1004
Chapter Eighteen LINE INTEGRALS
across a cross-section of the wire. The magnitude of B is constant along any circle which
is centered on and perpendicular to the z-axis. The direction of B is tangent to such circles.
Show that the magnitude of B , at a distal)Ce r from the z-axis, is given by
{
al
--
I\B II= 2;;;.
for r 2:: ro
--
21rr0 2
for r
<
1·0.
(b) A torus is a doughnut-shaped surface obtained by rotating around the z-axis the circle
(x - (3)2 + z2 = o:2 of radius a and center ((3, 0, 0), where O < a < (3. A toroidal solenoid
is constructed by wrapping a thin wire a large number, say N, times around the torus.
Experiments show that if a constant current I flows though the wire, then the magnitude
of the magnetic field is constant on circles inside the torus which are centered on and
perpendicular to the z-axis. The direction of B is tangent to such circles. Explain why, at
all points inside the torus,
11.B II=
aNI
21rr '
and 11.B II = 0 otherwise. [Hint: Apply Ampere's Law to a suitably chosen surface S with
boundary curve C. What is the net current passing through S?]
•�11....
4. Conservative Forces, Friction, and Gravity
The work done by a conservative force on an object does not depend on the path taken but only
on the endpoints-in other words, the vector field representing the force is path-independent.
Only for a conservative force is the work done stored as energy; for a non-conservative force,
the work may be dissipated.
(a) The frictional drag force f on an airplane flying through the air with velocity iJ is given
by
f = -cl\v
\\v.
The constant c is positive and depends on the shape of the airplane. One airplane takes off
from an airport located at the point (2, 0, 0) and follows the path
r(t) = (2cost)T + (2sint)j+ 3tk,
from t = 0 tot= 1r.
An identical airplane takes off from the same point and follows the path
-
2- 4t
r(t) =--tr+ 3tk
7r
also from t = 0 tot=
7r.
(i) Compute the total work done by the drag force f on each airplane.
(ii) Is the drag force conservative? Explain.
(b) Suppose the two airplanes of the preceding problem each have mass m, while the earth has
mass !YI. The gravitational force f acting on either airplane at a point with position vector
i" is given by
-
F (r) = -GMm
r
l\r\l3
,
where G is the gravitational constant.
(i) Compute the total work done by the gravitational force f on each airplane as it moves
from the point (2, 0, 0) to the point (-2, 0, 31r ).
(ii) Is the gravitational force conservative? Explain.
1006
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
19.1
19.1 THE IDEA OF A FLUX INTEGRAL
THE IDEA OF A��
FLUX-----INTEGRAL
---�------------------------------------
Flow Through a Surface
Imagine water flowing through a fishing net stretched across a stream. Suppose we want to measure
the flow rate of water through the net, that is, the volume of fluid that passes through the surface per
unit time.
Example1
1007
s
Direction of orientation:
Positive flow
Figure 19.2: An oriented surface showing
directions of positive and negative flow
A flat square surface of area A, in m , is immersed in a fluid. The fluid flows with constant velocity
2
v, in m/sec, perpendicular to the square. W rite an expression for the rate of flow in m3/sec.
Figure 19.3: Area vector A = n A of flat smface
with area A and orientation n
The Area Vector
!he flux through a fl �t surface depends both on the area of the surface and its orientation. Thus,
1s useful to represent its area by a vector as shown in Figure 19.3.
The area vector of a flat, oriented surface is a vector A such that
• The magnitude of A is the area of the surface.
• The direction of A is the direction of the orientation vector 11, .
The Flux of a Constant Vector Field Through a Flat Surface
Figure 19.1: Fluid flowing perpendicular
to a surface
Solution
Suppose the velocity vector fieid, v, of a fluid is constant and A is the area vector of a flat surface.
The flux through this surface is the volume of fluid that flows through in one unit of time. The
skewed box i': Figure l�.4 has cross-sectional area II.A II and height llv II cosB, so its volume is
(llv II cosB) IIA II= v ·A. Thus, we have the following result:
In one second a given particle of water moves a distance of llv II in the direction perpendicular to the
square. Thus, the entire bcidy of water moving through the square in one second is a box of length
llv II and cross-sectional area A. So the box has volume llv IIA m3, and
If v is constant and A is the area vector of a flat surface, then
Flow rate = llv IIA m 3/sec.
Flux through surface =
This flow rate is called the flux of the fluid through the surface. We can also compute the flux·
of vector fields, such as electric and magnetic fields, where no flow is actually taking place. If the
vector field is constant and perpendicular to the surface, and if the surface is flat, as in Example 1,
the flux is obtained by multiplying the speed by the area.
Next we find the flux of a constant vector field through a flat surface that is not perpendicular
to the vector field, using a dot product. In general, we break a surface into small pieces which are
approximately flat and where the vector field is approximately constant, leading to a flux integral.
v · A.
Orientation of a Surface
Before computing the flux of a vector field through a surface, we need to decide which direction of
flow through the surface is the positive direction; this is described as choosing an orientation.'
At each point on a smooth surface there are two unit normals, one in each direction. Choosing
an orientation means picking one of these normals at every point of the surface in a contin­
uous way. The unit normal vector in the direction of the orientation is denoted by ii . For a
closed smface (that is, the boundary of a solid region), we choose the outward orientation
unless otherwise specified.
We say the flux through a piece of sur ace is positive if the flow is in the direction of the
orientation and negative if it is in the opposite direction. (See Figure 19.2.)
Figure 19.4: Flux of iJ through a smface with area vector A is the volume of this skewed box
Example2
Water is flowing down a cylindrical pipe 2 cm in radius with a velocity of 3 cm/sec. Find the flux of
the velocity vector field through the ellipse-shaped region shown in Figure 19.5. The normal to the
ellipse makes an angle of e with the direction of flow and the area of the ellipse is 41rj (cosB) cm2 .
�-- A (where IIA 11 = 41r/ (cos B) cm2 )
---- iJ ( where llil 11 = 3 cm/sec)
f
1 Although we will not study them, there are a few surfaces for which this cannot be done. See page !012.
Figure 19.5: Flux through ellipse-shaped region across a cylindrical pipe
I
-I
1008
Solution
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
19.1 THE IDEA OF A FLUX INTEGRAL
There are two ways to approach this problem. One is to use the formula we just derived, which gives
Flux through ellipse= iJ ·A= llv IIIIAII cosB = 3(Area of ellipse) cosB
In computing a flux integral, we have to divide the surface up in a reasonable way, or the limit
might not exist. In practice this problem seldom arises; however, one way to avoid it is to define flux
integrals by the method used to compute them shown in Section 21.3.
= 3 (�) cosB = 127T cm 3 /sec.
cose
The second way is to notice that the flux through the ellipse is equal to the flux through the
circle perpendicular to the pipe in Figure 19.5. Since the flux is the rate at which water 1s flowrng
down the pipe, we have
Flux through circle=
Velocity
of water
x
Flux and Fluid Flow
If iJ is the velocity vector field of a fluid, we have
Rate fluid flows
through surface S
Area of
circle
a
If the vector field, F, is not constant or the smiace, S, is not flat, we divide the surface into
e
6:.A,
area
with
patch
particular
a
For
19.6.)
"'!._,
patchwork of small, almost flat pieces. (See Figure
pick a unit orientation vector n at a point on the patch and define the area vector of the patch, 6:.A ,
as
t:.A = n t:.A.
Example3
Flux through patch �
Flux through whole surface �
-yi + xj
B(x,y,z)=
x 2 +y 2
z
s
L
·
F 6:.A.
l�m
ll6A 11-+0
Thus, provided the limit exists, we make the following definition:
b.A
--!---
y
x
Solution
L F . t:.A.
If Sis a closed surface miented outward, we describe the flux through Sas the flux out of S.
Area b.A
2
y
Figure 19.8: The vector field
i +xi
B (x , y , z) = -Y_
x2+y2
The flux integral of the vector field F through the oriented surface S is
jS
z
s
2
L F · 6:.A,
As each patch becomes smaller and 116:.A II ----+ 0, the approximation gets better and we get
l�m
llll.A 11-+0
- -
i
F · 6:.A,
so, adding the fluxes through all the small pieces, we have
=
h v. dA
Find the flux of the vector field B (x, y, z) shown in Figure 19.8 through the square S of side 2
shown in Figure 19.9, oriented in the direction, where
(See Figure 19.7.) If the patches are small enough, we can assume that F is approximately constant
on each piece. Then we know that
r F . dA
Flux of iJ
through S
The rate of fluid flow is measured in units of volume per unit time.
