MATH 252-01: Multivariable Calculus II (82809)
JB-387, MW 4 PM - 5:50 PM
SYLLABUS Fall 2014
John Sarli
JB-326
[email protected]
909-537-5374
O¢ ce Hours: MW 11AM - 1PM, or by appt.
Text: Marsden/Tromba (W.H. Freeman and Company)
Vector Calculus (sixth edition)
Prerequisites: MATH 213 with a grade of "C" or better, and MATH 251
This a second course in multivariable calculus, covering the integral calculus of
vector functions. We will use the geometry of space as developed in di¤erential multivariable calculus to study the classical theorems of integration, which
generalize the fundamental theorem of calculus in one variable. (For review of
the matrix/vector representation of this geometry, refer to the syllabus notes for
MATH 251 on the website: www.math.csusb.edu/faculty/sarli/) This course
is a combination of basic theory and straightforward computation, primarily with
functions of two and three variables. The goal is to develop pro…ciency so that
applications to various branches of science and mathematics become accessible.
We will cover the second half of the text, Chapters 5 through 8, but the course
will follow my notes for MATH 252 on the above website. It is important that
you read both the text and my notes.
Grading will be based on two midterm exams, a cumulative …nal exam, and four
graded assignments (one from each chapter), weighted as follows: First Midterm
(15%), Second Midterm (25%), Final Exam (40%), Graded Assignments (20%).
To reinforce written communication skills the Graded Assignment solutions
should be clearly presented, either in a "bluebook" or sent to me as a PDF (do
not scan in handwritten work).
I will take attendance until the class roster has been …nalized, however there is no
attendance requirement for this class. Nonetheless, experience shows that regular
attendance and active participation signi…cantly improve performance.
A list of Suggested Exercises will be provided for each chapter. The exams
will be written at the level of these exercises though the format may not be
identical, mainly because the purpose of exams is to bring related ideas together
so that they can be used in relation to one another. I will list exercises that are
representative of a particular technique or concept, but you should attempt as
many similar exercises in the text as needed for understanding. In this way we
can avoid "practice exams" and other routines that use the time we need to cover
this material. It is your responsibility to bring questions to class that arise as you
work through exercises and notes.
After computing your total scores weighted according to the percentages above,
course grades will be assigned as follows:
A
91
A
86 90
B+
81 85
B
76 80
B
71 75
C+
66 70
C
61 65
C
51 60
D
45 50
F
< 45
Success in this course requires a balance of three activities:
1) Read the text and work the exercises regularly. Keep notes of your solutions.
If you have organized them e¢ ciently, bring them to the in-class exams.
2) Follow the lecture notes on my website and read the syllabus there. Bring
questions on these notes to class as they occur to you.
3) Participate in the class sessions as actively as you can. Lectures are more
useful to you if you use them to clarify ideas as we develop them.
Notes
2
1) I strongly suggest that you complete the CSU/UC Mathematics Diagnostic
Testing Project CR test within the …rst two weeks of the course. Flaws in fundamental skills often compromise progress at this advanced level and it is important
that you identify any such misunderstandings early. Go to mdtp.ucsd.edu and
scroll down to MDTP Web Based Tests. Select the CR test. The items will
appear one at a time and the test will be scored instantly once you submit it.
Your response to each item will be provided to you and I can help you interpret
the results. I do not record the test results and they do not a¤ect your course
grade in any way.
2) Mid-term exam dates are subject to change. Due dates for the graded exercises
will be set as we approach the end of each chapter of the text.
3) Learning Outcomes: Upon successful completion of this course, students
will be able to:
1.2 make connections between mathematical ideas verbally, numerically, analytically, visually, and graphically;
3.5 evaluate reasonableness of proposed results using estimation and context.
4) Please refer to the Academic Regulations and Policies section of your current
bulletin for information regarding add/drop procedures. Instances of academic
dishonesty will not be tolerated. Cheating on exams or plagiarism (presenting the
work of another as your own, or the use of another person’s ideas without giving
proper credit) will result in a failing grade and sanctions by the University. For
this class, all assignments are to be completed by the individual student unless
otherwise speci…ed.
5) If you are in need of an accommodation for a disability in order to participate
in this class, please let me know ASAP and also contact Services to
Students with Disabilities at UH-183, (909)537-5238.
Some important dates:
September 29: First day of class
October 1: Last day to add open classes over MyCoyote for Fall quarter
October 15: Fall CENSUS; last day to submit add/drop slips
3
October 22: First Exam
November 12: Second Exam
December 3: Last day of class
December 8: Final Exam (4-5:50)
Approximate course schedule:
Read through the text at the rate of approximately two sections per week. The
material in class will not always be presented in exactly the same order, but topics
will be covered according to the following general schedule.
First Week: Integrals over 2-dimensional regions and Cavalieri’s Principle
Second Week: Iterated integrals over 2-dimensional and 3-dimensional regions
Third Week: Fubini’s Theorem and Mean-Value Theorem on R2
October 22: First Exam
Fourth Week: Maps from R2 to R2 ; Change of variables
Fifth Week: Average values and center of mass
Sixth Week: Path integrals and line integrals
November 12: Second Exam
Seventh Week: Geometry of surfaces
Eighth Week: Surface integrals of scalar and vector functions
Ninth Week: Green’s Theorem and Stokes’Theorem
Tenth Week: Conservative …elds; Gauss’Theorem
4
Suggested Exercises for Chapter 5
The following exercises are not to be handed in. They represent skills required for
basic mastery.
5.1 (pages 269-271):
1; 2; 7
5.2 (pages 282-283):
1; 7; 11; 13; 14
5.3 (pages 288-289):
1; 3; 6; 11; 15
5.4 (pages 293-294):
2; 3; 5; 9
5.5 (pages 302-304):
1; 2; 7; 11; 13; 17; 25; 26; 27
First Graded Assignment
Due: October 20
To reinforce written communication skills the Graded Assignment solutions
should be clearly presented in a "bluebook" or provided in PDF format. Late
papers will not be graded.
First Graded Assignment. Do any one of the following:
Page 271: 12
2
2
2
Page 294: The top half of the ellipsoid xa2 + yb2 + zc2 = 1 is the graph of f (x; y) =
q
2
2
c 1 xa2 yb2 . Exercise 11 shows that the volume under the graph is 23 abc. Use
this result along with the answer to exercise 6 on page 289 to …nd the level set
for f consisting of the points (x0 ; y0 ) that satisfy Theorem 5 on page 292. Find
A(D0 )
, where D0 is the region bounded by the level set.
A(D)
Page 304: 30
Page 305: 37
5
Review of Coordinate Systems
In R2 there are two systems of coordinates that are used most often: rectangular
and polar. Rectangular coordinates are often called Cartesian coordinates - any
point is uniquely represented by an ordered pair of real numbers. Polar coordinates
are not unique - they represent a point in terms of its distance from the origin and
its angular position relative to a reference ray, usually the positive horizontal axis.
The relation between the rectangular description (x; y) and the polar description
(r; ) is given by
x = r cos
y = r sin
so there are in…nitely many choices of for any given point once we determine
r for that point. Further, we sometimes want to consider r to be the signed
2; 43 are both polar coordistance from the origin; for example, 2; 3 and
p
dinate descriptions of the point whose rectangular coordinates are 1; 3 . For
any given polar coordinate description (r; ) we obtain the rectangular description
(x; y) from the above equations. But to go in the other direction requires some
conventional choices to obtain a unique polar description. These conventions vary
by context, but the default is to take r 0 and 2 [0; 2 ). Then we can compute
p
r = x2 + y 2
and use basic trigonometry to …nd in terms of arctan xy . Even though r is now
uniquely determined, we need to be careful in computing because the range of
the inverse tangent function is
; , which only covers half of the plane. In
2 2
order to have 2 [0; 2 ) we set
y
= arctan , x > 0; y 0
x
y
= 2 + arctan , x > 0; y < 0
x
y
=
+ arctan , x < 0
x
=
, x = 0; y > 0
2
3
=
, x = 0; y < 0
2
6
The only point not covered by these conventions is the origin. Note that the polar
description of a point associates the point with a position vector with magnitude
r and direction . The origin would correspond to the zero vector, which has
magnitude 0 but no direction. Therefore, is not de…ned for the origin; we simply
write r = 0 as the polar description of the point whose rectangular coordinates
are (0; 0).
Polar coordinates are easily extended to R3 if we want to emphasize radial symmetry about the vertical axis:
x = r cos
y = r sin
z = z
The coordinates (r; ; z) are called cylindrical coordinates because the set of points
described by the equation r = a is a cylinder of radius a whose axis of symmetry
is the z-axis. Note also that the equation = is a half-plane bounded by the
z-axis if we only allow non-negative values for r, whereas it is a plane through the
z-axis if r can be a signed distance from the z-axis. The equation z = a is a plane
parallel to the xy-plane.
3
A third coordinate system commonly used
p in R , spherical coordinates, describes
2
2
points relative to their distance = x + y + z 2 from the origin. Spherical
coordinates are the mathematical formalization of latitude and longitude on a
globe of radius :
x =
y =
z =
sin cos
sin sin
cos
Here 2 [0; 2 ) has the same interpretation as in polar and cylindrical coordinates.
The angular coordinate is usually measured o¤ of the positive vertical axis,
so 2 [0; ]. If the point with rectangular coordinates (x; y; z) were projected
orthogonally into the xy-plane then its polar coordinates would be (r; ) where
r = sin
The equation
equation =
= a describes a sphere of radius a centered at the origin; the
is a half-plane bounded by the z-axis because there is generally
7
no advantage to allowing to be negative; the equation = is the branch of
a cone with vertex at the origin consisting of the points P such that the angle
!
!
between unit vector k and vector OP is . Since OP = we have
= arccos
!
k OP
because the range of the inverse cosine function is [0; ].
8
Double Integrals and Cavalieri’s Principle
We are mainly concerned with scalar and vector functions on R2 and R3 though
the methods we develop generally apply to domains of any dimension. We begin
with generalizing to R2 the idea of a continuous function on a closed interval in
R. This begins with the de…nition of a rectangle R in R2 as the product of two
intervals in R :
R = [a; b] [c; d]
f(x; y) : a x b; c
y
dg
First, suppose that f : R ! R is a non-negative function. Then the graph
of z = f (x; y) is a surface above R and we have a situation like that in singlevariable calculus of a curve above a closed interval. In that situation we motivated
the de…nition of the integral of f on the interval by the idea of area under the
curve. We want to proceed then with the idea of volume under the surface. The
actual concept of geometric measurement is very deep, but we can proceed as did
Archimedes under the assumption that our de…nitions should agree with intuitive
ideas. So, even though we do not yet have a rigorous de…nition of volume we
state:
The volume of the 3-dimensional region above R and under the
graph of f is written
ZZ
f dA
R
and is called the integral of f over R.
RR
RR
We write R f dA = R f (x; y)dxdy when we want to indicate how we will compute this integral using coordinates, as we will later. (See Examples 1 and 2 on
pages 264-265.) We need to develop a rigorous de…nition of the integral, which
will then apply to any continuous function, much as we did with Riemann sums
in single-variable calculus. First we look at volume V in more detail through a
principle that was developed by a contemporary of Galileo.
Cavalieri’s Principle. Let S be a solid and, for a
x
b let
Px be a family of parallel planes with S between Pa and Pb . If A (x)
is the area of Px \ S then
Z b
V (S) =
A(x)dx
a
9
We have stated this principle in modern form because integral calculus was being
developed around the time of Cavalieri’s death. Notice, however, that the idea
motivates techniques such as the ’shell’and ’disk’methods for …nding volumes.
In fact, the idea goes back to Euclidean geometry where the formulas for areas of
triangles and volumes of simple solids were …rst developed.
Example. A lamppost with an elliptical base is formed by translating the ellipse
vertically while rotating it about the axis perpendicular to its center. Find the
volume of the lamppost in terms of its height and the major and minor axes of
the ellipse. Does the volume depend on the rate of rotation?
