (σ (h (τ)))

¯̄̄̄dh

dτ

¯̄̄̄ ¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dτ =

Z βα

f (σ (h (τ)))dh

dτ

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dτ .


By the substitution rule,

Z βα

f (σ (h (τ)))dh

dτ

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dτ =

Z βα

f (σ (h (τ)))

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dh

dτdτ =

Z h(β)h(α)

f (t)

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dt

=

Z ba

f (t)

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dt.

Thus, Z βα

f (σ (h (τ)))

¯̄̄̄¯̄̄̄dσ

dτ

¯̄̄̄¯̄̄̄dτ =

Z ba

f (t)

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dt,

as claimed.

Similarly, if dh/dτ < 0,

Z βα

f (σ (h (τ)))

¯̄̄̄¯̄̄̄dσ

dτ

¯̄̄̄¯̄̄̄dτ =

Z βα

f (σ (h (τ)))

¯̄̄̄dh

dτ

¯̄̄̄ ¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dτ

= −Z βα

f (σ (h (τ)))dh

dτ

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dτ

=

Z αβ

f (σ (h (τ)))dh

dτ

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dτ

=

Z h(α)h(β)

f (σ (t))

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dt =

Z ba

f (σ (t))

¯̄̄̄¯̄̄̄dσ

dt

¯̄̄̄¯̄̄̄dt.

¥

Now we will prove Proposition 2 with regard to the effect of reparametrization on the lineintegral of a vector field. Recall the statement:

Assume that the curve C is parametrized by σ : [a, b]→ R2. If σ̃ : [α,β]→ R2 is an orientationpreserving parametrization of C, then

Z βα

F (σ̃ (τ)) · dσ̃ (τ)dτ

dτ =

Z ba


If σ̃ : [α,β]→ R2 is an orientation reversing parametrization of C, then

Z βα

F (σ̃ (τ)) · dσ̃ (τ)dτ

dτ = −Z ba


Proof

a) By the chain rule,

dσ (h (τ))

dτ=dh

dτ

dσ

dt(h (τ)) .


Therefore, Z βα

F (σ̃ (τ)) · dσ̃ (τ)dτ

dτ =

Z βα

F (σ (h (τ))) · dσ (h (τ))dτ

dτ

=

Z βα

F (σ (h (τ))) ·µdh

dτ

dσ

dt(h (τ))

¶dτ

=

Z βα

F (σ (h (τ))) · dσdt(h (τ))

dh

dτdτ

=

Z h(β)h(α)


=

Z ba

F (σ (t)) · dσdtdt,

by the substitution rule.b) As in part a)Z β

α

F (σ̃ (τ)) · dσ̃ (τ)dτ

dτ =

Z βα

F (σ (h (τ))) · dσdt(h (τ))

dh

dτdτ

=

Z h(β)h(α)


Since h (τ) is a decreasing function, we have h (α) = b and h (β) = a. Therefore,Z h(β)h(α)


Z ab

F (σ (t)) · dσdtdt = −

Z ba

F (σ (t)) · dσdtdt,

as claimed. ¥

Problems

In problems 1-5 evaluate the given integral of a scalar function with respect to arc length:

1. ZC

y3ds,

where C is parametrized byσ (t) =

¡t3, t

¢, 0 ≤ t ≤ 2.

2. ZC

xy2ds,

where C is parametrized by

σ (t) = (2 cos (t) , 2 sin (t)) , −π2≤ t ≤ π

2.

3. ZC

xeyds,

where C is the line segment traversed from (2, 1) to (4, 5) .

4. ZC

y

x2 + y2ds,


where C is the semicircle of radius 2 that is centered at (4, 3) and traversed from (6, 3) to (2, 3)in the counterclockwise direction.

5. ZC

(x− 2) e(y−3)(z−4)ds,

where C is the line segment from (2, 3, 4) to (3, 5, 7).

In problems 6-10 evaluate the line integral ZC

F·dσ

6.F (x, y) = x2i+ y2j

and C is the line segment from (1, 2) to (3, 4).

7.F (x, y) = yi− xj

and C is the part of the unit circle traversed from (0, 1) to (−1, 0) in the counterclockwisedirection.

