CHAPTER 16 VECTOR CALCULUS
16.1 Vector Fields
• Vector Field on R2
Let D be a set in R2 (a plane region). A vector field
on R2 is a function F that assigns to each point (x, y) in D a two-dimensional
vector F(x, y).
• Vector Field on R3
Let E be a subset of R3 . A vector field on R3 is a
function F that assigns to each point (x, y, z) in E a three-dimensional vector
F(x, y, z).
1
2
CHAPTER 16
• Example A vector field on R2 is defined by
F(x, y) = −yi + xj
Decribe F by sketching some of the vectors F(x, y).
(x, y)
F(x, y)
(x, y)
(1, 0)
(−1, 0)
(2, 2)
(−2, −2)
(3, 0)
(−3, 0)
(0, 1)
(0, −1)
(−2, 2)
(2, −2)
(0, 3)
(0, −3)
F(x, y)
CHAPTER 16
• Some other vector fields:
x
y
F(x, y) = p
i+ p
j
x2 + y 2
x2 + y 2
Newton’s Law of Gravitation: Gravitational force field
mM G
F(r) = −
r, r = hx, y, zi
krk3
3
4
CHAPTER 16
• The Gradient of a Scalar Field
Let f determine a scalar field and suppose
that f is differentiable. The gradient of f , denoted by ∇f , is the vector field
given by
∂f
∂f
i+
j
∂x
∂y
∂f
∂f
∂f
∇f (x, y, z) =
i+
j+
k
∂x
∂y
∂z
∇f (x, y) =
• Example Find the gradient vector field of f (x, y) = x2 y − y 3 .
CHAPTER 16
• Conservative Vector Field
5
A vector field F that is the gradient of a scalar
function f is called a conservative vector field. f is called the potential function
of the vector field F.
F = ∇f
• Example Let F be the force resulting from an inverse square law; that is
F(x, y, z) = −c
r
xi + yj + zk
=
−c
3 .
2
2
2
krk3
(x + y + z ) 2
Show that
c
f (x, y, z) = p
x2 + y 2 + z 2
is a potential function of F. Is F a conservative vector field?
Exercises
Page 1073
11, 13, 15, 17, 23.
6
CHAPTER 16
16.2 Line Integrals
• Let C be a smooth plane curve given parametrically by
x = x(t),
y = y(t),
a≤t≤b
where x0 and y 0 are continuous but not simultaneously zero on the interval (a, b).
b−a
. The curve C
n
is then divided into n subarcs Pi−1 Pi in which the points Pi corresponds to ti .
• Divide [a, b] into sub-intervals [ti−1 , ti ] of equal width ∆t =
>
• Let ∆si = Pi−1 Pi . Choose a sample point t∗i ∈ [ti−1 , ti ] and consider the
n
X
Riemann sum
f (x∗i , yi∗ ) ∆si where x∗i = x(t∗i ), yi∗ = y(t∗i ).
i=1
• Definition
The line integral of f along C is
Z
n
X
f (x, y) ds = lim
f (x∗i , yi∗ ) ∆si
C
if the limit exists.
n→∞
i=1
CHAPTER 16
7
• Theorem
Z b
Z
f (x, y) ds =
C
f (x(t), y(t))
q
[x0 (t)]2 + [y 0 (t)]2 dt
a
Z
• If f (x, y) ≥ 0,
f (x, y) ds represents the area of one side of the “fence” or
C
“curtain” whose base is C and whose height above the point (x, y) is f (x, y).
8
CHAPTER 16
Z
• Example Evaluate
(2 + x2 y) ds, where C is determined by the parametric
C
equations x = cos t, y = sin t, 0 ≤ t ≤ π. Also, show that the parametrization
√
x = x, y = 1 − x2 , −1 ≤ x ≤ 1, gives the same value.
CHAPTER 16
9
Z
• Example Evaluate
2x ds, where C consists of the arc C1 of the parabola
C
y = x2 from (0, 0) to (1, 1), followed by the vertical line segment C2 from (1, 1)
to (1, 2).