Let’s see how Cavalieri’s principle can be used to …nd the volume under the graph
of a non-negative function over a rectangle. Consider the following example:
f (x; y) = 5 2x2 3y 2
R = [ 1; 1] [0; 1]
5
2x2
3y 2
z=0
5
4
3
z
2
1
-1.0
0
0.0 0.0
0.5
1.0
-0.5
0.2
0.4
0.6
x
y
z=5
2x2
10
3y 2
0.8
1.0
5
4
3
z
2
1
-1.0
0
0.0 0.0
0.5
1.0
-0.5
0.2
0.4
0.6
x
y
Px0 for x0 =
0.8
1.0
1
5
f (x; y) 0 on R and we compute the volume of the solid S under the graph with
base R. Let Px0 be the plane x = x0 , the vertical plane parallel to the yz-plane
that passes through (x; 0; 0), and let A(x0 ) be the area of Px0 \ S. Then A(x0 ) is
the area under the curve z = 5 2x20 3y 2 over the y-interval [0; 1] and so
Z 1
A(x0 ) =
f (x0 ; y) dy
0
Z 1
=
5 2x20 3y 2 dy
0
= 4
2x20
We now let x0 vary over the interval [ 1; 1] and apply Cavalieri’s principle to
obtain the volume of S :
Z 1
V =
A(x)dx
1
Z 1
=
(4 2x2 )dx
1
20
=
3
11
We should expect to get the same result if we apply Cavalieri’s principle with Py0
being the plane y = y0 for 0 y0 1 :
Z
A(y0 ) =
Z
=
1
f (x; y0 ) dx
1
1
3y02 dx
1
26
3
=
2x2
5
6y02
Then
V
=
Z
1
A(y)dy
0
=
Z
1
0
(
26
3
6y 2 )dy
20
=
3
5
4
3
z
2
1
-1.0
0
0.0 0.0
0.5
1.0
-0.5
0.2
x
0.4
0.6
y
Py0 for y0 =
12
1
2
0.8
1.0
The fact that we obtained the same result suggests that volumes under graphs
can be computed using iterated integrals: integrate over a rectangle by integrating
with respect to one variable holding the other constant and then integrating the
resulting integrand with respect to the second variable. Our result also suggests
that the order we choose should not matter, at least where the calculation of
volumes is involved. It remains to be seen whether this is always true once we
de…ne the de…nite double integral precisely.
This de…nition follows the same procedure that was used for single-variable functions on a closed interval [a; b]. Now we assume that f : R2 ! R is continuous
on a closed rectangle R = [a; b] [c; d]. A continuous function on a closed set is
bounded, so we can partition R into sub-rectangles by dividing the intervals [a; b]
and [c; d] each into n equal sub-intervals:
1
(b
n
1
y =
(d
n
x =
A =
a)
c)
x y=
1
(b
n2
We can index the n2 sub-rectangles by Rjk for 0
cjk in each Rjk . The sum
n 1
X
f (cjk ) A
a) (d
c)
j; k
n
1 and choose a point
j;k=0
is a Riemann sum for f on R. As n ! 1 the sequence of Riemann sums will
converge to a limit, because f is continuous, and that limit will not depend on
how the points cjk are chosen at each stage. We say that f is integrable over R.
As long as f is bounded on R we can still construct Riemann sums and f might
be integrable. We will mostly deal with functions that are continuous, a su¢ cient
but not necessary condition for integrability: See Theorem 2 on page 274. We can
now use the notation
ZZ
f dA
R
for any function integrable over R and, if f (x; y)
0 we de…ne the volume of
the solid between R and the graph z = f (x; y) to be this integral. The usual
properties of the integral hold (see page 275), from which we obtain:
13
Theorem. If f is integrable over R then the volume under the
graph of jf j is greater than or equal to the absolute value of the integral
of f .
Proof. Since
jf j f jf j
the properties of the integral imply
ZZ
ZZ
jf j dA
f dA
R
i.e.,
RR
R
f dA
R
RR
R
ZZ
R
jf j dA
jf j dA.
The integrability condition for a bounded function on R (see the theorem on page
274 again) can be stated as follows:
( ) The set of points where f is discontinuous is on a …nite union
of graphs of continuous functions.
If ( ) holds we can usually compute the integral of f asan iterated integral in
either order:
Rb
Fubini’s Theorem. If a f (x; y)dx exists for each y 2 [c; d] then
RR
RdRb
Rd
f dA = c a f (x; y)dxdy. Similarly, if c f (x; y)dy exists for each
R
RR
RbRd
x 2 [a; b] then R f dA = a c f (x; y)dydx.
RR
Corollary. If f is continuous then R f dA can be computed as
an iterated integral in either order.
The proof in the continuous case uses the Mean-Value Theorem for single-variable
integrals (see page 277).
An example where Fubini’s Theorem does not apply. (See 17 on page
283.)
Let
x2 y 2
(x2 + y 2 )2
R = [0; 1] [0; 1]
f (x; y) =
14
The domain of f does not include (0; 0) and we cannot de…ne f
at the origin to make it continuous there. (Why?) However, we can
proceed with iterated integration as follows. First we compute
Z 1
f (x; y)dy
0
by treating x as a constant and letting y = x tan . Then
Z 1
Z 1
Z 1
dy
y 2 dy
2
f (x; y)dy = x
2
2 2
2
2 2
0
0 (x + y )
0 (x + y )
1
1
y
1
y
+ arctan
2
2
2 x +y
x
x
1
=
1 + x2
=
+
0
1
y
2
2 x + y2
1
y
arctan
x
x
1
0
Now
Z
0
1
Z
1
Z
1
dx
2
0 1+x
= [arctan x]10
f (x; y)dydx =
0
=
4
However, when we reverse the order of integration we …nd
Z
0
1
Z
1
f (x; y)dxdy =
0
=
=
Z
1
dy
2
0 1+y
[arctan y]10
4
The di¤erent results do not contradict Fubini’s Theorem because the theorem
does not apply in this case: f is not bounded on R because it approaches 1 as
(x; y) ! (0; 0).
15
20
10
1.0
0.8
0.6
y
0.4
0.2
z
0
0.0 0.0
0.2
0.4
0.6
0.8
1.0
x
-10
-20
z=
x2 y 2
(x2 +y 2 )2
on R
Integrals over Regions in R2
In Fubini’s Theorem, the hypothesis restricted the set of discontinuities to a …nite
union of graphs of continuous functions. Our theory of the integral should extend
beyond rectangular regions to other closed regions whose boundaries consist of
graphs of continuous functions. We begin with the concept of an elementary
region D. A region D in R2 is elementary provided it satis…es one or both of the
following conditions:
i) D = f(x; y) : x 2 [a; b] ; 1 (x) y
2 (x)g with 1 and 2 continuous on
[a; b]
ii) D = f(x; y) : y 2 [c; d] ; 1 (y) x
2 (y)g with 1 and 2 continuous on
[c; d]
If D satis…es i) it is called y-simple, if it satis…es ii) it is called x-simple, if it
satis…es both it is called simple.
Example. The region below is elementary but not simple. It is x-simple
but not y-simple.
16
y
8
7
6
5
4
3
2
1
-1.0
-0.5
D:1
0.0
0.5
y
1.0
1 + 2 ;0
1.5
x
2.0
1 + cos (y
2.5
3.0
x
1)
We do not want to rede…ne the integral over an elementary region. Instead, we
enclose D inside a rectangle R and extend f , assumed to be continuous on D, to
the function f by de…ning f to be 0 on RnD. Unless f is 0 on @D the function
f will not be continuous on R, but it will be integrable by Fubini’s Theorem
since the set of discontinuities is …nite union of graphs of continuous functions.
For example, if D is x-simple then
Z dZ b
ZZ
f (x; y) dxdy
f dA =
c
a
R
Z d Z 2 (y)
=
f (x; y) dxdy
c
1 (y)
In the above example, we could choose R = [0; 2]
f (x; y) =
[1; 1 + 2 ] and let
y, (x; y) 2 D
0, (x; y) 2 RnD
17
Then
ZZ
f dA =
D
Z
1+2
1
=
Z
Z
1+cos(y 1)
ydxdy
0
1+2
y (cos (y
1
= 2 ( + 1)
18
1) + 1) dy
If D is a simple region and f is continuous on D then we can set up the iterated
integral in either order. It is often easier to …nd the antiderivative for the …rst
integrand by using one order instead of the other.
Example. Let D be bounded by the
the line x = 2, and the
RR x-axis,
2
graph of x = y 2 for y 0. Evaluate D yex dA:
y
2.0
1.5
1.0
0.5
0.0
0.0
0.5
1.0
1.5
D is a simple region and
Z
ZZ
x2
ye dA =
D
0
=
Z
0
p
2.0
2
2Z
Z
2
y2
p
x
2.5
2
yex dxdy
2
yex dydx
0
The …rst iterated integral is di¢ cult to evaluate because
not an elementary anti-derivative. However,
Z 2
Z 2 Z px
1 x2
x2
xe dx
ye dydx =
0
0 2
0
2
1 x2
=
e
4
0
1 4
=
e
1
4
19
3.0
x
R
2
ex dx is
Mean-Value Theorem
In single-variable calculus, if f is continuous on [a; b] then there exists c 2 [a; b]
such that
Z b
f (x)dx = f (c) (b a) .
a
Suppose that f : R2 ! R is continuous on an elementary region D. Let
A (D) be the geometric area of D, A(D) > 0. A continuous function
on a closed set has a minimum value m and a maximum value M , and
so for all (x; y) 2 D
m f (x; y) M
and
mA(D)
ZZ
f dA
M A(D)
D
Further, since f is continuous on D it assumes every value between m
and M . Thus, there exists (x0 ; y0 ) 2 D such that
ZZ
1
f (x0 ; y0 ) =
f dA
A(D) D
and so
ZZ
f dA = f (x0 ; y0 ) A (D)
D
which is analogous to the single-variable theorem.
2
In the previous example, f (x; y) = yex and A(D) =
minimum and maximum values for f on D are
m = 0
p 4
M =
2e
and we found
ZZ
f dA =
D
1 4
e
4
1
8
< M A (D) = e4
3
20
R2p
xdx =
0
4
3
p
2. The
To …nd (x0 ; y0 ) in D to verify the Mean-Value Theorem we set
1 4
4p
2
e
1 =
2y0 ex0
4
3
and obtain all such points on the following curve:
y
2.0
1.5
1.0
0.5
0.0
0.0
0.5
y0 =
3
32
p
1.0
2 (e4
1) e
1.5
x20
2.0
x
in D
Regions in R3
This de…nition of the integral generalizes to functions f : Rn ! R for any dimension n. We will go as far as n = 3 because that will be su¢ cient to derive
the so-called classical theorems. In three dimensions the simplest regions are
rectangular parallelepipeds
B = [a; b]
[c; d]
[p; q]
and Riemann sums are formed from partitions into boxes of volume
V =
1
(b
n3
a) (d
21
c) (q
p)
by selecting a point cijk from each box. As before, if f is bounded on B and the
limit of these Riemann sums exists as n ! 1 (independent of the choice of cijk )
then f is integrable and this limit is the integral of f over B
ZZZ
f dV
B
Cavalieri’s Principle generalizes to any dimension once we have a concept of higherdimensional volumes. These triple integrals can also be computed by iterated integration. Note that there are now six possible orders of integration, and there are
now corresponding versions of Fubini’s Theorem. Again, we focus on elementary
regions in R3 in order to facilitate iterated integration. These are regions such
that one of the variables is bounded between graphs of continuous functions of
the other two, then one of the remaining variables is bounded between graphs of
continuous functions of the third. For example,
1
(x; y)
1 (x)
a
z
y
x
(x; y)
2 (x)
2
b
Regions that can be described this way for any order of the variables are called
symmetric elementary.