The Flux Integral
Flux through S=
1009
3
Figure 19.9: Flux of B through the
square S of side 2 in xy-plane and
oriented in J direction
Figure 19.10: A small patch of surface with
area lib.A 11 = b.xb.z
Con�ider a small rectangular_patch with area vector t:.A in S, with sides 6:.x and 6:.z so that
116:.A II = 6:.x 6:.z. Since 6:.A points in the J direction, we have t:.A = 6:.xt:.z. (See Fig­
ure 19.10.)
At the point (x, 0, z) in S, substituting y = 0 into B gives B (x, 0, z) = (1/x)i. Thus, we
have
Flux through small patch � B · t:.A = -j · (j 6:.xt:.z) = - 6:.x 6:.z.
x
x
Therefore,
i
-
Flux through smiace =
1B ·
S
dA =
- ( 1-) -
l�m
ll6A 11-+0
1
LB · t:.A=
lim
l\.x-+
o
L ! 6:.x 6:.z.
X
l\.z-+ 0
Figure 19.6: Surface S divided into small, almost flat piece�,
showing a typical orientation vector n and area vector b.A
,,
,
'.
V_§ctor field
1lF(x,y, z )
Figure 19.7: Flux of a vector field
through a curved surface S
This last expression is a Riemann sum for the double integral JR
1 :S x :S 3, 0 :S z :S 2. Thus,
Flux through surface=
{B
Js
· dA =
{ � dA=
JR x
!
Jo
2
i dA, where R is the square
3
{
11
�
x
dx dz = 2 ln 3.
The result is positive since the vector field is passing through the smia�e in the positive direction.
1010
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
Example 4
19.1 THE IDEA OF A FLUX INTEGRAL
Each of the vector fields in Figure 19.11 consists entirely of vectors parallel to the xy-plane, and is
constant in the z direction (that is, the vector field looks the same in any plane parallel to the x y ­
plane). For each one, say whether you expect the flux through a closed surface smTmmding the origin
to be positive, negative, or zero. In part (a) the surface is a closed cube with faces perpendicular to
the axes; in parts (b) and (c) the Ulface is a closed cylinder. In each case we choose the outward
orientation. (See Figure 19.12.)
W
Y
M
"- \
\
Y
�
I I/
l
\
-- x
+-·---"I---- x
I
Y
1 --
- - ,,...
__
.-
/--
��- x
\
I
I
I
Figure 19.13: Flux of E = q1' /117' 113 through the surface of a sphere of radius R centered at the origin
Solution
\
1
.
This vector field points radially outward from the origin in the same direction as n Thus' since ii,
is a unit vector,
E · t.A' =E · n t.A = IIE 11 t.A.
2
R
On the sphere, IIE II = q/ , so
Figure 19.11: Flux of a vector field through the closed surfaces whose cross-sections are shown in the xy-plane
z
z
x
-y
ii
r E . dA =
Js
- ii
y
x
r-
lsE
Figure 19.12: The closed cube and closed cylinder, both oriented outward
(a) Since the vector field appears to be parallel to the faces of the cube which are perpendicular to
they- and z-axes, we expect the flux through these faces to be zero. The fluxes through the two
faces perpendicular to the x-axis appear to be equal in magnitude and opposite in sign, so we
expect the net flux to be zero.
(b) Since the top and bottom of the cylinder are parallel to the flow, the flux through them is zero.
On the curved sU1face of the cylinder, iJ and t.A appear to be everywhere parallel and in the
same direction, so we expect each term iJ · t.A to be positive, and therefore the flux integral
fs iJ ·dA to be positive.
(c) As in part (b), the flux through the top and bottom of the cylinder is zero. In this case iJ and t.A
are not parallel on the round surface of the cylinder, but since the fluid appears to be flowing
inward as well as swirling, we expect each term iJ · t.A to be negative, and therefore the flux
integral to be negative.
Calculating Flux Integrals Using dA
= rt d A
n
For a small patch of surface t.S with unit normal and area t.A, the area vector is t.A =11, t.A.
The next example shows how we can use this relationship to compute a flux integral.
Examples
An electric charge q is placed at the origin in 3-space. The resulting electric field E (r) at the point
with position vector r is given by
-
E (r)= q
r
3
llrll '
Find the flux of E out of the sphere of radius R centered at the origin. (See Figure 19.13.)
l�m
ll6A 11-+0
LE . t.A = lim � _J_2 t.A= _J_ lim � t.A.
6A-+0 L..., R
R2 6A-+0 L...,
The last sum approximates the surface area of the sphere. In the limit as the subdivisions get finer
we have
Jim � t.A =Surface area of sphere.
6A-+OL...,
Thus, the flux is given by
�
Solution
1011
q
q
.
·dA- = 2}r�
Lt.A= R 2 ·(Surface area of sphere)= ;2 (41rR2 )=41rq.
R
0
This result is known as Gauss's law.
To compute a flux with an integral instead of Riemann sums, we often write dA = ii, dA as in
the next example.
Example6
Solution
Suppose S is the surface of the cube �ounded by the six planes x = ±1, y = ±1, and z = ±1.
Compute the flux of the electric field E of the previous example outward through S.
It is enough to compute the flux of E through a single face, say_the top face S1 defined by z =1,
where -1 S x S 1 and -1 S y S 1. By symmetry, the flux of E through the other five faces of S
must be the same.
On the top face, S1, we have dA = dA = k dx dy and
n
xi+ yJ + k
E (x
y 1) -- q-,--,--__::_::_
__
·,'
(x 2 +y2 +1)3/2 ·
The corresponding flux integral is given by
f
Js 1
E . dA = q
ff
1
-1
1
_1
ff
1
xi + yJ + f
. f d·x dy = q
2
2
3/
2
_1
(x +y +1)
Computing this integral numerically shows that
Flux through top face=
Thus,
Total flux of E out of cube=
1
_1
1
d .d
(x 2 +y2 +1)3/2 x Y·
f E . dA � 2.0944q.
ls1
{E
ls
· dA � 6(2.0944q) =12.5664q.
1012
19.1 THE IDEA OF A FLUX INTEGRAL
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
Example 5 on page 1010 showed that the flux of E through a sphere of radius R centered at
the origin is 41rq. Since 41r � 12.5664, Example 6 suggests that
z
(a)
z
(b)
y
x
Total flux of E out of cube= 41rq.
By computing the flux integral in Example 6 exactly, it is possible to verify that the flux of E
through the cube and the sphere are exactly equal. When we encounter the Divergence Theorem in
Chapter 20 we will see why this is so.
19. The triangular plate with vertices (1, 0, 0), (0, 1, O),
(0, 0, 1), oriented away from the origin.
In Exercises 20-22, find the flux of H
tlu·ough the surface S.
Notes on Orientation
20.
z
(c)
x
(d)
z
Two difficulties can occur in choosing an orientation. The first is that if the smface is not smooth,
it may not have a normal vector at every point. For example, a cube does not have a normal vector
along its edges. When we have a surface, such as a cube, which is made of a finite number of smooth
pieces, we choose an orientation for each piece separately. The best way to do this is usually clear.
For example, on the cube we choose the outward orientation on each face. (See Figure 19.14.)
-Y
x
(e)
S is the cylinder x2 + y2 = 1, closed at the ends by the
planes z
22. S is the disk of radius 1 in the plane x + y + z
oriented in upward.
23.
I
F' = 3i + 2; + k
25. H
J)-y
= -zi + xk.
26.
r) = r.
In Exercises 12-15, compute the flux of iJ = i + 2J - 3k
through the rectangular region with the orientation shown.
12.
The second difficulty is that there are some surfaces which cannot be oriented at all, such as the
Mobius strip in Figure 19.15.
ln Exercises 1---4, find the area vector of the oriented flat sur­
face.
1. The triangle with vertices (0, 0, 0), (0, 2, 0), (0, 0, 3) ori­
ented in the negative x direction.
2. Circular disc of radius 5 in the xy-plane, oriented up­
ward.
3. y = 10, 0 S x S 5, 0 S z S3, oriented away f om the
xz-plane.
r
4. y = -10, 0 S x S 5, 0 S z S 3, oriented away f om
the xz-plane.
r
5. Find an oriented flat surface with area vector 150J.
6. Let S be the diskof radius 3 perpendicular the the y-axis,
centered at (0, 6, 0) and oriented away from the origin. Is
(xi+ yJ) ·dA a vector or a scalar?
(0, 2, 2)
(2, 2, 2)
7. Compute
1
14.
( 4i
+ 5k ) · dA, where S is the square
'::. 2, 0)
8. Compute
1
(b) Away from the origin.
I
(0, 0, 4)
x
/
(2,2,0)
(0, 2, 0)
( , 0, 3)
0
(O, 2' 3)
�
y
(2, o, 0)
Toward the origin.
(b) Away from the origin.