10
CHAPTER 16
• Line Integrals with respect to x and y
Z
f (x, y) dx
definition
=
n→∞
C
Z
f (x, y) dy
lim
definition
=
lim
n→∞
C
Z
• Example Evaluate
n
X
i=1
n
X
theorem
f (x∗i , yi∗ )∆xi =
Z b
f (x(t), y(t))x0 (t) dt
a
f (x∗i , yi∗ )∆yi
i=1
theorem
Z b
=
f (x(t), y(t))y 0 (t) dt
a
y 2 dx + xdy, where
C
(a) C = C1 is the line segment from (−5, −3) to (0, 2);
(b) C = C2 is the arc of the parabola x = 4 − y 2 from (−5, −3) to (0, 2).
CHAPTER 16
11
• In general, a given parametrization determines an orientation of a curve C.
• C is the curve with the initial point A corresponds to the parameter t = a and
the terminal point B corresponds to the parameter t = b.
• −C denotes the curve consisting of the same points as C, but with the opposite
orientation.
• Theorem
Z
Z
f (x, y) dx = −
−C
f (x, y) dx
C
Z
Z
f (x, y) dy = −
−C
f (x, y) dy
C
12
CHAPTER 16
• Line Integrals in Three-Space
For a smooth curve r(t) in three-space given
parametrically by
x = x(t),
y = y(t),
Z b
Z
f (x, y, z) ds =
C
z = z(t),
a≤t≤b
q
f (x(t), y(t), z(t)) [x0 (t)]2 + [y 0 (t)]2 + [z 0 (t)]2 dt
a
Z b
=
f (r(t))kr0 (t)k dt
a
Z
• Example Evaluate
y sin z ds where C is the circular helix given by the
C
equations x = cos t, y = sin t, z = t, 0 ≤ t ≤ 2π.
CHAPTER 16
13
Z
• Example Evaluate
y dx + z dy + x dz, where C consists of the line segment
C
C1 from (2, 0, 0) to (3, 4, 5), followed by the vertical line segment C2 from (3, 4, 5)
to (3, 4, 0).
14
CHAPTER 16
• Line Integrals of Vector Fields – Work Done
Suppose that
F(x, y, z) = F1 (x, y, z)i + F2 (x, y, z)j + F3 (x, y, z)k
is a continuous force field on R3 . We wish to compute the work done by this
force in moving a particle along a smooth curve C: r(t), a ≤ t ≤ b. Then
Z b
Z
F · dr =
W =
C
0
Z
F(r(t)) · r (t) dt =
a
Z
F · T ds =
C
F1 dx + F2 dy + F3 dz
C
where T(x, y, z) is the unit tangent vector at the point (x, y, z) on C.
• Recall:
work done = force × distance
dr =
dr
× dt = r0 (t) dt
dt
T ds = dr
F · dr = hF1 , F2 , F3 i · hdx, dy, dzi = F1 dx + F2 dy + F3 dz
CHAPTER 16
15
• Example Find the work done by the force F(x, y) = x2 i − xyj in moving a
π
particle along the quarter-circle r(t) = cos ti + sin tj, 0 ≤ t ≤ .
2
16
CHAPTER 16
Z
• Example Evaluate
F · dr, where F(x, y, z) = xyi + yzj + zxk and C is the
C
curve r(t) = ti + t2 j + t3 k, 0 ≤ t ≤ 1.
Exercises
Page 1084
1, 3, 5, 7, 11, 13, 15, 19, 21, 39, 41.
CHAPTER 16
16.3 The Fundamental Theorem for Line Integrals
• Fundamental Theorem for Line Integrals
Let C be a piecewise smooth curve given parametrically by r = r(t), a ≤ t ≤
b. If f is continuously differentiable on an open set containing C, then
Z
∇f (r) · dr = f (r(b)) − f (r(a))
C
17
18
CHAPTER 16
Z
r
, and C is a piecewise
krk3
C
smooth curve from (3, 4, 12) to (2, 2, 0) that does not pass through the origin.
• Example Calculate
F(r) · dr, where F(r) = −c
CHAPTER 16
19
• We say that a set D is connected if any two points in D can be joined by a
piecewise smooth curve lying entirely in D.
• Example Which of the following sets are connected?
• A curve is called closed if its terminal point coincides with its initial point.
• Definition
Z
F(r) · dr is said to be independent of path in a con-
The line integral
C
nected set D if for any two points A and B in D, the line integral has the
same value for every path C in D that is from A to B.
20
CHAPTER 16
• Theorem
Z
The line integral
F · dr is independent of path in D ⊂ Rn (n ≥ 2) if and
C
Z
only if
F · dr = 0 for every closed path C in D.