Example. The mass of a solid is the integral of its density. Find
the mass of the solid bounded in the …rst octant by the coordinate
planes, the plane z = 1, and the surface x2 + y 2 z 2 = 1, if he density
is given at any point by (x; y; z) = xyz.
22
1.0
0.9
0.8
0.7
0.6
z
0.5
0.4
1.4
0.3
1.2
1.0
0.2
0.8
0.6
y
0.1
0.4
0.2 0.0
0.0 0.0
The mass is given by the integral
Z Z p
Z p
1
0
1+z 2
0.2
0.4
0.6
0.8
1+z 2 y 2
0
0
p
1+z 2
1Z
Z
1
=
yz 1
2 0 0
Z
1 1
2
=
z z 2 + 1 dz
8 0
7
=
48
23
(x; y; z) dxdydz
y 2 + z 2 dydz
1.0
x
1.2
1.4
Suggested Exercises for Chapter 6
The following exercises are not to be handed in. They represent skills required for
basic mastery.
6.1 (pages 313-314):
1; 3; 7; 11
6.2 (pages 326-328):
3; 5; 6; 17; 21; 23
6.3 (pages 337-338):
4; 5; 13; 16
Second Graded Assignment
Due: October 29
To reinforce written communication skills the Graded Assignment solutions
should be clearly presented in a "bluebook" or provided in PDF format. Late
papers will not be graded.
Second Graded Assignment. Do any one of the following:
Page 328: 33
Page 347: 10
Page 348: 12
Page 348: 21
24
Change of Variables in Rn
The substitution method for integrating f : R ! R over a closed interval is based
on transformation of the real line. Given the de…nite integral
Z x2
f (x)dx
x1
we can rewrite the integral as
Z
u2
f (x (u))
u1
dx
du
du
The Chain Rule allowed us to do this because, formally, we have a C 1 function
u 7! x(u)
that is one-to-one on the interval
I = [u1 ; u2 ]
which is mapped to the interval
I = [x1 ; x2 ]
A simple example is
Z
3
xe
x2
dx
2
which we expect to be a positive number since the integrand is positive on [2; 3].
The substitution u = x2 is really the transformation
p
u 7!
u
which is both C 1 and one-to-one on the interval
I = [ 9; 4]
which maps to the interval
The transformation u 7!
p
[3; 2]
u is a decreasing function of u and so x0 (u) < 0 :
25
3
x
2
1
0
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
u
[ 9; 4] 7! [3; 2]
Speci…cally,
dx
=
du
1
=
2x
1
p
2
u
and so
Z
4
9
dx
f (x (u)) du =
du
Z
2
xe
x2
dx
3
Z
4
p
1
p
du
2
u
9
Z 4
1
1
=
eu du =
e 9 e 4
2 9
2
p
This is a negative number because x =
u is decreasing and so the image of
I = [ 9; 4] was [3; 2] instead of I = [2; 3] as given. We can write the changeof-variable rule as
Z
Z
dx
f (x(u))
du = f (x)dx
du
I
I
=
ueu
if we want I and I to have the same orientation. Thus
26
Z
3
1
e 4 e 9 >0
2
2
as expected. The reason that we introduced the language of transformations is so
that we can be aware of what is needed to change variables in R2 or even higher
dimensions. We will mostly focus on R2 since the theory there generalizes.
xe
x2
dx =
R1p
Exercise. Compute 0 1
then by letting x(u) = cos u.
x2 dx …rst by letting x(u) = sin u and
The above discussion shows that the substitution procedure is not so much about
changing the given variable x as it is about changing another variable u into x.
The idea is that the integrand in terms of x has been altered from a simpler form
that it would have taken on another interval. This is the idea that we want to
apply to R2 where our regions of integration are two-dimensional and bounded by
graphs of continuous functions. Given an integral
ZZ
f (x; y)dA
D
we ask if D could be the image of another region D that may be easier to integrate
over.
Example. To …nd the area of a disk of radius a we could compute
ZZ
dA
D
where D is the disk of radius a centered at (0; 0). As an iterated
integral:
Z a Z pa2 x2
dydx
p
a
a2 x2
We know this integral equals a2 . Is there a simpler way to compute it? We ask if D can be realized as the image of another region
D that is perhaps a rectangle such as
[0; a]
[0; 2 ]
27
The transformation
T (x ; y ) = (x cos y ; x sin y )
maps D onto D because the vertical segment x = x0 is mapped
to the circle
(x0 cos y ; x0 sin y ) ; 0 y 2
(What does T do to the horizontal segment y = y0 ?) This suggests
that we can write
Z a Z pa2 x2
Z aZ 2
dydx =
f (x ; y ) dy dx
p
a
a2 x2
0
0
for some function f on D . For example, f (x ; y ) = x seems to
work. But why? This must somehow be determined by the transformation T .
We want to work with a transformation T : D ! D that is C 1 and, ideally, oneto-one. Note that T in the above example is not one-to-one as de…ned because
T (0; y ) = (0; 0) for any value of y . Also, T (x ; 0) = T (x ; 2 ) for any value of
x . We could restrict the domain of T to be (0; 0) [ (0; a] [0; 2 ) and then T
is both a one-to-one and onto map. Restrictions of this type, where we remove
subsets of D that are …nite unions of graphs of continuous functions, will be
allowed in our change-of-variables formula.
If D is a parallelogram then we can always …nd an a¢ ne transformation T that
maps a rectangle D onto D in a one-to-one manner. On R2 and a¢ ne transformation is de…ned by a matrix/vector action
T (x ; y ) =
a b
c d
x
y
which will be one-to-one precisely when ad
region bounded by the lines
y
y
y
y
=
=
=
=
+
=
x
y
bc 6= 0. For example, if D is the
x 1
x+1
1 2x
2 2x
28
p
q
y
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.5
1.0
1.5
x
-0.2
-0.4
D
The vertices of D are (1; 0) ; 13 ; 34 ; (0; 1) ; 23 ; 31 . Suppose we want D to be
the square region [0; 1] [0; 1]. We could …nd the a¢ ne transformation that takes
(0; 0) 7! 32 ; 31 and …xes (1; 0) and (0; 1):
T (x ; y ) =
1
3
1
3
4
3
2
3
x
y
+
2
3
1
3
Note that T (1; 1) = 13 ; 43 and T (D ) = D. Now A (D) = 32 , which is the
determinant of the matrix part of T times A (D ). This provides a clue to the
change-of-variables formula for any transformation T .
Exercise. Find the transformation T that takes (0; 0) 7! 23 ; 31
and interchanges (1; 0) and (0; 1). Show that T (D ) = D. What is
the determinant of the matrix part of T ?
The coordinates (x ; y ) are generic. They are used in di¤erential geometry to
suggest "pulling back" to reference domain. In practice we often use variables
29
that are easier to remember, such as (u; v) which remind us of the single-variable
case. The transformation T (x ; y ) = (x cos y ; x sin y ) is usually written
T (r; ) = (r cos ; r sin )
because we are mapping the polar coordinates r; to the corresponding Cartesian
variables.
Example. Let D = [0; 1] [0; 1] in the uv-plane. Find D = T (D )
in the xy-plane for the transformation
T (u; v) = u2
v 2 ; u2 + v 2
First, let’s …nd the images of the sides of the square since D is
likely to be bounded by these curves. We can express each side parametrically:
c1 (t)
c2 (t)
c3 (t)
c4 (t)
=
=
=
=
(t; 0) ; t 2 [0; 1]
(1; t) ; t 2 [0; 1]
(t; 1) ; t 2 [1; 0]
(0; t) ; t 2 [1; 0]
We have written the intervals for t to suggest starting at the origin
(0; 0) and moving around the square counter-clockwise. Thus, T (C1 )
is the curve in the xy-plane given by
t2 ; t2 ; t 2 [0; 1]
a straight line segment. Similarly, T (C2 ) ; T (C3 ) ; T (C4 ) are given
by
1
t2
t2 ; 1 + t2 ; t 2 [0; 1]
1; t2 + 1 ; t 2 [1; 0]
t2 ; t2 ; t 2 [1; 0]
30
y
2.0
1.8
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.2
0.4
0.6
0.8
-0.2
1.0
x
-0.4
D = T (D ) with T t; 21 , red, T p12 ; t , green, and T
blue.
1 1
t; t
2 2
+
1
2
,
The image is also a square whose boundary has also been traced
out counter-clockwise. However, this is not an a¢ ne transformation
since there are straight segments that map to parabolas.
In the above example we have A(D) = 2 and A(D ) = 1, and T is a one-to-one
map of D onto D. Note that
!
r
r
x+y
y x
T 1 (x; y) =
;
; (x; y) 2 D
2
2
Further,
2u
2u
DT (u; v) =
31
2v
2v
whose determinant is 8uv, and
Z 1Z
0
1
(8uv) dudv = 2
0
Now the derivative of an a¢ ne transformation is just its matrix part. This suggests
that a C 1 transformation T from D to D needs to be treated locally as the linear
transformation represented by its derivative DT .
Exercise. Let T : D ! D by T (u; v) = (v 2 u2 ; u2 + v 2 ). Compute det DT (u; v). In what direction does T (@D ) trace out @D?
Jacobian Determinants
A transformation T : R2 ! R2 has two coordinate functions
T (u; v) = (x (u; v) ; y (u; v))
so
@x
@u
@y
@u
DT (u; v) =
@x
@v
@y
@v
The Jacobian is the determinant of this matrix, written
@ (x; y)
@ (u; v)
which may be positive or negative depending on (u; v) but presumably zero only
at points that can be removed from D without a¤ecting the integral we want to
compute. For example, if D = [0; 1] [0; 2 ] and T (u; v) = (u cos v; u sin v) then
DT (u; v) =
cos v
sin v
u sin v
u cos v
@ (x; y)
= u
@ (u; v)
which is zero only on the points (0; v) and positive otherwise.
32
The derivative DT is used to …nd the a¢ ne linear approximation of T at a given
point (u0 ; v0 ) :
u
v
L (u; v) = T (u0 ; v0 ) + DT (u0 ; v0 )
u
v
= DT (u0 ; v0 )
p (u0 ; v0 )
q (u0 ; v0 )
For (u0 ; v0 ) =
1
;
2
p (u0 ; v0 )
q (u0 ; v0 )
+
(u0 ; v0 ) rx (u0 ; v0 )
(u0 ; v0 ) ry (u0 ; v0 )
x (u0 ; v0 )
y (u0 ; v0 )
=
u0
v0
in the polar coordinate example we have
L(u; v) =
=
1
2
0
1
0
+
0
1
0
0
u
v
1
2
u
v
1
2
+
1
2
0
1
2
This property of the derivative allows us to make use of the following fact:
If D is a closed, bounded region and D = T (D ) for an a¢ ne transformation T , then A (D) = j j A (D ) where is the determinant of
the matrix part of T .
a b
u
then D is the
c d
v
parallelogram spanned by the vectors v = ai + cj and w = bi + dj, so the area of
D is
jv wj = j(ad bc) kj = jad bcj
For example, if D = [0; 1]
[0; 1] and T (u; v) =
The a¢ ne linear approximation of a transformation T is therefore multiplying in…nitesimal areas around (u0 ; v0 ) by the absolute value of the Jacobian determinant
at this point. Thus, if D = T (D ) we would expect
ZZ
ZZ
@ (x; y)
dxdy =
dudv
@ (u; v)
D
D
33
There is an implicit assumption in this formula that D and D have ’compatible
orientations’in the same way the we assumed the intervals I and I had the same
orientation in the formula
Z
Z
dx
dx =
du
du
I
I
The concept of orientation needs to be developed further to formulate the classical
integration theorems. For now we will use examples where the orientation makes
intuitive sense.
Example. Let D = [0; 1]
[0; 1] and T (u; v) =
1 3
1 1
u
v
.