9. Let F' (x, y, z) = zi. For each of the surfaces in (a)-(e),
say whether the flux of F' through the surface is posi­
tive, negative, or zero. The orientation of the surface is
indicated by a normal vector.
f5 (sinx i+(y2 +z2 )J +y2 k)·dA whereSis a diskof
radius 1r in the plane x = 31r /2, oriented in the positive
x-direction.
J5(5i +5J +5k)·dAwhereSis a diskof radius 3 in
the plane x + y + z = 1,-oriented upward.
In Exercises 30-52, calculate the flux of the vector field
through the surface.
30.
z
31.
F' = 2f + 3.'f through the square of side 1r in the xyplane, oriented upward.
F' = 2f + 3.'f through the unit disk in the yz-plane,
centered at the origin and oriented in the positive x­
direction.
32. F' = xi + yJ + zk through the square of side 1.6 cen� Y
tered at (2, 5, 8), parallel to the xz-plane and oriented
(2, 2, 0)
x
away from the origin.
(2i + 3k) · dA, where S is the disk of ra­
dius 4 perpendicular to the x-axis, centered at (5, 0, 0)
and oriented
(a)
15.
/
of side length 3 perpendicular to the z-axis, centered at
(0, 0, -2) and oriented
(a) Toward the origin.
(
z
(2, 0, 4)
28.
29.
y
(xi + 4J) · dA where Sis the disk of radius 5 per­
r·
�
x
Exercises
c = xi
dA where S is the sphere of radius 3 centered at
27. fs
the origin.
(3, 2, 4)
z
:
Exercises and Problems for Section 19.1
1
13.
z
(0, o, 2)
1
= zxk
24.
pendicular to the x-axis, centered at (3, 0, 0) and oriented
toward the origin.
11. Repeat Exercise 9 with the vector field F' (
Figure 19.14: The orientation vector field 11 on the
cube surface determined by the choice of unit
normal vector at the point P
1
In Exercises 26--29, calculate the flux integral.
10. Repeat Exercise 9 with F' (x, y, z)
Figure 19.15: The Mobius strip
is an example of a
non-orientable surface
= 0 and z = 1 and oriented outward.
= 2 oriented in
Find the flux of the vector fields in Exercises 23-25 out of the
closed box OSx S1, 0 Sy S2, 0 S z S3.
z
x
= 2i + 3J + 5k
21. Sis the disk of radius 1 in the plane x
the positive x-direction.
1--=ry
I
1013
For Exercises 16--19 find the flux of the constant vector field
iJ = i - J + 3k through the given surface.
16. A diskof radius 2 in the xy-plane oriented upward.
17. A triangular plate of area 4 in the yz-plane oriented in
the positive x-direction.
18. A square plate of area 4 in the yz-plane oriented in the
positive x-direction.
JI4
33.
in a horizontal
F' = zk through a square of side
plane 2 units below the xy-plane and oriented downward.
34.
F' = -yi + and S is the square plate in the yz­
plane with corners at (0, 1, 1), (0, -1, 1), (0, 1, -1), and
(0, -1, -1), oriented in the positive x-direction.
35.
36.
xJ
F' = 7i + 6J + 5k andSis a diskof radius 2 in the yz­
plane, centered at the origin and oriented in the positive
x-direction.
F' = xi + 2yJ + 3zk and S is a square of side 2 in the
plane y
= 3, oriented in the positive y-direction.
1014
37. F = 7i+6J +5k and Sis a sphere of radius 1r centered
at the origin.
38.
F = -5i'' through the sphere of radius 2 centered at the
origin.
39. F = xi+Y.7 + (z2 + 3)k and Sis the rectangle z = 4,
0 :::; x :::; 2, 0 S y :::; 3, oriented in the positive z­
direction.
40.
19.1 THE IDEA OF A FLUX INTEGRAL
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
F = 6i + 7J
through triangle of area IO in the plane
x + y = 5, oriented in the positive x-direction.
45.
46.
47.
48.
F
2zi
+ xJ + xk through the rectangle x =
O S y :::; 2, 0 S
direction.
4,
z S 3, oriented in the positive x­
F = i + 2J tlu-ough a square of side 2 lying in the plane
x + y + z = 1, oriented away from the 01igin.
F = (x2 + y2 )k through the disk of radius 3 in the
xy-plane and centered at the origin and oriented upward.
F = cos(x2 + y2 )k through the disk x 2
oriented upward in the plane z = 1.
+ y2
S 9
41. F = 6i + x 2J - k, through the square of side 4 in the 49. F- = e'Y2+ • 2-i through the disk of radius 2 in the yzplane y = 3, centered on the y-axis, with sides parallel to
the x and z axes, and oriented in the positive y-direction.
42.
43.
F = (x + 3)i + (y + 5)J + (z + 7)k through the
rectangle x = 4, 0 S y :::; 2, 0 S z :::; 3, oriented in the
positive x-direction.
F = 7r through the sphere of radius 3 centered at the
origin.
44. F = -3r through the sphere of radius 2 centered at the
origin.
plane, centered at the origin and oriented in the positive
x-direction.
50.
F = -yi + xJ through the disk in the xy-plane with
radius 2, oriented upward and centered at the origin.
51. F =
r through the disk of radius 2 parallel to the xy­
plane oriented upward and centered at (0, 0, 2).
52.
F = (2 - x)i through the cube whose vertices include
the points (0, 0, 0), (3, 0, 0), (0, 3, 0), (0, 0, 3), and ori­
ented outward.
Problems
53. Let B be the surface of a box centered at the origin, with
edges parallel to the axes and in the planes x = ±1,
y = ±1, z = ±1, and let S be the sphere of radius I
centered at origin.
(a) Indicate whether the following flux integrals are pos­
itive, negative, or zero. No reasons needed.
(a) Ia xi · dA
(c) Is lx li · dA
(b)
(d)
56. (a) What do you think will be the electric flux tlu-ough
the cylindrical surface that is placed as shown in the
constant electric field in Figure 19.16? Why?
(b) What if the cylinder is placed upright, as shown in
Figure 19. 17? Explain.
Ia yi · dA
Is(Y - x)i · dA
(b) Explain with reasons how you know which flux in­
tegral is greater:
1
xi · dA.
or
l
xi · dA?
54. Suppose that E is a uniform electric field on 3-space,
so E (x, y, z) = at+ bj +ck, for all points (x, y, z),
where a, b, c are constants. Show, with the aid of sym­
metry, that the flux of E through each of the following
closed surfaces S is zero:
Figure 19.16
(a) S is the cube bounded by the planes x = ±1,
y = ±1, and z = ±1.
(b) Sis the sphere x2 + y2 + z2 = 1.
(c) S is the cylinder bounded by x2 + y2 = 1, z = 0,
and z = 2.
55. Water is flowing down a cylindrical pipe of radius 2 cm;
its speed is (3- (3/4)r2 ) cm/sec at a distance r cm from
the center of the pipe. Find the flux through the circu­
lar cross-section of the pipe, oriented so that the flux is
positive.
57. Let S be part of a cylinder centered on the y-axis. Ex­
plain why the three vectors fields F, G, and ii have the
same flux through S. Do not compute the flux.
F =xi+ 2yzk
G = :r:i + y sin xJ + 2yzk
ii =xi+ cos(x2 + z)J + 2yzk
r
(a) For what value of pis the flux a maximum?
(b) What is that maximum value?
61. Let S be the cube with side length 2, faces parallel to the
coordinate planes, and centered at the origin.
(a) Calculate the total flux of the constant vector field
= -i + 2J + k out of S by computing the flux
through each face separately.
(b) Calculate the flux out of S for any constant vector
field = ai + bJ + ck.
(c) Explain why the answers to parts (a) and (b) make
sense.
- .
E (x,
y, z)
v
62. Let S be the tetrahedron with vertices at the origin and at
(1, 0, 0), (0, 1, 0) and (0, 0, 1).
(a) Calculate the total flux of the constant vector field
= -i + 2J + k out of S by computing the flux
through each face separately.
(b) Calculate the flux out of S in part (a) for any con­
stant vector field .
(c) Explain why the answers to parts (a) and (b) make
sense.
v
·v
63. Let P(x, y, z) be the pressure at the point (x, y, z) in a
fluid. Let F (x, y, z) = P(x, y, z)k. Let S be the sur­
face of a body submerged in the fluid. If S is oriented
inward, show that Is F · dA is the buoyant force on the
body, that is, the force upward on the body due to the
pressure of the fluid surrounding it. [Hint: F · dA =
P(x, y, z)k · dA = (P(x, y, z) dA) · k.]
64. A region of 3-space has a temperature which varies from
point to point. Let T(x, y, z) be the temperature at a point
(x, y, z). Newton's law of cooling says that grad T is
proportional to the heat flow vector field, F , where F
points in the direction in which heat is flowing and has
magnitude equal to the rate of flow of heat.