C
CHAPTER 16
• Independence of Path Theorem
Let F(r) be continuous on an open connected set D ⊂ Rn (n ≥ 2). Then
Z
the line integral
F(r) · dr is independent of path in D if and only if
C
F(r) = ∇f (r) for some scalar function f ; that is, if and only if F is a
conservative vector field on D.
21
22
CHAPTER 16
• Summary
Let F(r) be continuous on an open connected set D ⊂ Rn (n ≥ 2). Then
the following conditions are equivalent:
1. F = ∇f for some f , i.e., F is conservative.
Z
2.
F(r) · dr is independent of path in D.
ZC
3.
F(r) · dr = 0 for every closed path in D.
C
• A simple curve is a curve that does not intersect itself anywhere between its
endpoints.
• A simply-connected domain is a connected domain where one can continuously shrink any simple closed curve into a point while remaining in the domain.
For two-dimensional regions, a simply connected domain is one without holes
in it.
CHAPTER 16
23
• Theorem
Let F = F1 (x, y)i + F2 (x, y)j be a vector field on an open simply-connected
domain D ⊂ R2 . Suppose that F1 and F2 have continuous first-order partial
derivatives. Then F is conservative if and only if
∂F1
∂F2
=
.
∂y
∂x
• Remark: To have
∂F2
∂F1
=
. ”
∂y
∂x
domain D need not be open simply-connected.
“ F is conservative implies
• Extension to three-space:
Let F = F1 i + F2 j + F3 k be a vector field on an open simply-connected
domain D ⊂ R3 . Suppose that F1 , F2 and F3 have continuous first-order
partial derivatives. Then F is conservative if and only if
∂F1
∂F2
=
,
∂y
∂x
∂F1
∂F3
=
,
∂z
∂x
∂F2
∂F3
=
.
∂z
∂y
24
CHAPTER 16
• Example
(a) Determine whether
F = 4x3 + 9x2 y 2 i + 6x3 y + 6y 5 j
is conservative. If yes, find its potential function.
Z
(b) Calculate
C
0 ≤ t ≤ π.
F(r)·dr, where C is the curve given by r(t) = et sin ti+et cos tj,
CHAPTER 16
• Example Show that
F = (ex cos y + yz) i + (xz − ex sin y) j + xyk
is conservative. Hence, find f such that F = ∇f .
25
26
CHAPTER 16
• Conservation of Energy
• Consider a continuous force field F that moves an object along a path C given
by r(t), a ≤ t ≤ b, where r(a) = A is the initial point and r(b) = B is the
terminal point of C.
• According to Newton’s Second Law of Motion: F(r(t)) = mr00 (t), where m is
the mass of the object.
1
1
• Work Done, W = mkv(b)k2 − mkv(a)k2 = K(B)−K(A)
2
2
(Kinetic Energy)
• In physics, the potential energy, P , of an object is defined as F = −∇P .
• Work Done, W = P (r(a)) − P (r(b)) = P (A) − P (B)
(Potential Energy)
• Work Done, W = K(B) − K(A) = P (A) − P (B)
• Law of Conservation of Energy: P (A) + K(A) = P (B) + K(B)
• This is the reason why the vector field is called conservative.
Exercises
Page 1094
1, 5, 7, 9, 11, 13, 15, 17, 19, 35.
CHAPTER 16
27
16.4 Green’s Theorem
• The positive orientation of a simple closed curve C refers to a single counterclockwise traversal of C. Thus if C is given by the vector function r(t),
a ≤ t ≤ b, then the region D is always on the left as the point r(t) traverses C.
• Green’s Theorem
Let C be a positively oriented, piecewise smooth, simple closed curve that
forms the boundary of a region D in the xy-plane. If P (x, y) and Q(x, y)
have continuous partial derivatives on an open region that contains D, then
ZZ Z
∂Q ∂P
−
dA =
P dx + Qdy.
∂x
∂y
C
D
• Remark: The notation
I
P dx + Qdy
C
is sometimes used in place of
R
C P dx + Qdy to indicate the positive orientation
of C. Another notation for the positively oriented boundary curve of D is ∂D,
so we can write
ZZ D
∂Q ∂P
−
∂x
∂y
Z
P dx + Qdy
dA =
∂D
28
CHAPTER 16
• Example Let C be the boundary of the triangle with vertices (0, 0), (1, 2) and
(0, 2). Calculate
I
4x2 ydx + 2ydy
C
(a) by the direct method;
(b) by Green’s Theorem.