Then D is the parallelogram
y
2.0
1.5
1.0
0.5
0.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
x
D = T (D )
RR
R1R1
dudv = 0 0 j1 1 3 1j dudv = 2 = A(D).
RR
RR
The evaluation D dxdy = D dxdy = 2 requires that we set up the
integral over D to be compatible with the one over D , but unlike the
one-variable case there is more that one way to do this even within
We have
D
@(x;y)
@(u;v)
34
the same order of integration:
Z 1Z x
Z 3 Z 1 x+ 2
Z
3
3
dydx +
dydx +
1
x
3
0
Z
1
0
Z
x
1
1
x
3
dydx +
Z
3
1
x
3
1
Z
4
dydx +
1
x+ 23
3
1
x+ 23
3
dydx = 2
x 2
3
1
x
3
Z
Z
4
3
Z
x 2
dydx = 2
1
x+ 23
3
In the …rst calculation we wrote the limits as increasing functions
of the respective variables. In the second we wrote them as decreasing
functions. If we had written one of the integrals with increasing limits
and the other with decreasing then we would have obtained 2.
The proper way to make sense of orientation in Rn for any n involves the theory
of di¤erential forms. For n = 2 or n = 3 we can keep track of orientation by
visualizing how T transforms the region D into D. In the above example, as we
move around @D counter-clockwise note that we trace out @D in the clockwise
direction. We sat that T has reversed orientation, which is re‡ected in the fact
@(x;y)
in
that the determinant is negative. When we used the absolute value of @(u;v)
the integral over D we assumed we would set up the integral in each plane with
compatible limits. For elementary regions this is usually not di¢ cult.
Theorem. Let T : D ! D be C 1 and and assume T is oneto-one except perhaps on a …nite union of graphs of a single variable.
Then
ZZ
ZZ
@ (x; y)
dudv
f (x; y)dxdy =
f (x (u; v) ; y (u; v))
@ (u; v)
D
D
The theorem can be generalized to integrable functions on Rn . For n = 3 we have
C 1 transformations T : W ! W between solid regions. The absolute value of
the determinant of DT (u; v; w) at a point (u0 ; v0 ; w0 ) tells us how the a¢ ne linear
approximation alters volumes and yields the volume di¤erential over W . A¢ ne
transformations now take the form
0
1 0
1
u
p1
T (u; v; w) = (aij ) @ v A + @ p2 A
w
p3
35
where (aij ) is a 3 3 constant matrix and p1 ; p2 ; p3 are constants. For closed
bounded regions we have
V (W ) = jdet (aij )j V (W )
Transformation on R3 that are not a¢ ne usually arise from coordinate systems
adapted to certain symmetries of a situation, such as cylindrical and spherical
change of coordinates.
Cylindrical. Let T (u; v; w) = (u cos v; u sin v; w). Then T (W ) =
R for W = [0; 1] [0; 2 ] ( 1; 1), and
3
@ (x; y; z)
=u
@ (u; v; w)
0
The volume di¤erential on the uvw-space is then
ududvdw
Spherical. Let T (u; v; w) = (u sin w cos v; u sin w sin v; u cos w).
Then T (W ) = R3 for W = [0; 1] [0; 2 ] [0; ], and
@ (x; y; z)
=
@ (u; v; w)
u2 sin w
0
The volume di¤erential on the uvw-space is then
u2 sin wdudvdw
Example. Let W be the region inside the unit ball but above the
plane z = a 2 [ 1; 1]. Find V (W ) …rst by using the cylindrical transformation and then by using the spherical transformation.
1) We will use the conventional cylindrical coordinate variables
u = r; v = ; w = z. Then
ZZZ
Z 2 Z p1 a2 Z p1 r2
dV =
rdzdrd
W
0
= 2
Z
a
0
p
1 a2
r
p
1
0
=
3
(a + 2) (a
36
1)2
r2
a dr
2) The conventional spherical coordinate variables are u = ; v =
;w = :
ZZZ
Z 2 Z arccos a Z 1
2
dV =
sin d d d
W
0
=
2
3
=
3
a
cos
0
arccos a
a3 tan sec2
sin
d
0
2
3
=
Z
1
a + a3 tan2 arccos a
2
(a + 2) (a
1
1)2
The second integration was more intricate than the …rst but it is
not always easy to foresee this from the geometry of the region.
1.0
0.5
-1.0
1.0
-0.5
z
0.5
0.0
0.0 0.0
y
-0.5
0.5
x
-0.5
-1.0
W with a =
37
1
2
-1.0
1.0
For a = 21 , the region W in 1) is enclosed by
1.0
0.8
0.6
w
0.4
6
0.2
4
2
0.0
0.0 0
v
0.2
0.4
0.6
0.8
u
1.0
and the region W in 2) is enclosed by
1.5
1.0
w
0.5
1.5
0.5
1.0
u
0.0
0 0.0
2
4
v
6
Note that this problem is most easily solved using Cavalieri’s Principle, since
A (z) = (1 z 2 ) and so
38
V (W ) =
Z
1
1
z 2 dz =
a
39
1
(a
3
1)2 (a + 2)
Study Guide for First Exam
Cavalieri’s Principle for …nding volumes of solids
Iterated integration over elementary regions in R2 and R3
Mean-Value Theorem for double integrals
Transformations:
Determining images D = T (D ) in R2
One-to-one and onto maps
Jacobian determinant
A¢ ne transformations
Change of Variables theorem in R2 and R3 including application to polar, cylindrical, and spherical coordinates
40
A Classic Problem
(complete solution for Extra Credit)
The intersection of the two cylinders
x2 + z 2 = 1
y2 + z2 = 1
is a bounded solid W consisting of sixteen congruent pieces (two in each octant).
The volume of W is easy to compute in Cartesian coordinates
V (W ) = 16
Z
1
0
Z
0
x
Z
p
1 x2
0
dzdydx =
16
3
Find the volume of the solid bounded by the three cylinders
x2 + z 2 = 1
y2 + z2 = 1
x2 + y 2 = 1
In my opinion there is a slight advantage to using cylindrical coordinates, but
Cartesian coordinates also work well. How do the volumes of these two solids
compare to the volume of the unit sphere?
41
Suggested Exercises for Chapter 7
The following exercises are not to be handed in. They represent skills required for
basic mastery.
7.1 (pages 356-358):
1; 5; 9; 11; 17; 25; 27
7.2 (pages 373-375):
1; 3; 5; 13; 17
7.3 (pages 381-383):
1; 2; 4; 5; 7; 11; 17; 19; 23
7.4 (pages 391-392):
1; 3; 4; 5; 9; 17; 19; 25
7.5 (pages 398-400):
1; 5; 7; 9
7.6 (pages 411-413):
1; 3; 5; 7; 19; 21
Third Graded Assignment
Due: November 17
To reinforce written communication skills the Graded Assignment solutions
should be clearly presented in a "bluebook" or provided in PDF format. Late
papers will not be graded.
Third Graded Assignment. Do any one of the following:
Page 382: 12
Page 398: 15
Page 399: 22
Page 424: 20
42
Path Integrals
For a function f : R ! R we are familiar with the integral over a closed interval
[a; b]. We can write this integral
Z
b
f (t)dt
a
and think of the interval as a path in R1 : c(t) = t; a
t
b. One interpretation of the integral is the signed area of the graph of f over this path. This
interpretation generalizes to f : Rn ! R over a C 1 path c. Most of our applications will be for n = 2 or 3 so we will formalize the de…nition for paths in R3 ,
c(t) = (x (t) ; y (t) ; z (t)) for t 2 [a; b].
The idea is to use the derivative of c to replace dt with the tangent di¤erential to
the curve C. Since f is a scalar-valued function this di¤erential uses the length of
the tangent vector as the limit of small lengths of segments of C near the point
c(t). The path integral of f over c is
Z
b
a
f (c(t)) jc0 (t)j dt
p
Rb
Recall that jc0 (t)j = c0 (t) c0 (t) and that a jc0 (t)j dt is the arc length of the
curve. As long as the composition f (c(t)) is continuous these formulas are independent of the chosen parameterization, which we will assume to be regular:
c0 (t) 6= 0 for t 2 (a; b).
Example. Let c(t) = (cos t; sin t; t2 ) for t 2 [0; 2 ]. Find the
arc length of this curve and compute the path integral of f (x; y; z) =
x2 + y 2 + z 2 .
The arc length is the path integral of the constant function 1 :
Z 2 p
1 + 4t2 dt
0
=
p
16
2
+1+
p
1
ln 4 + 16
4
40:41
43
2
+1
2
On the path, f (x; y; z) = cos2 t + sin2 t + (t2 ) = 1 + t4 , so the path
integral of f is
Z
2
1 + t4
p
1 + 4t2 dt
0
=
381 + 32
384
20647
2
+ 2048
4
p
16
2
+1+
p
129
ln 4 + 16
512
40
30
20
10
-1
-1.0 -0.8
-0.6
-0.4
0
0.00
-0.2
0.2
0.4
1
c(t) = (cos t; sin t; t2 )
Arc-Length Di¤erential
The arc length formula de…nes a function
Z t
s(t) =
jp0 ( )j d
a
where p (t) is any regular path. The derivative of s is
s0 (t) = jp0 (t)j
44
0.6
0.8
1.0
2
+1
Then
Z
b
0
a
jp (t)j dt =
Z
s(b)
ds = s(b)
0
Given a parameterization p(t), we can reparameterize the curve by arc length by
setting
c(s) = p(t)
By the Chain Rule
c0 (s)s0 (t) = p0 (t)
p0 (t)
jc0 (s)j =
=1
s0 (t)
For this reason a path c is said to parameterized by its arc length s (constant
speed equal to 1) provided
jc0 (s)j = 1
Example. Consider two parameterizations of the
p …rst-quadrant
portion of the circle centered at the origin of radius 2.
p(t) =
c(t) =
1 t
1+t
p
;p
; t 2 [ 1; 1]
1 + t2
1 + t2
p
p
t
t
2 cos p ; 2 sin p
; t 2 0; p
2
2
2
For the second path we have
c0 (t) =
sin
t
p
2
; cos
t
p
2
jc0 (t)j = 1
Rt
0
so this is a parameterization by arc length: s(t) =
d = t, and in particular, s
p
2
=
45
p
2
Rt
0
jc0 ( )j d =
, the length of the curve. For
the …rst path we have
p0 (t) =
1
t
3
2
;
1
t
3
!
(t2 + 1) (t2 + 1) 2
p
2
jp0 (t)j = 2
t +1
Z t p
p 1
2
s(t) =
d
=
2
+ arctan t
2+1
4
1
p
Thus s(1) = p2 and s(0) = 2 4 as expected; note that p(0) =
(1; 1). The arc-length di¤erential is
p
2
dt
ds = 2
t +1
and so the path integral of a function f can be written
p
Z 1
Z
1 t
1+t
2
f p
f ds =
;p
dt
2
1 + t2
1 + t2 t + 1
1
C
p
Exercise. Since s = 2 41 + arctan t we have t = tan
tan
tan
ps
2
ps
2
1
+1
ps
2
4
=
. Use this to …nd the explicit arc-length parameterization
c(s) = p(t). What is the interval for s?
Line Integrals
Line integrals are used to integrate vector …elds over paths. This generalizes the
vector concept of work done by a constant force F in moving a particle through a
straight-line displacement d :
W =F d
We use reasoning similar to the construction of the path integral of a scalarvalued function. Given a path c we think of the tangent vector c0 (t) as providing
a limiting displacement for segments of the curve C described by c. The force at
t should be used to compute the limiting value of work
F (c(t)) c0 (t)
46
If F is continuous on the C 1 path c : [a; b] ! Rn the line integral of F along the
path is
Z b
F (c(t)) c0 (t)dt
a
which can be abbreviated
Z
F ds
C
since the integral does not depend on the regular parameterization chosen for C.