Rate of heat
loss from l,V
=k
1
(gradT)·dA.
s
E , in the xy-plane, given
+ yj
= 2>- xi
x2 +y2
.
(b) Compute the flux of E outward through the cylinder
x 2 + y2 = R2 , for O S z S h.
66. An infinitely long straight wire lying along the z-axis
carries an electric current I flowing in the k direction.
Ampere's Law in magnetostatics says that the current
gives rise to a magnetic field B given by
I -yi + xJ
B- (x, Y, z) = - 2
21r x + y2
v
(a) Suppose F = k grad T for some constant k. What
is the sign of k?
(b) Explain why this form of Newton's Jaw of cooling
makes sense.
(c) Let W be a region of space bounded by the surface
S. Explain why
Figure 19.17
65. The z-ax:is carries a constant electric charge density of>­
units of charge per unit length, with A > 0. The resulting
electric field is E .
(a) Sketch the electric field,
58. Find the flux of F = r /llr 11 3 out of the sphere of radius
R centered at the origin.
2
59. Find the flux of F = i'' / I lrl1 out of the sphere of radius
R centered at the origin.
60. Consider the flux of the vector field F = /I Ir I Ip for
p ::". 0 out of the sphere of radius 2 centered at the origin.·
1015
(a) Sketch the field B in the xy-plane.
(b) Suppose S1 is a disk with center at (0, 0, h), radius
a, and parallel to the xy-plane, oriented in the k di­
rection. What is the flux of B through S1 ? Is your
answer reasonable?
(c) Suppose S2 is the rectangle given by x = 0, a :::;
y :::; b, 0 :::; z :::; h, and oriented in the -i direc­
tion. What is the flux of B through S2 ? Does your
answer seem reasonable?
67. An ideal electric dipole in electrostatics is characterized
by its position in 3-space and its dipole moment vector
p. The electric field D , at the point with position vector
of an ideal electric dipole located at the origin with
dipole moment p is given by
r,
fJ (r) = 3 (r · r)r 5
llf 11
iI .
llr 113
Assume p = pk, so the dipole points in the k direction
and has magnitude p.
(a) What is the flux of D through a sphere S with center
at the origin and radius a > O?
(b) The field D is a useful approximation to the elec­
tric field E produced by two "equal and opposite"
charges, q at 2 and -q at 1, where the distance
llr 2 - r 1 II is small. The dipole moment of this con­
figuration of charges is defined to be q(r 2 - r 1).
Gauss's Law in electrostatics says that the flux of
E through S is equal to 41r times the total charge
enclosed by S. What is the flux of E through S if
the charges at 1 and 2 are enclosed by S? How
does this compare with your answer for the flux of
fJ through s if iI ·= q(r 2 - r 1)?
r
r
r
r
1016
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
19.2 FLUX INTEGRALS FOH GRAPHS, CYLINDERS, AND SPHERES
How do we divide S into small pieces? One way is to use the cross-sections off with x or y
constant and take the patches in a wire frame representation of the surface. So we must calculate the
area vector of one of these patches, which is approximately a parallelogram.
Strengthen Your Understanding
In Problems 68-69, explain what is wrong with the statement.
68. For a certain vector field F and oriented surface S, we
have Is F · dA = 2i - 3] + k.
69. If S is a region in the xy-plane oriented upwards then
Isft. dA > o.
81. If Is F · dA > Is G
points on the surface S.
F 1 = 2T - 3J - 4k
F2 = T - 2J + 1f
F3 = -1T +sJ +6k
F'4 = -11i +4J - 5k
F's= -5i + 3] + 5k
70. A nonzero vector field F such that Is F · dA = 0,
where S is the triangular surface with corners (1, 0, 0),
(0, 1, 0), (0, 0, 1), oriented away from the origin.
71. A nonconstant vector field F (x, y, z) and an oriented
surface S such that Is F · dA = 1.
Are the statements in Problems 72-81 true or false? Give rea­
sons for your answer.
z
(a)
72. The value of a flux integral is a scalar.
73. The area vector A of a flat, oriented smface is parallel to
the surface.
74. If S is the unit sphere centered at the origin, oriented
outward and the flux integral Is F · dA is zero, then
ft = o.
75. The flux of the vector field F = i through the plane
x = O, with O :::; y :::; 1, 0 :::; z :::; 1, oriented in the i
direction is positive.
76. If Sis the unit sphere centered at the origin, oriented out­
ward and F = xi+ yJ + zk = r, then the flux integral
Is F · dA is positive.
77. If Sis the cube bounded by the six planes x = ±1, y =
±1, z = ±1, oriented outward, and F = k, then
Isft .
(c)
(b)
L
/
x
I
19.2
· dA = 2 Is G · dA.
)­
-Y
J)-
[l6A 11-tO
L F . 6.A.
y.
+f(x, y)k.
ar - + fxk'
y
Suppose Sis the graph of the differentiable function z = f(x, y), oriented upward, and that F is
a smooth vector field. In Section 19.1 we subdivided the surface into small pieces with area vector
6.A and defined the flux of F through S as follows:
l�m
vx x v
To find iJ x and iJ y , notice that a point on the surface has position vector i =xi+ yJ
Thus, a cross-section of S with y constant has tangent vector
Flux of a Vector Field Through the Graph of z = f ( x, y)
=
Figure 19.19: Parallelogram-shaped patch in
the tangent plane to the surface
Consider the patch of surface above the rectangular region with sides 6.x and 6.y in the xy ­
plane shown in Figure 19.18. We approximate the area vector, 6.A, of this patch by the area vector
of the corresponding patch on the tangent plane to the surface. See Figure 19.19, This patch is the
parallelogram determined by the vectors iJ x and iJ y, so its area vector is given by
6.A ::::::
In Section 19.1 we computed flux integrals in certain simple cases. In this section we see how to
compute flux through surfaces that are graphs of functions, through cylinders, and through spheres.
r F . dA
fx
Figure 19.18: Surface showing coordinate patch and tangent
vectors r x and r y
z
FLUX INTEGRALS FOR GRAPHS, CYLINDERS, AND SPHERES
ls
Coordinate patch
x
x
x
w.
·iJ y
)-
F . dA = - Ls F . dA.
2
rY
z
I
79. If S1 is a rectangle with area I and S2 is a rectangle with
area 2, then 2Is 1 F · dA = Is F · dA.
According to the geometric definition of the cross product on page 745, the vector iJ x w has
magnitude equal to the area of the parallelogram formed by iJ and tu and direction perpendicular
to this parallelogram and determined by the right-hand rule. Thus, we have
Area vector of parallelogram = A = iJ x
z
y
(e)
dA = o.
80. If F = 2G, then Is F
The Area Vector of a Coordinate Patch
z
(d)
z
78. If S is an oriented surface in 3-space, and -S is the
same surface, but with the opposite orientation, then
Is
· dA then IIF II > IIG II at all
82. For each of the surfaces in (a)-(e), pick the vector field
F 1, F2, F 3, F4, F's, with the largest flux through the
surface. The surfaces are all squares of the same size.
Note that the orientation is shown.
In Problems 70-71, give an example of:
1017
rx =- = i
ax
and a cross-section with x constant has tangent vector
ar = -
r
r y = ay
J
-
+fy k .
The vectors x and iJ x are parallel because they are both tangent to the surface and parallel
to the xz-plane. Since the x-component of r x is i and the x-component of iJ x is (6.x)i, we have
iJ x = (6.x)r x· Similarly, we have iJ y
(6.y)r y· So the upward-pointing area vector of the
parallelogram is
This is our approximation for the area vector 6.A on the surface. Replacing 6.A, 6.x, and 6.y by
dA, dx and dy, we write
dA = (-fx i -fy ] +
dxdy .
f)
1018
19.2 FLUX INTEGRALS FOR GRAPHS, CYLINDERS, AND SPHERES
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
The Flux of F Through a Surface Given by a Graph of z
z
= f(x, y)
F · dA =
l
F(x, y,.f(x,y)) · (-fxi - fy ]
Coordinate patch
R
Suppose the surface Sis the part of the graph of z= J(x,y) above a region R in the xy-plane,
and suppose S is oriented upward. The flux of F through S is
ls
+ k) dxdy.
1019
z
f=
y
Example1
Compute fs F · dA where F(x,y, z) = zk and Sis the rectangular plate with corners (0, 0, 0),
(1, 0, 0), (0, 1, 3), (1, 1, 3), oriented upward. See Figure 19.20.
z
(0,1,3)
·x
y
Figure 19.21: Outward-oriented
cylinder
Figure 19.22: Coordinate patch with area b.A on
surface of a cylinder
Replacing 6A, D.z, and 66 by dA, clz, and dB , we write
dA= (cos B i + sineJ) RdzdB .