CHAPTER 16
• Example Calculate
I
sin x
3y − e
p
4
dx + 7x + y + 1 dy
C
where C is the circle x2 + y 2 = 9.
29
30
CHAPTER 16
I
• Example Evaluate
y 2 dx + 3xydy, where C is the boundary of the semian-
C
nular region D in the upper half-plane between the circles x2 + y 2 = 1 and
x2 + y 2 = 4.
What if D is the annular region between x2 + y 2 = 1 and x2 + y 2 = 4?
• Green’s Theorem can be extended to apply to regions with holes, that is, regions
that are not simply-connected.
CHAPTER 16
31
Z
−yi + xj
• Example If F(x, y) = 2
, show that
F · dr = 2π for any positively
x + y2
C
oriented simple closed path that encloses the origin.
Exercises
Page 1101
3, 7, 9, 17.
32
CHAPTER 16
16.5 Curl and Divergence
• The operator ∇ is a vector operator
∇=
∂
∂
∂
i+ j+ k
∂x
∂y
∂z
• Curl
Let F = F1 i+F2 j+F3 k be a vector field for which the first partial derivatives
exist. Then
i
j k
∂ ∂ ∂
curl F =∇ × F =
∂x ∂y ∂z
F1 F2 F3
∂F3 ∂F2
∂F1 ∂F3
∂F2 ∂F1
=
−
i+
−
j+
−
k
∂y
∂z
∂z
∂x
∂x
∂y
• Example Given
F(x, y, z) = x2 yzi + 3xyz 3 j + x2 − z 2 k
Find curl F.
CHAPTER 16
• Theorem
curl (∇f ) = 0
33
34
CHAPTER 16
• Theorem
1.
If F is a conservative vector field, then curl F = 0.
2.
If F is a vector field defined on all of R3 whose component functions have
continuous partial derivatives and curl F = 0, then F is a conservative
vector field.
• Example Determine whether the vector field
F = xzi + xyzj − y 2 k
is conservative.
CHAPTER 16
35
• Example Determine whether the vector field F = y 2 z 3 i + 2xyz 3 − 2yz j +
3xy 2 z 2 − y 2 + 6z k is conservative. If yes, find a potential function for the
vector field.
36
CHAPTER 16
• Divergence
Let F = F1 i+F2 j+F3 k be a vector field for which the first partial derivatives
exist. Then
div F =∇ · F
∂
∂
∂
=
i + j + k · (F1 i + F2 j + F3 k)
∂x
∂y
∂z
=
∂F1 ∂F2 ∂F3
+
+
∂x
∂y
∂z
• Example Let
F(x, y, z) = x2 yzi + 3xyz 3 j + x2 − z 2 k
Find div F.
Exercises
Page 1109
1, 13, 17.
CHAPTER 16
37
16.6 Parametric Surfaces and Their Areas
• Given
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k,
(u, v) ∈ D,
the set of points of r(u, v) is called a parametric surface.
• Example Describe the surface defined parametrically by
r(u, v) = 2 cos ui + vj + 2 sin uk
38
CHAPTER 16
• Example Find parametric equations for the following surfaces.
(a) The plane that passes through the point P0 and contains two nonparallel
vectors a and b.
(b) The sphere x2 + y 2 + z 2 = a2 .
CHAPTER 16
• Example Find parametric equations for the following surfaces.
(a) The cylinder x2 + y 2 = 4, 0 ≤ z ≤ 1.
(b) The elliptic paraboloid z = x2 + 2y 2 .
p
(c) The surface z = 2 x2 + y 2 .
39
40
CHAPTER 16
• Tangent Planes
Suppose a parametric surface S is traced out by a vector
function
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k
at a point P0 with position vector r(u0 , v0 ).