Example. Consider again the path c(t) = (cos t; sin t; t2 ) for t 2
[0; 2 ] and let F(x; y; z) = (x; y; z). Then
Z
Z 2
F ds =
cos t; sin t; t2 ( sin t; cos t; 2t) dt
C
0
Z 2
= 2
t3 dt = 8 4
0
If we think of F as the force equal to the position vector in space
then 8 4 is the amount of work done by F in moving a particle along
the spiral C from (1; 0; 0) to (1; 0; 4 2 ). For the same path, the work
done by the force
F(x; y; z) = (xy; yz; xz)
is
Z
=
2
Z0 2
(cos t) 2t3
sin2 t + t2 sin t dt
t2 (2t + sin t) cos tdt = 23
2
0
We can covert a line integral into a path integral by setting
c0 (t)
= T(t)
jc0 (t)j
47
Then
Z
F ds =
C
=
Z
Za
b
F (c(t)) c0 (t)dt
f ds
C
where f is a scalar function whose values on C are given by
F (c(t)) T(t)
the tangential component of the force along c.
48
Di¤erential Form Notation
A vector …eld on Rn has n component functions each of which is a function of n
variables. For n = 3 it is conventional to write these functions as
f (x; y; z) = (P (x; y; z) ; Q (x; y; z) ; R (x; y; z))
We then write
Z
Z
Z
F ds = P dx + Qdy + Rdz =
C
b
P
a
C
dx
dy
dz
+Q +R
dt
dt
dt
dt
where c : [a; b] ! R3 is a path with image C. The expression P dx + Qdy + Rdz is
an example of a di¤erential form. The framework of di¤erential forms uni…es the
fundamental integral theorems of calculus. We will not develop this framework in
full detail but it is the preferred notation in advanced calculus for keeping track of
the orientation of higher-dimensional regions of integration. We should be more
speci…c about orientation for line integrals. The claim that the line integral over
C is independent of the chosen path c is based on a few conditions:
1) c is regular;
2) c is piecewise C 1 ;
3) c traces out C in a given direction.
Condition 3) needs to be stated mathematically. For example, if c were changed
so that we start at c(b) and end at c(a) then we would expect the line integral
Ra
Rb
to change sign, just as b f (x)dx =
f (x)dx for a single-variable function on
a
[a; b]. In general, we can address this situation in terms of reparameterization.
Suppose we have an interval [a; b] over which we want to compute a line integral
but we have a parameterization of C that uses another interval
c : [a1 ; b1 ] ! R3
If h : [a; b] ! [a1 ; b1 ] is one-to-one and C 1 then p = c h : [a; b] ! R3 is a path
that reparameterizes C.
Example. Given c : [a1 ; b1 ] ! R3 we can change to a parameterization over the unit interval [0; 1] by
h(t) = a1 + (b1
49
a1 ) t
So, if we parameterize the unit circle by (cos t; sin t) with t 2 [0; 2 ]
and let h(t) = 2 t then
p(t) = (cos 2 t; sin 2 t) ; t 2 [0; 1]
is a familiar reparameterization of the unit circle. Note that p0 (t) =
c0 (h (t)) h0 (t) = ( 2 sin 2 t; 2 cos 2 t) so jp0 (t)j = 2 whereas jc0 (t)j =
1.
A notational caution. There are competing conventions as to what is meant by
an interval [a; b]. Is it the set of points x with a x b or is it an oriented domain
that traces out the points from a to b? In the second interpretation moving from
b to a would be the interval [b; a], though this is the same set of points,
R a and this
is the viewpoint that is often used in single-variable calculus, that is, b f (x)dx is
Rb
the integral of f over [b; a] and a f (x)dx is the integral over [a; b]. The language
of di¤erential forms is designed to remove this ambiguity by shifting the issue of
orientation to the way in which we parameterize the set. An important example
helps to clarify this idea. Let h(t) = a + b t. Then h(a) = b and h(b) = a, but
we choose to write this map as
h : [a; b] ! [a; b]
that is, our interpretation of domains and ranges is as sets of points. If c :
[a; b] ! R3 then p = c h : [a; b] ! R3 is a reparameterization sometimes called
cop , the opposite path to c, because it traces out C from c(b) to c(a). We have
p0 (t) = c0 (h (t)) h0 (t) = c0 (a + b t), so the tangent vectors have changed sign
because h0 (t) = 1. We say that p reverses orientation. More generally, let
p = c h be any reparameterization as described above. Then,
Z
Z b
F ds =
[F (c (h(t))) c0 (h (t))] h0 (t)dt
a
p
=
Z
h(b)
F (c(u)) c0 (u)du, where u = h(t)
h(a)
R
R
If p preserves orientation then h(a) = a1 and h(b) = b1 so p F ds = c F ds. But
R
R
if p reverses orientation then h(a) = b1 and h(b) = a1 so p F ds =
F ds.
c
50
Why wasn’t orientation an issue with path integrals? We had
Z
Z b1
f (c(t)) jc0 (t)j dt
f ds =
a1
c
and if p = c h : [a; b] ! R3 then
Z
Z b
f ds =
f (c (h(t))) jc0 (h (t))h0 (t)j dt
a
p
Again let u = h(t). Then
Z
Z
f ds =
h(b)
h(a)
p
f (c(u)) jc0 (u)j
du
dt
dt
If p preserves orientation then h(a) = a1 ; h(b) = b1 and du
dt = du, so
dt
R
f ds. However, if p reverses orientation then h(a) = b1 ; h(b) = a1 but
c
du, so
Z a1
Z
f (c(u)) jc0 (u)j du
f ds =
p
=
Z
R
f ds =
dt =
p
du
dt
b1
b1
f (c(u)) jc0 (u)j du
Za1
=
f ds
c
Thus, the line integral is an oriented integral but the path integral is not.
Gradient Fields
Consider the vector …eld that assigns the position vector to every point in space
F(x; y; z) = (x; y; z)
and compute the line integral along a piecewise C 1 path c : [a; b] ! R3
Z b
(x(t); y(t); z(t)) c0 (t)dt
a
Z b
=
[x(t)x0 (t) + y(t)y 0 (t) + z(t)z 0 (t)] dt
a
51
This integral is easy to evaluate since every term is of the form
to
1 2
b
x (t) + y 2 (t) + z 2 (t) a
2
1
[c(b) c(b) c(a) c(a)]
=
2
R
udu. It is equal
The only thing we needed to know about the path was its values at the endpoints
of the integral. This suggests that the vector …eld only has to "see" the endpoints
of the curve C. Most vector …elds do not have this property, but gradient …elds
do:
If f : R3 ! R is C 1 then
R
c
rf ds = f (c (b))
f (c (a)).
The proof of this fact is identical to the computation in the above example (see
page 367). What is f in that example? In this case it is easy to guess
f (x; y; z) =
Note that in fact
1
2
1 2
x + y2 + z2
2
b
[x2 (t) + y 2 (t) + z 2 (t)]a = f (c (b))
f (c (a)). This indepen-
dence of path property for line integrals of gradient …elds will be studied in greater
detail when we develop the classical integral theorems. Note that this property is
di¤erent from independence of parameterization where we were working with different path description of the same geometric curve. Here we are saying that one
can connect the endpoints by di¤erent curves without a¤ecting the line integral
of a gradient …eld.
Simple Curves
We have generally assumed that c : [a; b] ! R3 is one-to-one though this was not
strictly necessary to de…ne the line integral. The assumption avoids behavior in
the trajectory such as changing direction, though this may be an important aspect
of certain phenomena. It also avoids certain geometric behavior such as the curve
C crossing over itself.
52
Example. Let C be given by the curve
y
-1.0
-0.8
-0.6
-0.4
0.2
-0.2
0.2
0.4
0.6
0.8
1.0
x
-0.2
-0.4
-0.6
-0.8
-1.0
Any single path description of C cannot be one-to-one since the trajectory will pass through the origin twice. For example,
c(t) = (cos t cos 2t; sin t cos 2t) ; t 2 [0; ]
starts the trajectory at (1; 0) and ends at ( 1; 0) but reaches (0; 0)
3
sin 3t 21 sin t; 32 cos 3t
at both t = 4 and t = 34 . Note that c0 (t) =
2
so
p
p
c0
=
2;
2
4
p p
3
c0
=
2; 2
4
1
2
cos t
The image of a path with c(a) = c(b) is called a closed curve. If c is one-to-one
on [a; b) then the image is a simple closed curve. Suppose that the integral curves
for a vector …eld F are described by the paths
ca (t) = a2 cos t cos 2t; a2 sin t cos 2t ; t 2 [0; 2 ]
These curves are closed but not simple. They could be ‡ow lines for a magnetic
force …eld.
53
y
2
1
-2
-1
1
2
x
-1
-2
Since an integral curve (‡ow line) for a vector …eld is a path de…ned by
c0 (t) = F(c(t))
the line integral of F along such a path is
Z b
2
jc0 (t)j dt
a
In this example, the line integral along a ‡ow line is
Z 2
5 3
4
cos 4t dt = 5a4
a
2
2
0
Surfaces
A large class of surfaces in R3 can be described as graphs of functions f : R2 ! R.
By de…nition such a surface has the property that any line perpendicular to the
domain plane intersects the surface in at most one point. This property does not
hold for most surfaces: In general there is no plane such that any line perpendicular
54
to that plane intersects the surface in at most one point. A sphere is a basic
example. In order to use calculus to study surfaces we can ask whether the surface
can be divided into graphs of functions. For example, a sphere is the union of two
hemispheres each of which can de represented as a graph. (This was the same
approach for a curve in the plane that could not be represented as a graph of a
single function.) The theory of manifolds in di¤erential geometry is based on this
idea, but a great deal of theory is needed to make it precise. First it is necessary
to develop the concept of parameterization in more detail.
Example. Let S be the sphere of radius
S : x2 + y 2 + z 2 =
0
centered at the origin
2
0
Let
x(u; v) =
y(u; v) =
z(u; v) =
cos u sin v
0 sin u sin v
0 cos v
0
Every point on S is (x(u; v); y(u; v); z(u; v)) for some choice (u; v) 2
R , in fact, for in…nitely many choices. Can we restrict the domain D
of (u; v) so that the function : D ! R3 is one-to-one? As before,
when spherical coordinates were introduced, we could let D = [0; 2 )
[0; ]. For purposes of integration it is preferable to work with closed
domains, so if we want to be one-to-one we will usually except, as
usual, points on the boundary of D.
2
As with curves, parameterization of surfaces requires involves two questions:
1) Given a surface S, perhaps in Cartesian form, can we …nd
and D to
describe it parametrically, preferably with C 1 coordinate functions?
2) Given a parameterization , can we understand its image surface even it
we cannot visualize it, and perhaps …nd a Cartesian description?
It helps to keep some uncomplicated examples in mind, one of which we have
already studied in di¤erential vector calculus.
55
Planes. A plane in R3 with Cartesian equation ax + by + cz = d is
easily parameterized using the fact that n = ai+bj+ck is orthogonal to
(x x0 ) i + (y y0 ) j + (z z0 ) k where (x0 ; y0; z0 ) is any given point
on the plane and (x; y; z) varies among the points of the plane. In
di¤erential vector calculus we posed the problem the other way around:
Given a normal vector and a known point …nd the Cartesian equation
of the plane. Given the Cartesian equation, one way to …nd is as
follows. First we …nd two independent vectors x1 and x2 orthogonal
to n. Since n is not the zero vector at least one of a; b; c is not zero,
for example, we might have b 6= 0. Let
x1 = bi
x2 = cj
aj
bk
Then x1 x2 = bn, so both x1 and x2 are orthogonal to n and
these two vectors span a plane through the origin with normal vector
n. This plane is
= fux1 + vx2 : u; v 2 Rg
Now let (x0 ; y0; z0 ) be any point such that ax0 + by0 + cz0 = d. Then
+x0 i+y0 j+z0 k, is the plane ax+by+cz = d and so it is parameterized
by
(u; v) = (bu + x0 ; au + cv + y0 ; bv + z0 )
= (x(u; v); y(u; v); z(u; v))
Working with a given is generally easier than …nding from geometric information, but quadric surfaces are a good source of examples where a a choice of
is suggested by the geometry.