8.F (x, y) = ln (y) i− exj

and C is parametrized byσ (t) =

¡ln (t) , t3

¢, 1 ≤ t ≤ e.

9.F (x, y, z) = cos (y) i+ sin (z) j+ xk

and C is parametrized by

σ (t) = (cos (t) , t, t) ,π

4≤ t ≤ π

2

10.F (x, y) = xyi+ (x− y) j

and C = C1 + C2, where C1 is the line segment from (0, 0) to (2, 0) and C2 is the line segmentfrom (2, 0) to (3, 2).

In problems 11 and 12 evaluate ZC

F ·Tds

11.F (x, y) = −2xyi+ (y + 1) j

and C is the part of the circle of radius 2 centered at the origin traversed from (−2, 0) to (0, 2)in the clockwise direction.

12.F (x, y, z) = ex+yi+ xzj+ yk

and C is the line segment from (1, 2, 3) to (−1,−2,−3) .

In problems 13 and 14


a) Express the line integral ZC

F·dσ

in the differential form notation,b) Evaluate the expression that you obtained in part a).

13.

F (x, y) = −xyi+ 1x2 + 1

j

and C is parametrized by

σ (t) =¡t, t2

¢, −4 ≤ t ≤ −1

14.F (x, y) = yi− xj

and C is the part of the unit circle that is traversed from (0,−1) to (0, 1) in the counterclockwisedirection.

In problems 15-18 evaluate the given line integral:

15. ZC

3x2dx− 2y3dy

where C is the part of the unit circle traversed from (1, 0) to (0, 1).

16. ZC

exdx+ eydy,

where C is the part of the ellipse x2 + 4y2 = 4 traversed from (0, 1) to (2, 0) in the clockwisedirection.Hint: Parametrize C by a function of the form (a cos (θ) , b sin (θ)).

17. ZC

− sin (x) dx+ cos (x) dy,

where C is the part of the parabola y = x2 traversed from (0, 0) to¡π,π2

¢.

18. ZC

x3dx+ y2dy + zdz,

where C is the line segment from the origin to (2, 3, 4).

14.3 Line Integrals of Conservative Vector Fields

In this section we will see that the line integral of the gradient of a scalar function on a curveis simply the difference between the values of the function at the endpoints of the curve. Thus,such a line integral does not depend on the particular path that connects two given points. Wewill discuss conditions under which a vector field is the gradient of a scalar function.

14.5. SURFACE INTEGRALS 239

-5

0

5

z

5

y0

2

x-5 1

0

8.Ω (x, θ) = (x cos (θ) , x sin (θ) , ex) ,

where 0 ≤ x ≤ 2 and 0 ≤ θ ≤ 2π,P = Ω (1,π/2)

Note that S is part of the surface of revolution that is obtained by revolving the graph of z = ex

about the z-axis.

2

y0

2

-2

4z

x

6

0-22

14.5 Surface IntegralsIn this section we will compute the area of a parametrized surface and define the integrals ofscalar functions and vector fields on such a surface. We will consider only orientable surfaces,as we discussed in Section 14.4.

Surface Area

We will motivate the definition of the area of a surface. Assume that the surface M is parame-trized by the function Φ : D ⊂ R2 → R3, such that

Φ(u, v) = (x (u, v) , y (u, v) , z (u, v)) , (u, v) ∈ D.

We assume that Φ is smooth, i.e., the tangent vectors

∂Φ

∂u(u, v) =

∂x

∂u(u, v) i+

∂y

∂u(u, v) j+

∂z

∂u(u, v)k

and∂Φ

∂v(u, v) =

∂x

∂v(u, v) i+

∂y

∂v(u, v) j+

∂z

∂v(u, v)k

are continuous on D, and the normal vector

N (u, v) =∂Φ

∂u(u, v)× ∂Φ

∂v(u, v) 6= 0


for each (u, v) ∈ D. Let (u, v) be a point in the interior of D and assume that ∆u and ∆v arepositive numbers that are small enough so that the rectangle D (u, v,∆u,∆v) determined bythe points (u, v) and (u+∆u, v +∆v) is contained in the interior of D. The vectors

∆u�

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