• Suppose C1 is the curve obtained by setting u = u0 , then the tangent vector to
C1 at P0 is:
rv =
∂x
∂y
∂z
(u0 , v0 )i + (u0 , v0 )j + (u0 , v0 )k
∂v
∂v
∂v
• Suppose C2 is the curve obtained by setting v = v0 , then the tangent vector to
C2 at P0 is:
ru =
∂x
∂y
∂z
(u0 , v0 )i +
(u0 , v0 )j +
(u0 , v0 )k
∂u
∂u
∂u
• If ru × rv 6= 0, then the surface S is called smooth (it has no corners).
• For a smooth surface, the tangent plane is the plane that contains the tangent
vectors ru and rv .
• The vector ru × rv is a normal vector to the tangent plane.
CHAPTER 16
• Example Find the equation of the tangent plane to the surface
r(u, v) = u2 i + v 2 j + (u + 2v)k
at the point (1, 1, 3).
41
42
CHAPTER 16
• Surface Area
The surface area of the parametric surface
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k,
is
ZZ
kru × rv k dA
A(S) =
D
(u, v) ∈ D
CHAPTER 16
• Example Find the surface area of the sphere of radius a.
43
44
CHAPTER 16
• For the special case of a surface S with equation z = f (x, y), where (x, y) lies
in D and f has continuous partial derivatives, then
ZZ
A(S) =
s
1+
D
∂z
∂x
2
+
∂z
∂y
2
dA
CHAPTER 16
45
• Example Find the surface area of the paraboloid z = x2 + y 2 that lies under
the plane z = 9.
Exercises
Page 1120
13, 15, 17, 19, 23, 25, 35, 39, 41, 47, 49.
46
CHAPTER 16
16.7 Surface Integrals
• Let the surface S has a parametrization r(u, v) for (u, v) in D = [a, b] × [c, d].
• Partition D into sub-rectangles Rij by dividing [a, b] into m equal sub-intervals,
and dividing [c, d] into n equal sub-intervals.
• This results in a corresponding partition of the surface S into mn pieces Sij .
Choose a sample point u∗ij , vij∗ in Rij , and let Pij∗ x∗ij , yij∗ , zij∗ = r(u∗ij , vij∗ ) be
the corresponding point on Sij .
• The surface integral of f over S is defined as
ZZ
f (x, y, z) dS = lim
m,n→∞
S
where ∆Sij is the area of Sij .
m X
n
X
i=1 j=1
f r(u∗ij , vij∗ ) ∆Sij
CHAPTER 16
• The area of Sij :
∆Sij ≈ kru (u∗ij , vij∗ ) × rv (u∗ij , vij∗ )k∆u∆v
• Theorem
For a parametric surface S given by r(u, v) where (u, v) ∈ D,
ZZ
ZZ
f (x, y, z) dS =
f (r(u, v))kru (u, v) × rv (u, v)k dA
S
D
• In particular
If S is the graph of function z = f (x, y) where (x, y) ∈ D, then
s
2 2
ZZ
ZZ
∂z
∂z
f (x, y, z) dS =
f (x, y, f (x, y)) 1 +
+
dA
∂x
∂y
S
D
• Observe that
ZZ
ZZ
kru (u, v) × rv (u, v)k dA = A(S)
1 dS =
S
D
47
48
CHAPTER 16
ZZ
• Example Compute
S
x2 dS where S is the unit sphere x2 + y 2 + z 2 = 1.
CHAPTER 16
49
ZZ
• Example Evaluate
2
z dS, where S is the surface whose sides S1 are given
S
2
by the cylinder x + y = 1, whose bottom S2 is the disk x2 + y 2 ≤ 1 in the
plane z = 0, and whose top S3 is the part of the plane z = 1 + x that lies above
S2 .
50
CHAPTER 16
• Oriented Surface
To define surface integrals of vector fields, we need to rule
out nonorientable surfaces such as the Möbius strip, which has only one side.
• From now on, we consider only orientable (two-sided) surfaces. There are two
possible orientations for any orientable surface.
• For the parametric surface r(u, v), the unit normal vector of one orientation is:
n=
ru × rv
kru × rv k
• For the surface z = f (x, y), the unit normal vector is
∂z
∂z
i− j+k
∂x
∂y
n= s
2 2
∂z
∂z
1+
+
∂x
∂y
−
Since the k-component is positive, this gives the upward orientation of the
surface.
and the opposite orientation has unit normal vector −n.
CHAPTER 16
51
• For a closed surface E, the convention is that the positive orienatation is
the one for which the normal vectors point outward from E, and inward-pointing
normals give the negative orientation.