2
2
2
Ellipsoids. The equation xa2 + yb2 + zc2 = 1 describes an ellipsoid
centered at the origin with axes parallel to the coordinate axes. A
standard choice for comes from a straightforward generalization of
spherical coordinates:
x(u; v) = a cos u sin v
y(u; v) = b sin u sin v
z(u; v) = c cos v
56
with D = [0; 2 ]
[0; ], for example.
Hyperboloids. The analogous parameterization for these surfaces
uses hyperbolic sines and cosines.
2
2
x2
+ yb2 zc2 = 1 :
a2
x(u; v) = a cosh u cos v
y(u; v) = b cosh u sin v
z(u; v) = c sinh u
20
10
-20
-20
-10
z
-10
0
0 0
y10 -10
20
10
x
20
-20
(4 cosh u cos v; 3 cosh u sin v; 2 sinh u)
x2
a2
+
y2
b2
z2
c2
=
1:
x(u; v) = a sinh u cos v
y(u; v) = b sinh u sin v
z(u; v) =
c cosh u
57
4
2
3
2
-2 0
1 00
2
-1
-2
-3
-2
-4
(4 sinh u cos v; 3 sinh u sin v; 2 cosh u)
Paraboloids can generally be positioned as graphs of functions, such as z = x2 +y 2
and z = x2 y 2 . Note that z = f (x; y) always has a natural parameterization of
the form
x(u; v) = u
y(u; v) = v
z(u; v) = f (u; v)
Tangent Planes
is regular at a point (u0 ; v0 ) on its image surface provided it determines a
tangent plane at that point. This plane must have a normal vector which means
that must determine an independent pair of tangent vectors to curves through
(u0 ; v0 ):
@
@u
@
=
@v
Tu =
Tv
58
If Tu Tv 6= 0 at (x0 ; y0; z0 ) =
this point has normal vector
(u0 ; v0 ) then the tangent plane to the surface at
n = Tu
Example.
Tv
(u; v) = (u cos v; u sin v; u2 + v 2 ) ; D = R2 :
Tu = (cos v; sin v; 2u)
Tv = ( u sin v; u cos v; 2v)
Tu
2 u2 cos v
Tv =
v sin v; u2 sin v + v cos v;
1
u
2
Tu Tv = 0 only if u = 0, but in that case v must be 0 also. Thus
is regular (there is a tangent plane) except at (0; 0) = (0; 0; 0).
Note that S contains the positive z-axis, the image points (0; v0 ) =
(0; 0; v02 ). The tangent plane at one of these points has a normal vector
parallel to sin v0 i cos v0 j if v0 6= 0 so the tangent planes contain the
z-axis and rotate as v0 increases. What are the tangent planes at
(u0 ; 0) = (u0 ; 0; u20 ), u0 6= 0?
400
300
200
100
-15
-10
-5
0
-15
-10
-5
00
5
5
10
10
15
S : (u cos v; u sin v; u2 + v 2 )
59
15
The surface S can be pieced together from the z-axis (u = 0) along
with graphs z = f (x; y). First, if v = (2k + 1) 2 then
(u; v) =
0; u; u2 + (2k + 1)2
2
4
This is a family of parabolas in the yz-plane. Each one of these
parabolas belongs to the boundary of the graph of
z = x2 + y 2 + arctan
y
+k
x
2
+ k ; 2 + k , and
which is the image of on Dk = Rnf0g
2
these graphs are joined at the parabolas. What do sections of this
surface by planes parallel to the xy-plane look like?
We will generally make the usual assumptions about parameterized surfaces unless
stated otherwise:
: D ! R3 is C 1 and one-to-one except possibly on @D, and
elementary region, and (D) = S is regular except possibly at a …nite
number of points.
Area of S
Integrals over surface regions require an area di¤erential just as integrals over
portions of curves required an arc-length di¤erential. For curves described by
paths the tangent vector was used to construct the di¤erential. For surfaces
parameterized by we saw that the tangent vectors Tu and Tv determine the
tangent plane at a point from the normal vector Tu Tv . Since jTu Tv j is the
area of the parallelogram spanned by these two tangent vectors we can construct
an area di¤erential from the multiples uTu and vTv based on the idea that
jTu Tv j u v approximates the area of the image under of a corresponding
small rectangle in the uv-plane. In the limit,
jTu
Tv j u v ! jTu
60
Tv j dudv
Consequently,
(D) = S then
A(S) =
ZZ
D
jTu
Tv j dudv
is the surface area of S. Note the similarity to the arc-length integral. In fact,
jTu Tv j can be expressed in terms of Jacobian determinants since
Tv =
Tu
@ (y; z)
i
@ (u; v)
@ (x; z) @ (x; z)
j+
k
@ (u; v) @ (u; v)
Example. Consider a portion of the surface in the previous example
u cos v; u sin v; u2 + v 2 ;
h
i
D = [0; 1]
0;
2
(u; v) =
We have
jTu
Tv j =
p
4u4 + u2 + 4v 2
and so
A(S) =
Z
0
1
Z
0
1
2
Z
1
u2
=
4 0
3:0261
p
4u4 + u2 + 4v 2 dvdu
p
Z 1p
2 + u2 + 4u4
1
+
2
p
du +
4u4 + u2 +
4u + 1 ln
4 0
u2 + 4u4
As with arc-length integrals, surface integrals often require non-elementary antiderivatives. Consider the same surface as a portion of the graph
z = x2 + y 2 + arctan2
y
x
with the graph parameterization
(u; v) = u; v; u2 + v 2 + arctan2
v
;
u
where D is the portion of the unit disk in the …rst quadrant. In this case
61
2 du
2v
v
arctan
2
+v
u
2u
v
0; 1; 2v + 2
arctan
2
u +v
u
2v
2u
v
v
2u;
arctan
arctan
2
2
2
2
u +v
u
u +v
u
Tu =
1; 0; 2u
Tv =
Tv =
Tu
Thus
RR
D
jTu
Z
u2
2v; 1
Tv j dudv =
1
0
Z
p
1 u2
0
q
4 arctan2
v
u
+ (u2 + v 2 ) (4u2 + 4v 2 + 1)
p
dvdu
u2 + v 2
and converting to polar coordinates we obtain
A(S) =
Z
0
1
Z
2
0
p
4r4 + r2 + 4 2 d dr
as with the …rst parameterization.
3
2
z
1.0
0.8
1.0
1
0.6
0.8
0.6
0.4
y
0.2
0.4
0.2
0
0.0 0.0
A(S)
3:0261
62
x
Surface Integrals
We want to integrate a function over a surface in its domain. Again, the focus is
on scalar …elds (f : R3 ! R) and vector …elds (F : R3 ! R3 ). The de…nition of
these integrals is based on the area di¤erential provided by a parameterization,
just as path and line integrals were developed from the arc-length di¤erential:
ZZ
ZZ
f dS =
f ( (u; v)) jTu Tv j dudv
ZZ S
Z ZD
F dS =
F ( (u; v)) (Tu Tv ) dudv
D
Example. Find the average temperature on the surface S of a ball of
radius a with center at the origin if the temperature at each point is
T (x; y; z) = e z .
We can use
(u; v) = (a cos u sin v; a sin u sin v; a cos v) ;
D = [0; 2 ] [0; ]
so that
Tu Tv =
a2 cos u sin2 v; a2 sin u sin2 v; a2 sin v cos v
jTu Tv j = a2 sin v
On S, T (x; y; z) = e a cos v so
ZZ
Z 2 Z
2
T dS = a
S
0
a cos v
e
sin vdvdu
0
= 2 a ea e
= 4 a sinh a
a
Since A(S) = 4 a2 the average temperature is
ZZ
1
1
T dS = sinh a
A(S) S
a
63
Let F(x; y; z) =
We have
rT (x; y; z). Evaluate
F(x; y; z) =
F ( (u; v)) =
ZZ
F dS =
=
0; 0; e z
0; 0; e a cos v
Z 2 Z
2
a
e
0
RR
a cos v
F dS on this sphere.
sin v cos vdvdu
0
4 (a cosh a
64
sinh a)
Study Guide for Second Exam
Paths
Arc Length
R
Integral of scalar …eld (path integral): R C f ds
Integral of vector …eld (line integral): c F ds
Surfaces
Parameterization of S
Standard parameterizations of spheres and other quadric surfaces
Graph parameterization
Tangent vectors Tu ; Tv
Tangent plane at a point on S RR
RR
Surface integrals of scalar …elds: RRS f dS (for example, A(S) = S dS)
Surface integrals of vector …elds:
F dS
Note. Focus on the basic computational aspects of the above ideas. Since both
types of trigonometric functions are frequently used to parameterize curves and
surfaces be familiar with the following facts:
d sinh t
dt
d cosh t
dt
cosh2 t sinh2 t
cosh2 t + sinh2 t
2 cosh t sinh t
= cosh t
= sinh t
= 1
= cosh 2t
= sinh 2t
Page 398: 7
The vertices of the tetrahedron are (0; 0; 0), (0; 0; 1), (1; 0; 0), (1; 1; 0)
65
1.0
0.8
0.6
z
0.4
0.2
0.0
0.0 0.0
0.2
0.6
0.4
0.2
0.8
1.0
y
0.4
0.6
x
0.8
1.0
The tetrahedron is a surface S composed of four pieces:
y=0:
(u; v) = (u; 0; v) ; Tu Tv = j
f ( (u; v)) = 0 so integral on this face is zero
z=0:
R 1(u;
R uv) = (u; v; 0)1 ; Tu
uvdvdu = 8
0 0
Tv = k
x=y:
= (u; u; v) ; Tu pTv = i
p(u;R v)
1R1 u 2
1
2 0 0 u dvdu = 12
2
j
x+z =1:
= (u; v; 1 u)p; Tu Tv = i + k
p(u;R v)
1Ru
2 0 0 uvdvdu = 18 2
ZZ
p
1
1p
1p
1
xydS = 0 + +
2+
2=
3+5 2
8 12
8
24
S
66
Orientation
Given two regular parameterizations of a surface S the change-of-variables process
shows that the integral of a scalar …eld is the same for both. This why we write
ZZ
f dS
S
similar to the path integral over a curve which also did not depend on parameterization. For the integral of a vector …eld over S it is reasonable to assume
that the orientation of the parameterization matters. This is true, but we need
to determine what orientation for surfaces means. The intuitive idea that can
be made mathematically precise is that at any point the surface divides space
into two half-spaces and a normal vector at that point is directed toward one
or the other half-space. The simplest case is when the surface is a plane, which
divides all of R3 into two open half-spaces. The normal vector n we choose to
de…ne the plane can be replaced by n for the same purpose, but if we want
an orientation on the plane we choose one or the other. For surfaces that are
"closed" (such as spheres) there is sometimes a preferred orientation depending
on whether we want the normal vector to point inside or outside. The precise definition of inside/outside requires topology but for spheres and other basic surfaces
the intuitive interpretation su¢ ces. Once this preferred orientation is chosen (it
is conventional to choose the outward normal to de…ne positive orientation of a
closed surface) then any parameterization (u; v) can be classi…ed as orientationpreserving or orientation-reversing depending on whether Tu Tv points to the
outside or inside. This is why we do not want Tu Tv = 0 at any point so that
its orientation cannot change. For some surfaces it is not possible to …nd so
that Tu Tv never vanishes. The Möbius band is an example. Such surfaces
are called non-orientable. The concept of orientation-preserving/reversing is best
understood by comparing one parameterization to another. We saw that we were
able to re-parameterize a given path and decide whether the process changed the
orientation of the curve. The same idea applies to surfaces: If parameterizes
S on domain D and the transformation T : D ! D then
T : D ! S is a
reparameterization of S. We can compare the direction of the normal vectors at
any point provided by these two parameterizations to see if they point in the same
direction. This idea is developed further in a course on di¤erential geometry. For
our purposes it su¢ ces to look at some common examples.