This gives the following result:
The Flux of a Vector Field Through a Cylinder
The flux of F through the cylindrical surface S, of radius R and oriented away from the
z-axis, is given by
x
Figure 19.20: The vectrn:: field F = zk on the rectangular surface S
Solution
ls
We find the equation for the plane S in the form z = f(x, y). Since f is linear, with x-slope equal
to O and y-slope equal to 3, and f(0, 0)= 0, we have
clA=(-fx i - fy ] + k)dxdy=(Oi - 3J +k)dxdy = (-3J + k)dxdy.
The flux integral is therefore
ls
F · dA =
11 11
3yk · (-3J + k) dxdy=
11 11
h
F(R,B,i) · (cos Bi +sineJ) RdzdB,.
where Tis the Bz-region corresponding to S.
z= J(x,y) = o+ox+3y=3y.
Thus, we have
F ·dA=
Example2
Compute fs F · dA where F(x, y, z) = yJ and Sis the part of the cylinder of radius 2 centered
on the z-axis with x � 0, y � 0, and O :::; z :::; 3. The surface is oriented toward the z-axis.
z
3ydx dy=1.5.
Flux of a Vector Field Through a Cylindrical Surface
.x
Consider the cylinder of radius R centered on the z-axis illustrated in Figure 19.21 and oriented
away from the z-axis. The coordinate patch in Figure 19.22 has surface area given by
./
6A ,:::o R666z.
The outward unit normal
n ==
n points in the direction of xi +yJ, so
xi+yj
llxi + yJII
Rcos Bi +R sinBJ
. 6..,.
=cos 6..,.i + sm
J.
R
Therefore, the area vector of the coordinate patch is approximated by
6A =n6A,:::o (cos Bi +sineJ)R6z66.
Figure 19.23: The vector field F
Solution
= yJ on the surface S
In cylindrical coordinates, we have R =2 and F = yJ =2sineJ. Since the orientation of S is
toward the z-axis, the flux through S is given by
r- -
rr
-
-
-
JsF ·dA =- J 2sin6j ·(cos Bi +sin6j)2dzd6=-4
r ro 3 sin Bdzd6=-31r.
}0
/2
.
J
2
1020
19.2 FLUX INTEGRALS FOR GRAPHS, CYLINDERS, AND SPHERES
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
Flux of a Vector Field Through a Spherical Surface
Consider the piece of the sphere of radius R centered at the origin, oriented outward, as illustrated
in Figure 19.24. The coordinate patch in Figure 19.24 has surface area given by
�A� R sin¢�¢�8.
Example3
Solution
2
The outward unit normal n points in the direction of 1'
Find the flux of F
outward.
ls ls
F
-,
_,
�A �n �A= ; �A= (sin¢cos8i +sin¢sin8] +cos¢k)R sin¢�¢�8.
ll II
Example4
2
�e by d.A, d¢, and dB, we write
27r
r/2 8sin¢cos2 ¢d¢d8
Thus, we obtain the following result:
y
I
I
\
The flux of F through the spherical surface S, with radius R and oriented away from the
origin, is given by
=
f. 1 1:: 11
;
\
I
,�
'
\
dA
I
I
, I I I
I
F(R,8,¢). (sin¢cos8i +sin¢sin8] +cos¢k) R sin¢d¢d8,
2
.
I
I
\
t
I
I
-� -�-, ......,,,... "
" ,'
... �
" \
t
,, .,, I I
I
where Tis the 8¢-region corresponding to S.
3.
r)
The Flux of a Vector Field Through a Sphere
ls lls
=
l61r
The magnetic field iJ due to an ideal magnetic dipole,µ, located at the origin is defined to be
- _
fJ,
3(µ . r
B (r) = +
3
5
.
ll f 11
ll f 11
I
=
- cos3 ¢ 1r/2
) 1 )
3
</>=D
= 21r (8 (
Figure 19.25 shows a sketch of iJ in the plane z = 0 for the dipoleµ = i. Notice that iJ is similar
to the magnetic field of a bar magnet with its north pole at the tip of the vector i and its south pole
at the tail of the vector i.
Compute the flux of iJ outward through the sphere S with center at the origin and radius R.
d.A = _!____ dA = (sin¢cos8i + sin¢sin8] +cos¢k) R2 sin¢d¢d8.
llrll
f ·dA
e and ¢, with O ::; e ::; 21r and O ::;
2cos¢k · ( sin¢cos8i +sin¢sin8] +cos¢k)4sin¢d¢d8
lo lo
Therefore, the area vector of the coordinate patch is approximated by
Replacing �A, �¢. and
·dA =
= !
7°1
n = - = sin¢cos8i + sin¢sin8j +cos<pk.
llrll
-I
= zk through S, the upper hemisphere of radius 2 centered at the origin, oriented
The hemisphere S is parameterized by spherical coordinates
¢ ::; 1r /2. Since R = 2 and F = zk = 2 cos ¢k, the flux is
= xi +y] +zk, so
1021
.
' ,
I
,,
x
\
I
I
,,
.
_,
I
I ,
I I \ "
...
I
I
I
\
I
I
. . -
Figure 19.25: The magnetic field of a dipole, i, at the ongm:
Solution
ls ls·
ls � ls
Since i
-i
B = Iii ll3
ls (-�;I�:
· 1' = x and llrII =Ron the sphere of radius R, we have
iJ
.d
A
=
=
3(
(- ; 3 + �;
11 1
1
1�·11:
dA =
r
;I
� ).I
1 1;�
4
dA =
ls
dA =
�4
+
+
3(i . r)r
3 (i
Iii ll5
rll
i ;iJ�
1
2
)
dA
x dA.
But the sphere S is centered at the origin. Thus, the contribution to the integral from each positive
x-value is canceled by the contribution from the coITesponding negative x-value; so fs x dA = 0.
Therefore,
B · dA = 4
1- R21s xdA = 0.
s
Exercises and Problems for Section 19.2
x
Figure 19.24: Coordinate patch with area t.A on surface of a sphere
Exercises
In Exercises 1-4, find the area vector dA for the surface
z = f(x,y), oriented upward.
1. f(x,y) = 3x - 5y
2. f(x, y) = 8x + 7y
3. f(x,y) = 2x2 - 3y2
4. f(x,y) =xy+y2
In Exercises 5-8, write an iterated integral for the flux of
F through the surface S, which is the part of the graph of
z = f(x,y) cmTesponding to the region R, oriented upward.
Do not evaluate the integral.
= lOi + 20J + 30k
f(x,y) = 2x - 3y
5. F(x,y,z)
R: -2 S
xS
3, 0 Sy S 5
6. F(x,y,z)=zi+xJ +yk
f(x, y)
= 50 - 4x + lOy
R: 0 S x S 4, 0 S y S 8
7. F(x,y,z)=yzi+xyJ +xyk
f(x, y) = cos x + sin 2y
R: Triangle with vertices (0, 0), (0, 5), (5, 0)
8. F(x,y,z)
= cos(x + 2y)J
= xe 3Y
R: Quarter disk of radius 5 centered at the origin, in
quadrant I
In Exercises 9-12, compute the flux of F through the surface
S, which is the part of the graph of z = f(x, y) corresponding
to region R, oriented upward.
9. F(x,y,z) = 3i - 2J + 6k
f(x, y) = 4x - 2y
R: 0 S x S 5, 0 S y S 10
= i - 2J + zk
f(x, y) = xy
10. F(x, y,z)
R: 0 S
x S 10, 0 S
y S 10
11. F(x,y,z)=cosyi+zJ +k
f(x,y) = x 2 + 2y
R: O S x S 1, 0 S y S 1
12. F(x,y,z)=xi+zk
= zJ + 6xk
R: Triangle with vertices (-1, 0), (1, 0), (0, 1)
In Exercises 13-16, write an iterated integral for the flux of F
through the cylindrical surface S centered on the z-axis, ori­
ented away from the z-axis. Do not evaluate the integral.
In Problems 30-45 compute the flux of the vector field
through the surface S.
19. F(x,y,z)=:ryzJ +:re z k
30. F = zk and S is the portion of the plane x + y + z = 1
S: radius 10, x 2 0, y 2 0, 0 S z S 3
S: radius 2, 0 S y S x, 0 S z S 10
= xyi + 2zJ
S: radius 1, x 2 0, 0 Sy S 1/2, 0 S z S 2
In Exercises 21-24, write an iterated integral for the flux of F
through the spherical surface S centered at the origin, oriented
away from the origin. Do not evaluate the integral.
F (x, y, z) = z i
2
In Exercises 25-27, compute the flux of F through the spher­
ical surface S centered at the origin, oriented away from the
origin.
-
16. F (x,y,z)
-y
35.
36.
S: radius 20, x 2 0, y 2 0, z 2 0
+ zk
38.