• Surface Integrals of Vector Fields
If F = F1 i + F2 j + F3 k is a continuous
vector field defined on an oriented surface S given by r(u, v), where (u, v) ∈ D
with unit normal vector n, then the surface integral of F over S is
ZZ
ZZ
F · dS =
S
F · n dS
S
This integral is also called the flux of F across S.
• Theorem
ZZ
ZZ
F · dS =
S
If z = f (x, y), then
ZZ
D
ZZ
F · dS =
S
F · (ru × rv ) dA
F·
D
∂z ∂z
− ,− ,1
∂x ∂y
dA
52
CHAPTER 16
• Example Find the flux of the vector field F(x, y, z) = zi + yj + xk across the
unit sphere x2 + y 2 + z 2 = 1.
CHAPTER 16
53
ZZ
• Example Evaluate
F · dS, where F = yi + xj + zk, and S is the boundary
S
of the solid region enclosed by z = 1 − x2 − y 2 and the plane z = 0.
Exercises
Page 1132
5, 9, 13, 17, 21, 23, 27, 31.
54
CHAPTER 16
16.8 Stokes’ Theorem
• Let S be an oriented surface with a continuously varying unit normal n. Assume
that the boundary C is oriented consistently with n. This means that if you
walk in the positive direction around C with your head pointing in the direction
of n, then the surface will always be on your left.
• Stokes’ Theorem
Let S be an oriented piecewise-smooth surface that is bounded by a simple,
closed, piecewise-smooth boundary curve C with positive orientation. Let
F be a vector field whose components have continuous partial derivatives on
an open region in R3 that contains S. Then
Z
ZZ
F · dr =
curl F · dS
C
S
• The positive oriented boundary curve of the oriented surface S is often written
as ∂S, then we have
ZZ
Z
F · dr =
∂S
curl F · dS
S
CHAPTER 16
55
• Example Verify Stokes’ Theorem for F = yi − xj + yzk if S is the paraboloid
z = x2 + y 2 with the circle x2 + y 2 = 1, z = 1 as its boundary by independently
calculating
Z
(a)
F · dr
C
ZZ
curl F · dS
(b)
S
56
CHAPTER 16
Z
• Example Evaluate
F · dr, where F = xyi + yzj + zxk, and C is the triangle
C
with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1), oriented counterclockwise as viewed
from above.
CHAPTER 16
57
ZZ
• Example Evaluate
curl F · dS, where F(x, y, z) = xzi + yzj + xyk and S is
S
part of the sphere x2 + y 2 + z 2 = 4 that lies inside the cylinder x2 + y 2 = 1 and
above the xy-plane.
58
CHAPTER 16
• Example Let D be simply connected. Show that if curl F = 0 in D, then F is
conservative there.
Exercises
Page 1139
1, 5, 7, 9, 17.
CHAPTER 16
59
16.9 The Divergence Theorem
• The Divergence Theorem (Gauss’s Theorem)
Let E be a simple solid region and let S be the boundary surface of E,
given with positive (outward) orientation. Let F = F1 i + F2 j + F3 k be a
vector field such that F1 , F2 and F3 have continuous partial derivatives on
an open region that contains E. Then
ZZ
ZZZ
F · dS =
div F dV
∂E
E
• The boundary surface S consists of three pieces:
1.
The bottom surface S1 .
2.
The top surface S2 .
3.
The vertical surface S3 , which lies above the boundary curve of D. It might
happen that S3 does not appear as in the case of a sphere.
60
CHAPTER 16
• Example Verify Gauss’s Theorem for
F = yi + xj + zk
and
E = (x, y, z) : x2 + y 2 ≤ 1, 0 ≤ z ≤ 1 − x2 − y 2
by independently calculating
ZZ
(a)
F · dS
Z∂EZ Z
(b)
div F dV
E
CHAPTER 16
61
• Example Compute the flux of the vector field F = x2 yi + 2xzj + yz 3 k across
the surface of the rectangular solid E determined by
0 ≤ x ≤ 1,
Exercises
Page 1145
0 ≤ y ≤ 2,
3, 5, 7, 11.
0≤z≤3
62
CHAPTER 16
16.10 Summary
• Fundamental Theorem of Line Integral
• Green’s Theorem
• Stoke’s Theorem
• Divergence Theorem