67
Example.
(u; v) = (cos u sin v; sin u sin v; cos v) ;
D = [0; 2 ] [0; ]
parameterizes the unit sphere and
Tu
Tv =
cos u sin2 v; sin u sin2 v; sin v cos v
= ( sin v) (cos u sin v; sin u sin v; cos v)
The vector (cos u sin v) i + (sin u sin v) j + (cos v) k is the directed
line segment from the origin to the point (u; v) on the sphere, which
points outward. Since v 2 [0; ] the factor ( sin v) 0, so Tu Tv
points inward, toward the origin. Now let T : D ! D by
T (u; v) = (v; u)
Then
and
T (u; v) = (cos v sin u; sin v sin u; cos u) on [0; ]
Tu
[0; 2 ]
Tv = (sin u) (cos v sin u; sin v sin u; cos u)
which points outward since sin u 0. The parameterization
reverses the orientation of . Note that
T
0 1
1 0
DT =
which has negative determinant.
Assuming we work with regular parameterizations and a continuous vector …eld
de…ned on S we have:
Theorem. Let S be parameterized by 1 and let
parameterization. Then
ZZ
ZZ
F dS =
F dS
2
depending on whether
tation of 1 .
2
be a re-
1
2
preserves (+) or reverses ( ) the orien-
68
Application. Gauss’ Law. The reason why the concept of orientation is sometimes ascribed to the surface itself is due to the history of physical laws whose
mathematical description requires surface integrals. When the vector …eld is an
electric …eld E then
ZZ
E dS
S
is the ‡ux of the …eld over the surface S. When S has an inside and an outside, as
does a sphere, Gauss’Law says that the ‡ux is the net charge Q enclosed by S. It
is conventional to give S a positive orientation by assigning the outward normal
vector at each point. At points on S the …eld E may be directed toward the
outside, toward the inside, or may be tangential to the surface. As an example,
suppose
E(x; y; z) = xz; yz; z 2 1
and let S be the unit sphere with parameterization (u; v) = (cos v sin u; sin v sin u; cos u)
on [0; ] [0; 2 ]. Then Tu Tv = (sin u) (cos v sin u; sin v sin u; cos u) and
E ( (u; v)) = cos u sin u cos v; cos u sin u sin v;
sin2 u
and so
E ( (u; v)) (Tu
Tv ) = 0
Therefore the net charge enclosed by S is Q = 0. This is because E ( (u; v)) =
(sin u) Tu and so the electric …eld is tangent to S at every point on the sphere.
69
Suggested Exercises for Chapter 8
The following exercises are not to be handed in. They represent skills required for
basic mastery.
8.1 (pages 437-439):
7; 11; 13; 19; 23
8.2 (pages 450-452):
5; 7; 11; 17; 23
8.3 (pages 459-461):
1; 3; 11; 13; 17; 19; 23
8.4 (pages 474-476):
1; 3; 5; 13
Fourth Graded Assignment
Due: December 8
To reinforce written communication skills the Graded Assignment solutions
should be clearly presented in a "bluebook" or provided in PDF format. Late
papers will not be graded.
Fourth Graded Assignment. Do any one of the following:
Page 438: 18
Page 452: 34
Page 461: 24
Page 474: 14
Page 491: 18
70
Green’s Theorem
The idea of the Fundamental Theorem of Calculus is that the integral of a di¤erential form over a region is equal to the integral of a related di¤erential form over
the boundary of the region. How these di¤erential forms are related is explained
by a generalization of the relation between derivatives and anti-derivatives. For
regions in R2 this relation is described by Green’s Theorem. Consider the following
example.
Example. Let F(x; y) = (x2 + y 2 ; x2 y 2 ) and let D be the region
bounded above by the unit circle and below by the x-axis. Then @D
is the union of paths
c1 (t) = (cos t; sin t) ; t 2 [0; ]
c2 (t) = (t; 0) ; t 2 [ 1; 1]
With this parameterization @D is traversed in the counter-clockwise
direction and
Z 1
Z
Z
Z
t2 dt
(cos t cos 2t sin t) dt +
F ds+
F ds =
c1
1
0
c2
=
2
2+ =
3
4
3
Now consider the di¤erential form
@Q
@x
@P
@y
dxdy
where P (x; y) = x2 + y 2 and Q(x; y) = x2 y 2 . The integral of this
form over D is
ZZ
2
(x y) dxdy
D
Z Z 1
4
= 2
r2 (cos
sin ) drd =
3
0
0
71
This example seems to indicate that the line integral of the vector …eld F over
@P
the boundary of D is equal to the integral of the scalar function @Q
over D
@x
@y
itself. Note however that the sign of the line integral depends on the orientation
of the path we choose to describe @D whereas the scalar integral over D does not.
Therefore, in order to state the underlying theorem it is necessary to observe the
usual conditions:
1) D is a simple (or piecewise simple) region;
2) @D = C positively (counter-clockwise) oriented (when C is positively oriented we write C + and when C is negatively oriented we write C );
3) F is a C 1 vector …eld (P and Q are C 1 functions).
With these hypotheses Green’s Theorem is the statement
ZZ
Z
@Q @P
dxdy
P dx + Qdy =
@x
@y
D
@D
Corollary. If @D is a simple closed curve then
Z
1
xdy ydx
A(D) =
2 @D
Proof. Let F(x; y) = ( y; x) so that @Q
@x
Theorem the given line integral is equal to
ZZ
1
2dxdy = A(D)
2 D
@P
@y
= 2. By Green’s
The vector …eld F(x; y) = ( y; x) occurs frequently in vector analysis of the
plane. Note that it is obtained by assigning to each point represented by xi + yj
the counter-clockwise rotation of this vector by 2 . The proof of Green’s Theorem
follows from looking at the forms P dx and Qdy individually and integrating over
D as an elementary region. Looking at D as a y-simple region with C1+ ; C2+
72
parameterized by (x;
Z bZ
a
1 (x))
2 (x)
1 (x)
and (x;
@P
dydx =
@y
=
2 (x)),
Z
respectively, we have
b
[P (x;
2 (x))
a
Z
Z
P dx
C2+
=
Z
=
1 (x))]
P dx
C1+
P dx
Z
P dx
C1+
C2
Z
P (x;
P dx
@D
The last step follows because as we traverse @D counter-clockwise we move from
+
+
RC1 to C2 and back to C1 via vertical segments x = b and x = a along which
P dx = 0 since x is constant. Similarly, looking at D with bounding curves
x
y d we …nd
1 (y)
2 (y) and c
Z
c
d
Z
2 (x)
1 (x)
@Q
dxdy =
@x
Z
Qdy
@D
where this time no minus sign is introduced because the counter-clockwise movement implies a positive orientation for the parameterization (y; 2 (y)) and a negative orientation for the parameterization (y; 1 (y)). Adding the integrals yields
Z
ZZ
@Q @P
P dx + Qdy =
dxdy
@x
@y
@D
D
Application.
R The area of the region D bounded by the ellipse
y2
1
+ b2 = 1 is 2 @D xdy ydx. A counter-clockwise parameterization
of @D is
(a cos t; b sin t) ; t 2 [0; 2 ]
x2
a2
Then
Z
1
xdy
2 @D
Z
1 2
ydx =
( b sin t; a cos t) ( a sin t; b cos t) dt
2 0
Z
ab 2
sin2 t + cos2 t dt = ab
=
2 0
73
The connection between Green’s Theorem and other forms of the FTC is made
by interpreting the integrand in terms of vector operators. If we view the vector
…eld in R3 by letting R(x; y; z) = 0
F(x; y; z) = (P; Q; 0)
then the curl of F is
r
F=
0; 0;
@Q
@x
@P
@y
because the functions P and Q do not depend on z. This says that the integrand
in the integral over D is
(r F) k
The fact that k is the normal vector to the plane containing D is the key to Stokes’
Theorem.
Divergence
The divergence of a planar vector …eld F(x; y) = (P; Q) is
r F=
@Q
@P
+
@x
@y
The line integral in Green’s Theorem is
Z b
Z
Z
P dx + Qdy =
F ds =
F (c(t)) c0 (t)dt
@D
a
@D
for a positively-oriented parameterization c(t) = (x(t); y(t)). The tangent vector
c0 (t) = (x0 (t); y 0 (t)) points along the counter-clockwise direction of motion along
the curve @D and there are two normal directions to c0 (t). The one obtained by
rotating the tangent vector clockwise through 2 will point away from D. The unit
vector in this direction is called the outward unit normal
n=
(y 0 (t); x0 (t))
jc0 (t)j
74
Note that
F (c(t)) n = [P (c(t)); Q(c(t))]
and so the path integral
Z
Z b
[P (x(t); y(t))y 0 (t)
F nds =
@D
Za
=
Qdx + P dy
(y 0 (t); x0 (t))
jc0 (t)j
Q (x(t); y(t))x0 (t)] dt
@D
which by Green’s Theorem is equal to
ZZ
ZZ
@Q
@P
+
dxdy =
r Fdxdy
@x
@y
D
D
This is an alternate form of Green’s Theorem that is often used in the solution of
partial di¤erential equations:
ZZ
Z
div FdA
F nds =
D
@D
It is equivalent to our other formulation
ZZ
Z
(curl F k) dA
F ds =
@D
D
Stokes’Theorem
Suppose that the region D in Green’s Theorem were mapped into R3 in such a
way that the image of @D is a simple closed curve. We would expect the line
integral of a vector …eld F over this curve to be related to the integral of some
vector …eld over the surface that is the image of D. A good guess would be that
this vector …eld is the curl of F because we were able to write the statement of
Green’s Theorem in terms of the curl of the vector …eld. We can ensure that the
image of @D is a simple closed curve by using a one-to-one mapping. This will
result in a parameterization with
(D) = S
(@D) = @S
75
with @S a simple closed curve. Then
ZZ
Z
(r F) dS =
F ds
@S
This is Stokes’ Theorem, a more general form of the FTC. Since both of the
integrals in the theorem are oriented we must make sure that the orientation of
the boundary curve "matches" the orientation of the surface. In case the surface
S is the graph of a function over D this would mean that a counter-clockwise
orientation for @S corresponds to an "upward" normal to S.
Example. Let F (x; y; z) = (y; z; x) and let S be the upper unit
hemisphere. Then @S is the unit circle in the xy-plane. We can
think of S as the image of D = 0; 2
[0; 2 ] under (u; v) =
(cos v sin u; sin v sin u; cos u) and @S becomes (cos v; sin v; 0), v 2 [0; 2 ].
Then
Z
Z 2
F ds =
(sin v; 0; cos v) ( sin v; cos v; 0) dv
@S
0
Z 2
=
sin2 v dv =
0
ZZ
Since r
(r
F is the constant vector …eld ( 1; 1; 1) we have
F) dS =
Z
0
=
2
Z
Z
0
2
0
2
( 1; 1; 1) (cos v sin u; sin v sin u; cos u) (sin u) dudv
Z
2
cos v sin2 u + sin v sin2 u + cos u sin u dudv =
0
Example. Let S be the graph of f (x; y) = x2 y 2 over the domain
D = [0; 1] [0; 1]. Then S can be given the graph parameterization
Tx
Tx
Ty
Ty
x; y; x2 y 2
= (1; 0; 2x)
= (0; 1; 2y)
= ( 2x; 2y; 1)
76
Let F (x; y; z) = (xy; yz; xz). Then
Z
1
0
Z
1
0
r F (x; y; z) = ( y; z; x)
r F ( (x; y)) =
y; y 2 x2 ; x
r F ( (x; y)) (Tx Ty ) = 2x2 y + 2xy x 2y 3
1
2xy 2x2 y x + 2y 3 dxdy =
6
Now @D = C1 [ C2 [ C3 [ C4 with the positively-oriented parameterization
c1 (t)
c2 (t)
c3 (t)
c4 (t)
t; 0; t2 ; t 2 [0; 1]
1; t; 1 t2 ; t 2 [0; 1]
t; 1; t2 1 ; t 2 [1; 0]
0; t; t2 ; t 2 [1; 0]
=
=
=
=
Then
Z
0
1
4
2t dt +
Z
0
1
3
t
t dt
Z
1
2t
4
0
2
2t + t dt
Z
0
Z
F ds =
@S
1
t3 dt =
1
6
The boundary of S = (@D) receives its orientation by moving
counter-clockwise around the boundary of the square in the xy-plane.