In Exercises 28-29, compute the flux of = zk through the
rectangular region with the orientation shown.
(2,0,4)
-
= x2 yzJ--, +z3-k
S: radius 2, between the xy-plane and the
paraboloid z = x2 + y2
In Exercises 17-20, compute the flux of F tlu·ough the cylin­
drical surface S centered on the z-axis, oriented away from
the z-axis.
I
(0,0,3)
39.
(0,0,4)
t�
40.
(0, 2, 0)
y
2,3 1
(2,0,0)
Figure 19.26
In Problems 46-47, compute the flux of F through the cylin­
drical surface in Figure 19.27, oriented away from the z-axis.
F =
ln(x2 )i + ex J + cos(l - z)k and Sis the part
of the surface z = -y + 1 above the square O S x S 1,
O S y S 1, oriented upward.
z
F = 5i + 7J + zk andSis a closed cylinder of radius
3 centered on the z-axis, with -2 S z S 2, and oriented
outward.
s
F = xi + yJ + zk andSis a closed cylinder of radius
2 centered on the y-axis, with -3 S y S 3, and oriented
outward.
� y
(2,2,0)
x
F =
xi + yJ and S is the part of the surface
z = 25 - (x 2 + y2 ) above the disk of radius 5 centered
at the origin, oriented upward.
41.
(2,2,0)
29.
F = -yJ + zk and S is the part of the surface
z = y2 + 5 over the rectangle -2 S x S 1, 0 Sy S 1,
oriented upward.
triangle R with vertices (0, 0), (0, 2), (1, 0).
S: radius 1, above the cone cf>= 1r/4.
z
x
1,
= 3xi + yJ + zk and Sis the part of the surface
z = -2x - 4y + 1, oriented upward, with (x, y) in the
27_. F(x,y,z) =xi+ Y1
28.
= 2xJ + yk and S is the part of the surface z
+ 1 above the square O S x S 1, 0 S y S
y
Figure 19.27
37. F
S: radius 4, entire sphere
14. F(x,y,z) = xi + 2yJ + 3zk
S: radius 10, 0 S z S 5
-
32. F
25. F(x,y,z) = zi
26. F(x,y,z) = yi - xJ
= (x-y)i+zJ +3xk andSis the part of the plane
= x + y above the rectangle OS x S 2, 0 S y S 3,
oriented upward.
34.
S: radius 2, x 2 0
'"
24. f ( x, y, z) = e k
S: radius 3, y 2 0, z S O
S: radius 10, x 2 0, y 2 0, 0 S z S 5
-
z
S: radius 5, entire sphere
23.
y
31. F
33.
+ 3zk
z
1
that lies in the first octant, oriented upward.
oriented upward.
21. F (x,y,z) = i + 2J + 3k
S: radius 10, z 2 0
F (x, y, z) = i + 2J + 3k
2x
15. F (x,y,z) = z i + e _j + k
S: radius 6, inside sphere of radius 10
F
18. F(x,y,z)=yi+xzk
20. F(x,y,z)
1023
Problems
S: radius 5, y 2 0, 0 S z S 20
v
f(x,y)=x+y+2
13.
17. F (x,y,z)
22. F(x,y,z)=xi+ 2yJ
f(x, y)
··�·11111,
19.2 FLUX INTEGRALS FOR GRAPHS, CYLINDERS, AND SPHERES
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
1022
F = cos(x2 + y2 )k andSis as in Exercise 38.
47. F
= xzi + yzJ + z3 k
In Problems 48-51, compute the flux of F through the spher­
ical surface, S.
48.
F = zk and S is the upper hemisphere of radius 2 cen­
tered at the origin, oriented outward.
above the rectangle O S x S 2, 0 S y S 3, oriented
downward.
49.
F = r and S is the part of the surface z =
50.
and S is the spherical cap given by
+ z 2 = 1, z 2 0, oriented upward.
F = z 2k and S is the upper hemisphere of the sphere
x2 + y 2 + z2 = 25, oriented away from the origin.
F = xi + yJ + zk and S is the surface of the sphere
x 2 + y2 + z2 = a2, oriented outward.
F = r and S is the part of the plane x + y + z = 1
above the disk x2 +
42. F
43. F
y2
x2
S 1, oriented downward.
+ yk and S is the
= 9, z 2 0, oriented upward.
xzi
x2 + y2 + z2
+y
2
hemisphere
=
2
x2
52. Compute the flux of F = xi+yJ +zk over the quarter
+ k through the sur­
face S given by x = sin y sin z, with O S y S 1r/2,
O S z S 1r/2, oriented in the direction of increasing x.
53. Compute the flux of F =xi+ J
44. F = yi + J - xzk and Sis the surface y =
+z ,
with x2 + z2 S 1, oriented in the positive y-direction.
F = x2i + y2J + z2k
51.
F = yi - xJ + zk
x2 + y 2
cylinder S given by x 2+y 2 = 1, 0 S x S 1, 0 S y S 1,
OS z S 1, oriented outward.
-xzi - yzJ + z k and S is the cone z
x2 + y2 for O S z S 6, oriented upward.
J
45.
46. f =xi+ y_j
2
and Sis the oriented triangular
surface shown in Figure 19.26.
J
54. Compute the flux of F = (x + z)i + + zk through
the surface S given by y = x2 + z2, with O S y S 1,
x 2 0, z 2 0, oriented toward the xz-plane.
1024
sphere of radius p. In addition, E s atisfie� Gauss's
L aw for any simple closed surface S enclosmg a vol­
ume Hf:
55. Let ft = (xzeYz)i + xzJ + (5 + x 2 + y2 )k. C alcu­
late the flux of ft through the disk x2 + y2 :S 1 m the
xy-plane, oriented upward.
56. Let fi = (e y + 3z + 5)i + (1:_ + 5z + 3)J + (3z +
e"'Y)k. C alculate the flux of H through the square of
side 2 with one vertex at the origin, one edg e along the
positive y- axis, one edge in_the 3z-plane with x > 0,
z > 0, and the normal ii = i - k .
x
j. E
"'Y
s
p :Sa
p > a.
(a) Describ e the charge distribution in words.
_
(b) Find the electric field E due too. Assume th!t E
can be written in spherical coordin ates as E
E(p)ep, where eP is the unit outward normal to the
Strengthen Your Understanding
with th
In Problems 59-60, explain what is wrong
e
HIIJID
statement.
cylindrical co­
59. Flux outward through the cone, given in
the formula
using
d
e
ut
comp
ordinates by z = r, can b e
dB.
Rdz
BJ)
dA = (cosBi +sin
ard, the formula
60. For the surface z = f(x, y) oriented upw
dA
gives ii
. dA
= k J, odV,
w
k a constant.
e with density (in
58. Electric charge is distributed in spac
3
s by
coulomb/m ) given in cylindrical coordinate
57. Electric charge is distributed in space with density (in
coulomb/m3 ) given in spherical coordinates by
_ 80 (a constant)
o(p, rp, B) - {
0
1025
19.3 THE DIVERGENCE OF A VECTOR FIELD
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
= ii dA = (- fxi - Jy] + k) dxdy
= - f,,i - fy] + k and dA = dxdy.
In Problems 61-62, give an ex ample of:
o (r' () ' z)
=
{
80 (a constant)
0
if r
>a
THE DIVERGENCE OF A VECTOR FIELD
Imagine that the vector fields in Figures 19.29 and 19.30 are velocity vector fields describing the
flow of a fluid. 2 Figure 19.29 suggests outflow from the origin; for example, it could represent the
expanding cloud of matter in the big-bang theory of the origin of the universe. We say that the origin
is a source. Figure 19.30 suggests flow into the origin; in this case we say that the origin is a sink.
In this section we use the flux out of a closed surface surrounding a point to measure the outflow
per unit volume there, also called the divergence, or flux density.
y
y
nds only o� z;
66. The vector field, ft, in Figu r� 19.28 depe
an mcreasmg
is
that is, it is of the form g(:,)k, �here g
.
s the flux of
esent
r
p
e
r
function. The integral fs F dA
ft through this rectangle, S, oriented upward.e?In each of
the following c ases, how does the flux chang
x-direction,
(a} The rectangle is twice as wide in the
0
0,
(2,
), (2,1, 3),
n,
origi
with new corners at the
.
(0,1,3).
. ar.e at
corne1s
(b) The rectangle is moved so that its
.
(1, 0,0), (2, 0,0), (2, 1,3), (1, 1, 3)
rd:
(c} The orientation is changed to downwa
t its new cor­
a
th
so
,
e
siz
in
d
e
tripl
(d} The rectangle is
, (0, 3, 9).
,
(
,
9)
0
3,
0,
,
3
)
(
3
origin,
e
th
at
ners are
\
""\
"-.."
II/
.---,,,.-
' ,..__
\
(a) Describe the charge distribution in words.
(b} Find the electric field E due to o. Assume th'.1_t E
can b e written in cylindrical coordinates as E
E(r)e,., where e,. is the unit m�ward vector to t�e
cylinder of radius r, and that E satisfies Gauss s
Law (see Problem 57).
�ce z_= f (x, y)
61. A function J(x, y) such that, for the_surf
+ k )dxdy ·
j
+
i
oriented upwards, we have dA = (
� radius l O cen­
62. An oriented surface S on the cyli�1der
= 600, where
dA
·
F
s
f
t
a
th
tered on the z-axis such
--- ' '
x
/
// I
/II
Figure 19.29: Vector field
showing a source
///
/
\
\
,,,.-.---
x
""-..
\""
Figure 19.30: Vector field
showing a sink
Definition of Divergence
To measure the outflow per unit volume of a vector field at a point, we calculate the flux out 6f a
small sphere centered at the point, divide by the volume enclosed by the sphere, then take the limit
of this flux-to-volume ratio as the sphere contracts around the point.
Geometric Definition of Divergence
F
The divergence, or flux density, of a smooth vector field , written div ff , is a scalar-valued
function defined by
J8 ff. dA
divF(x,y,z)= lim
Volume-+0 Volume of S
Here Sis a sphere centered at (x, y, z), oriented outward, that contracts down to (x, y, z) in
the limit.
The limit can be computed using other shapes as well, such as the cubes in Example 2.
Cartesian Coordinate Definition of Divergence
ft =xi+ yJ.
If
false ? Give rea­
Are the statements in Problems 63-65 true or
sons for your answer.
F
= F1i + F2J + F3k, then
8F1
div.F = -OX
above a '.S
63. If S is the part of the graph of f lying
ce area
surfa
has
S
then
x :S b, c ::; y ::; d,
J; + J3 + 1dxdy.
fed
J: -J
(x, Y) oriented
64. If A (x, y) is the area vector for z = f
= - f (x, Y)
upward and B (x, y) is !h e area vecto� for z
y).
(x,
-B
=
oriented upward, then A (x, y)
2
2
2
ented outward
65. If S is the sphere x + y _+ z = 1 ori
erpend1cular to
p
is
z)
y,
(x,
F
n
e
th
0,
=
and fs ft . dA
xi+ yJ + zk at every point of S.
if r :Sa
19.3
8F2
8F3
+- +--.
8y
oz
The dot product formula gives an easy way to remember the Cartesian coordinate definition,
888and suggests another common notation for divF, namely 'v ·F . Using 'v = - i
- j - k,
0x
8y
0z
we can write
+
Figure 19.28
+
div ff= 'v· ff
2 Although not all vector fields represent physically realistic fluid flows, it is useful to think of them in this way.
-----
'-¥--­
.
1026
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
Example1
Solution
19.3 THE DIVERGENCE OF A VECTOR FIELD
T herefore, the flux t hrough the cube is given by
1 v.
Calc ulate the divergence of F(f)= f at th e origin
(a) Using the geometric definition.
(b) Using the Cartesian coordinate definition.
(a) Using the m ethod of Example 5 on page 1010, y ou ca n calculate the flux of F out of the sphere
of radius a, centered at the origin; it is 47ra3. So we have
(b) In Cartesian coordinates, F (x, y,z) =xi+ YJ + zk, so
8
8
8
div F = �(x) + �(y) + �(z) = 1 + 1 + 1 = 3.
uz
uy
ux
dA =
r
. dIVV (2,2,0) =
a
8
8
. d. 1vv = � (-x) + �(O) + -(0) = -1 + O + O= -1
ux
uy
oz
-- - .
-- - .
-- - .
-- -- - !/
-- - .
-- - .
-- - .
v
v
(a) (i) T he vector fie ld = -xi is parall el to the x-axis and is shown in the xy-plane in Fig­
ure 19.31. To comput e the flux density at (0, 0, 0), we use a cube S1, centered at the origin
with ed ges paralle l to the axes, of length 2c. Then the flux through the faces perpendicular
to the y- and z-axes is zer o (because the vector field is parallel to these faces). On the faces
perpendicu lar to the x-axis, the vector fie ld and the outward normal are parallel but point
in opposite directions. On the face at x= c, where v = -ci and .6.A. = IIA. Iii, we have
(0, 0, 0)
On the face at x= -c, where v= ci and D.A. = -IIA Iii, the dot product is still negative:
r
.
r
.
v dA
v dA +
}Face x=c
}Face x=-c
3
2
= -c · Area of one face + (-c) · A re a of other face =-2c(2c) = -8c .
dA =
Thus,
v·
dA
J
(-sc3)
s
=lim --3 =-1.
lim
Volume--,O Volume of cube c--,O (2c)
Since the vector field points inward toward the yz-plane, it makes sense t hat the divergence
is negative at the origin.
(ii) Take S2 to be a cube as before, but centered this time at th e point (2,2,0). See Figure 19.31.
As before, the flux through the faces perpendicular to the y- and z-axes is zero. On the face
divv(0,0,0)=
at x = 2 + c,
v . .6.A = -(2 + c) 11.6.A 11·
On the face at x= 2 - c with outward normal, the dot p roduct is positive, and
v . .6.A = (2- c) 11.6.AII·
+
+
+
+
+
+
--• s
.... -.... -.... -.... -.... --
( 2, 2, 0 )
....
�
x
Figure 19.31: Vector field v = -xi in the xy-plane
v . .6.A = -cll.6.AII·
r v.
y
81
v . .6.A = -c 11.6.A11·
}S1
.
3
fs v dA = lim -8c
1·Im
- -3 ) = -1· ·
(
Yolurne--,O Volume of cube c--,O (2c)
.!P
(a) Using the geomet ric definition, find the divergence of = -xi at : (i) (0, 0,0) (ii) (2,2,0).
(b) Con firm that the co ordinate definition gives the same results.
T herefore, the flux through the cube is given by
.
�! though '.he vect r field is flow_ing away from the point (2,2,O) on the left, this outflow
�
is mailer. ma mtud_t; than the mflow on the right, so the net outflow is negative.
�
.
�
(b) Smee v = -xi + 0.7 +Ok, t he formula gives
T he next example shows that the divergence can be negative if there is net inflow to a point.
Solution
r
.
v dA +
v dA
}Face x=2 -c
}Face x=2+c
= (2 - c) · Area of one face - (2 + c) · Area of other face = -2c(2c)2 =
-Sc3
Then, as before ,
S2
47ra3
Flux
lim 43 = lim 3= 3.
div F (0,0,0) = lim --- = a--,O
a--,O
a--,O Volume
37ra
Example2
1027
Why Do the Two Definitions of Divergence Give the Same Result?
T_h� g�ometric defini tion d�fin s di: F as the flux density of F . To see wh y the coordinate defi­
�
m 10n is also t?e flux density, 1magme computing the fl ux out of a small box-shaped s urface at
�
(xo,Yo, zo), wit? si_ des of leng th D.x, Doy, and Doz par allel to the axes. On S 1 (the back face of the
_ I:,1gure 19.32, �here x= xo), the outward normal is in the negative x-direction so
bo! shown 111
. F is approximately consta nt on S , we have
dA = -dydz i . Assummg
1
s
1s, F.
dA =
r F.
ls,
(-i)dydz � -Fi(xo,Yo,zo)
s2
F . dA =
dydz
= -Fi (xo,Yo, zo) · A rea of S1 = -F1(x0, y0, zo) Doy Doz.
Oi:i, S 2, the f ce where x = xo + D.x, th e
�
dA = dydz i . T herefore,
1
r
ls,
r
ls2
outward normal
points in
F . idydz � Fi (Xo + D.x,Yo, Zo)
r
Js2
the
dydz
positive x-direction, so
= Fi (xo + D.x,Yo, zo) · A rea of S 2 = Fi (x0 + D.x,y0, zo) Doy Doz.
1028
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
19.3 THE DIVERGENCE OF A VECTOR FIELD
""' \
55 (Bottom)
x
Thus,
Figure 19.32: Box used to find div F at (xo, yo, zo)
r
ls,
F . dA
+
r
ls2
F . dA � Fi(x o + 6.x, Yo, z o)6.y6.z - Fi(x o, Yo, zo)6.y6. z
_ A (x o+ 6.x, Yo, z o) F1 (x o, Yo, zo)
6.x6.y6.z
6.x
oFi
� -6.x6.y6.z .