Note that Tx Ty = k (fx i + fy j).
77
Normal vectors at
1 1
; ;0
2 2
Example. We saw that the surface
D = [0; 1]
0; 2 is part of a helicoid:
and
1 1 1
; ;
3 6 12
(r; ) = (r cos ; r sin ; ) on
1.5
1.0
z
0.5
0.0
0.0 0.0 0.2
0.2
0.4
y
0.4
0.6
0.8
x
1.0
78
0.6
0.8
1.0
We have
Tr = (cos ; sin ; 0)
T = ( r sin ; r cos ; 1)
T = (sin ; cos ; r)
Tr
This portion of the surface is bounded by the piecewise smooth
curve that has positive orientation corresponding to the boundary of
the quarter sector of the unit circle traversed counter-clockwise. The
parameterization of @S = C1 [ C2 [ C3 [ C4 could be
c1 (t) = (t; 0; 0) ; t 2 [0; 1]
h
c2 (t) = (cos t; sin t; t); t 2 0;
; t 2 [1; 0]
2
h
i
c4 (t) = (0; 0; t) ; t 2
;0
2
c3 (t) =
0; t;
Let F(x; y; z) = (z; x; y). Then r
ZZ
(r
F) dS =
S
Z
2
0
Z
2
i
F = (1; 1; 1) and so
1
(sin
cos + r) drd =
0
4
and
Z
F ds =
@S
Z
F ds+
c1
Z
F ds+
c2
Z
F ds+
c3
Z
F ds
c4
Note that the integrands for C1 ; C3 ; C4 are each zero so
Z
c2
F ds =
Z
2
cos2 t + sin t
0
t sin t dt =
4
The proof of Stokes’Theorem is based on Green’s Theorem and the equality of
mixed partial derivatives. The proof when S is a graph is easy to read (page 441)
and the proof for general parameterizations is very similar.
79
Corollary to Stokes’ Theorem. Let S be a surface that forms the boundary
1
of a region in R3 (so
RR S has no boundary). If F is a C vector …eld on a region
containing S then S (r F) dS = 0.
Proof. Let C be a closed curve on S and let S = S1 [S2 such that @S1 = @S2 =
C. If S is oriented by the normal vector in the integrand then the orientations on
C consistent with Stokes’Theorem are opposite for S1 and S2 . Therefore
ZZ
ZZ
(r F) dS =
(r F) dS
S
S1 [S2
Z
Z
F ds+
F ds = 0
=
C+
C
As an example, let S be the unit sphere and let F (x; y; z) = (y; z; x), as above.
With S oriented by this parameterization the normal vector is directed away from
the center of the sphere (outward normal). Let C be the unit circle in the xyplane with the parameterization c(v) =(cos v; sin v; 0) which is positively oriented
for the upper hemisphere S1 but negatively oriented for the lower hemisphere S2 .
ZZ
Z 2 Z
2
cos v sin2 u + sin v sin2 u + cos u sin u dudv =
(r F) dS =
0
S
0
ZZ 1
Z 2 Z
(r F) dS =
cos v sin2 u + sin v sin2 u + cos u sin u dudv =
S2
0
2
ZZ
(r F) dS =
+ =0
S
Path Independence
Many C 1 vector …elds on R3 that occur in physical theories have domains that omit
a single point or even a …nite set of points. If such a …eld F (x; y; z) = (P; Q; R)
is the gradient of a scalar …eld f with the same domain then r F = 0. The
veri…cation is a consequence of the equality of mixed partial derivatives. The fact
that the converse is also true is related to a deeper principle that is usually stated
as a list of four equivalent conditions that characterize conservative vector …elds.
80
Let c1 and c2 be paths for two simple curves in the domain of F
that share endpoints. Then c1 followed by c2 is
R a path for a simple
closed curve. Suppose F has the property that C F ds = 0 for any
simple closed curve C within the domain of F. Then apparently
Z
Z
F ds =
F ds
c1
c2
that is, the line integral of F is independent of the path taken from
one point to another; its value only depends on the endpoints. We saw
this before when computing the line integral of a gradient …eld. Here
is the analysis of what is happening.
Suppose (0; 0; 0) is in the domain of F and let C be any simple
curve in the domain of F joining it to an arbitrary point (x; y; z). If
the line integral of F is independent of the path taken from one point
to another then
Z
f (x; y; z) = F ds
c
de…nes a function f whose values we can compute by choosing any
path, such as the segment from (0; 0; 0) to (x; 0; 0) followed by the
segment from (x; 0; 0) to (x; y; 0) and then the segment from (x; y; 0)
to (x; y; z), for which we …nd
Z x
Z y
Z z
f (x; y; z) =
P (t; 0; 0) dt +
Q (x; t; 0) dt +
R (x; y; t) dt
0
0
0
By the FTC the partial derivative of f with respect to z is R(x; y; z)
because the …rst two integrals do not depend on z. Similarly, the
segment from (0; 0; 0) to (0; 0; z) followed by the segment from (0; 0; z)
to (x; 0; z) and then the segment from (x; 0; z) to (x; y; z) gives
Z z
Z x
Z y
f (x; y; z) =
R (0; 0; t) dt +
P (t; 0; z) dt +
Q (x; t; z) dt
0
0
from which we …nd
@f
= Q (x; y; z)
@y
81
0
Finally, the segment from (0; 0; 0) to (0; y; 0) followed by the segment from (0; y; 0) to (0; y; z) and then the segment from (0; y; z) to
(x; y; z) gives
Z x
Z z
Z y
P (t; y; z) dt
R (0; y; t) dt +
Q (0; t; 0) dt +
f (x; y; z) =
0
0
0
from which we …nd
@f
= P (x; y; z)
@x
Thus, path independence implies
F = rf
and since the curl of a gradient is 0 we have
r
F=0
Now suppose C is any simple closed curve in the domain of F and
let S be a surface in the domain of F with @S = C and oriented so
that we can apply Stokes’Theorem:
Z
ZZ
F ds =
(r F) dS
C
S
R
Thus, the fact that r F = 0 implies C F ds = 0. This was our
original assumption so apparently all four conditions are equivalent.
The equality of mixed partial derivatives also shows that r F = 0 is a necessary
condition for F = (P; Q; R) to be the curl of some vector …eld. Here the converse
~ R
~ with
is true provided F is C 1 on all of R3 . In fact, if G = P~ ; Q;
Z z
Z y
~
P (x; y; z) =
Q(x; y; t)dt
R(x; t; 0)dt
0
0
Z z
~
Q (x; y; z) =
P (x; y; t)dt
0
~ (x; y; z) = 0
R
then r F = 0 implies F = r
G.
82
Gauss’Theorem
Another form of the FTC states that for a surface S, such as a sphere, that bounds
a region W in R3 the integral of a C 1 vector …eld over S with outward normal is
equal to the integral of its divergence over W :
ZZ
ZZZ
F dS =
(r F) dV
S=@W
W
The proof of Gauss’ Theorem for elementary regions is a generalization of the
case where W is a solid rectangular box. Let F(x; y; z) = (P; Q; R) so r F =
Px + Qy + Rz and suppose W = [0; 1] [0; 1] [0; 1]. Then @W is the union of
six unit-square faces with normal vectors i; j; k. On the face S with normal
k we have
ZZ
Z 1Z 1
F dS =
[P (x; y; 1); Q(x; y; 1); R(x; y; 1)] kdxdy
S
0
0
Z 1Z 1
=
R(x; y; 1)dxdy
0
0
The other …ve surface integrals are computed similarly with the result
ZZ
Z 1Z 1
F dS =
[R(x; y; 1) R(x; y; 0)] dxdy
@W
0
0
Z 1Z 1
+
[Q(x; 1; z) Q(x; 0; z)] dxdz
0
0
Z 1Z 1
+
[P (1; y; z) P (0; y; z)] dydz
0
0
The volume integral is
ZZZ
Z 1Z 1Z 1
(Px + Qy + Rz ) dzdydx
(r F) dV =
W
0
0
0
Z 1Z 1Z 1
Z 1Z
=
(Px + Qy ) dzdydx +
0
0
0
0
1
[R(x; y; 1)
R(x; y; 0)] dxdy
0
since R is a z anti-derivative for Rz . Similarly
Z 1Z 1Z 1
Z 1Z 1Z 1
Z 1Z 1
(Px + Qy ) dzdydx =
Px dzdydx +
[Q(x; 1; z) Q(x; 0; z)] dxdz
0
0
0
0
0
0
0
0
Z 1Z 1
Z 1Z 1
=
[P (1; y; z) P (0; y; z)] dydz +
[Q(x; 1; z) Q(x; 0; z)
0
0
83
0
0
Thus, we have used Fubini’s Theorem to prove
ZZZ
ZZ
(r F) dV =
W
F dS
@W
The proof for any rectangular box is identical, and the proof for any elementary
region W is similar but we need to use function limits for the integrals instead of
constants (see pages 464-465).
Gauss’ Theorem applies to any region W that is a …nite union of elementary
regions. Often a single parameterized surface will bound such a region. Consider
the torus
(u; v) = ((2 + cos v) cos u; (2 + cos v) sin u; sin v)
Tu Tv = (cos v + 2) (cos v cos u; cos v sin u; sin v)
(u; v) 2 [0; 2 ] [0; 2 ]
The vector Tu
Tv is an outward normal for this surface.
-3
-3
1
-2
-1
z
0
0 0
y1 -1
-2
-1
1x
2
2
3
3
Torus: z 2 = (r
1) (3
r)
The vector …eld F(x; y; z) = (x; y; z) has constant divergence
r F=3
84
so
ZZZ
(r F) dV
W
= 3V (W )
ZZ
1
V (W ) =
F dS
3 @W
Z Z 2
1 2
=
(2 + cos v) (1 + 2 cos v) dudv
3 0
0
Z 2
2
=
(2 + cos v) (1 + 2 cos v) dv = 4 2
3 0
To compute V (W ) without Gauss’Theorem we could use the transformation
x(p; u; v)
y(p; u; v)
z(p; u; v)
p
=
=
=
2
(2 + p cos v) cos u
(2 + p cos v) sin u
p sin v
[0; 1]
where p is the distance from the center of a circle obtained from a vertical section
for a given value of u. Then
@ (x; y; z)
= p (2 + p cos v)
@ (p; u; v)
so
V (W ) =
=
ZZZ
Z
0
2
dV
W
Z
0
2
Z
1
p (2 + p cos v) dpdudv = 4
2
0
Corollary to Gauss’Theorem.
If F(x; y; z) is tangent to @W
ZZZ
for every (x; y; z) 2 @W then
(r F) dV = 0.
W
RR
RR
Proof.
F dS = @W (F n) dS where n is normal to the
@W
surface at the point where F is evaluated. If F at each point is tangent
to the surface then F n = 0.
85
Study Guide for Final Exam
Cavalieri’s Principle
Average value and Mean-Value Theorem for integrals of scalar functions
Change-of-Variable Theorem (Jacobian)
Path integrals and surface integrals of scalar functions:
Arc length and surface area
Line integrals and surface integrals of vector …elds:
Green’s Theorem
Stokes’Theorem
Gauss’Theorem
86