OX
/
Solution
//
//
/
//
Thus, adding these contributions, we have
x
\
""'
(a) The components of E are
E =
o
(
o
(
oz
Since the vol ume of the box is 6.x 6.y 6.z, the flux density is
0F3
0F2
oF 1
6.x 6.y6.z + -6.x6.y6.z + �6.x6.y6.z
uz .
oy
OX
�
6.x6.y6.z
oA
0F3
oFi
=-+-+.
oz
oy
ox
Example3
is
-
E
i"
= fif\lP
i" = xi + yJ + zk , i" i O ·
(a) Find a formula for div E.
_
(b) Is there a val ue of p for which E is divergence-free? If so, find it.
+
(x2+ y2+ z2)p/2 i (x2+ y2+ z2)p/2 J+ (x2+ y2+ z2)p/2 k.
x
.,.
y
-;
z
1
pz 2
(x2 + y2+ z2)p/2 ) = (x2 + y2+ z2)p/2 - (x2+ y2+ z2)(p/2)+1 ·
z
. div E =
2+ z2)
3
p(x2 + y__
-"----'---____::
...:....__
(x2+ y2+ z2)p/2 (x2+ y2+ z2)(p/2)+1
3-p
3-p
(x2+ y2+ z2)p/2
lir IIP.
Magnetic Fields
An important class of div ergence-free vector fields is the magn
etic fields. One of Maxwell's Laws
of Electromagnetism is that the magnetic field B satisfie s
' F = 0 everywhere that F is
said to be dive,gence free or sol.eno1'da,l 1'f dIV
F igure 19.33 shows, for three values of the constant p, the vector field
p=3
(b) The divergence is zero when p = 3, so F (i") = i" /llf"ll 3 is a divergence-free vector field.
Notice that the div ergence is zero even though the v ectors point outward from the origin.
_
Divergence-Free Vector Fields
F
//
/ I
p l
=
Figure 19.33: The vector field E (1°") = r
/1'77 ff P for p = 0, 1, and 3 in the xy-plane
p=O
So
0F3
0F2
oFi
Total flux through S � - 6.x 6.y 6.z+ � 6.x 6.y 6.z+ !Cl 6.x 6.Y 6.z .
uz
uy
ox
A vector field
defi ne d.
I /
x
1
px2
2
2
ox (x + y + z2) p/2 ) = (x2+ y2+ z2)p/2 - (x2+ y2+ z2)(p/2)+1
o
y
2
1
PY
(
2
2
oy (x + y + z2) p/2 ) = (x2+ y2+ z2)p/2 - (x2+ 2+ z2)(p/2)+1
y
0F3
A
- 6.x 6.y LJ.Z.
oz
s
y
//
We compute the partial derivatives
By an analogous argument, the contribution to the flux from S3 and S4 (the surfaces perpendicular
to the y-axis) is approximately
0F2
- 6.x 6.y 6. z ,
oy
and the contr ibution to the flux from Ss and S5 is approximately
Total flux through
Volume of box
y
1029
div B
Example4
=
0.
An infi nitesimal current loop, similar to that shown in F ig ure 19.34, is called a magnetic dipole. Its
magnitude is described by a constant vector fl,, called the dipole moment. Th e magnetic field due to
a magnetic dipole with moment fl, is
-
fl,
B = - lif 11
3+
3(µ. r)i"
lif 115 '
r i o.
•·
1030
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
Show that div B
(a)
Area= a
Y
(b)
"'-\\II!/
"' '/1 //
���·--�x
=
To show that div B
0 we can use the following version of the product rule for the divergence: if
g is a scalar function and F is a vector field, then
div(g.F)
= (gradg) · F + g div F.
Problems
.
div
__ r
iv µ . r l
5
=d
lf1 1
.
((- -)-) ( - )
µ ·r r
5
llf11
r
__
.
r
= g ad(µ r). llf Jl5
-
.
__
.
v
r) di
ple 3 on page 1028, we have
From Problems 67 and 68 on page 796 and Exam
grad
(1)
3
llfll
-3r
grad(µ . r)
=llfll 5 '
=
µ,
div
(-)
r
+ (1,i
ll
5
f11
B
µ+3gra d(µ . r) .
3µ r 6µ . r
+-----
- rad ( ; 3) .
= g
11 1 11
.
. fl,
3r
=
-5
llrll
= 0.
llf11
5
11
:
5
11
-2 .
2. div((2sin(xy)+tanz)i+(tany)J +(e +Y )k)
3. Which of the following two vector fields, sketched in the
xy-plane, appears to have the greater divergence at the
origin? The scales are the same on each.
I
I
1 I I
1. J e I J
centered at the point (1, 2, 3).
19. Let F be a vector field with div F
Estimate Js F · dA where Sis
(a)
y
����
'--'-' \ \
.........._,,,
\
.,......../// I
/////
H10
(II)
I////
I///.,,/
\
,,
x
. . . , , ,.
:::::!:::::
'' '
I
I
\ \ ''-'--
0��) ��\�
I
I
•
/ I I
I
I
,t'
;'
I I I I / /
/
.....................
.. ' '
I
= (-x+y)i + (y+z)J + (-z+x)k
7. F(x,y,z)
8. F(x,y)
= (x 2 - y2 )i + 2xyJ
= 3x2 i - sin(xz)(i + k)
9. F (x, y, z)
10. F
= (ln(x 2+l)i +(cosy) J +(xye z ) k)
=a
-
12. F (x,y)
13. f (r)
=
-
x i"
-yi + xJ
= ---"-------"---­
x2 + 2
y
i" - 1"' 0
- llf -
r a II '
1"' -1-. 1"' a
'
(b) Use !he geometric definition and part (a) to find
div F at �e point (a, b, c).
(c) Find div F using partial derivatives.
S1 (Back)
z
S4 (Back)
\
y
(i) A sphere of radius 0.1 centered at (2, 0, O).
......
\ \ '\..
x
(a) div F at (4, 5, 2)
(b) The flux' of F out of a sphere of radius 0.2 centered
at (4, 5, 2).
= 2xi - 3yJ + 5zk
through a box with four of its corners at the points
(a, b, c), (a+w, b, c), (a, b+w, c), (a, b, c+w) and
edge length w. See Figure 19.35.
21. (a) Find the flux of F
Ss (Bottom)
x
x2 + y 2 _ z.
20. The flux of F out of a small sphere of radius 0.1 centered
at (4, 5, 2), is 0.0125. Estimate:
y
'' \
= -yi + xJ
6. F(x, y) =-xi+ yJ
11. f (r)
(ii) A box of side 0.2 with edges parallel to the axes
and centered at (0, 0, 10).
(b) The P0nt (2, 0, 0) is called a source for the vector
field F; the point (0, 0, 10) is called a sink. Ex­
plain the reason for these names using your answer
to part (a).
(I)
5. F (x,y)
18. A smooth vector �eld F has div F (1, 2, 3) = 5. Esti­
mate the flux of F out of a small sphere of radius 0.01
+ 3(µ . r) div ( 11 :115 )
Exercises
2
.:::.:::.:::J.:::.:::.:::
r
17. A small sphere of radius 0.1 surrounds the point
(2, 3, -1). The flux of a ve_<:tor field G into this sphere
1s 0.000041r. Estimate div G at the point (2, 3, -1).
Exercises and Problems for Section 19.3
x2
... ...
In Exercises �-13, _find t�e divergence of the vector field.
(Note: = xi +yj +zk .)
16. Draw two vector fields that have zero divergence every­
where.
l lf115
Are the quantities in Exercises 1-2 vectors or scalars? Calcu­
late them.
2 2
2
z 1. div((x + y)i + (xye )j - ln(x + y )k )
-
15. Draw two vector fields that have negative divergence ev­
erywhere.
.
Putting these results together gives
div
•-
14. Draw two vector fields that have positive divergence ev­
erywhere.
5
r ) =llf 11
( llrll
5
y
-
____________....__ x
i i ilti·
I I
I
0, we have
(See Problem 36 on page 1032.) Thus, since divµ =
and
...
/Ill\\"'(c)
y
-
::;;1:::
Figure 19.34: A cun-ent loop
Solution
�I�t� ���
4. For each of the following vector fields, sketched in the
xy-plane, decide if the divergence is positive, zero, or
negative at the indicated point.
= 0.
1031
19.3 THE DIVERGENCE OF A VECTOR FIELD
Figure 19.35
= (3x + 2)i + 4xJ + (5x + l)k. Use the
method of Exercise 21 to find div F at the point (a, b, c)
by two different methods.
22. Suppose F
v
23. Use the geometric definition of divergence to find div
= -2r. Check that you get the
at the origin, where
same result using the definition in Cartesian coordinates.
v
24. (a) Letf(x,y)=axy+ax2 y+y3 .Finddivgradf.
(b) If possible, choose a so that div grad f = O for all
x,y.
25. Let
F
= (9a2 x + 10ay 2 )i + (10z3 - 6ay)J
2
2
(3z + 10x + 10y )k.
Find the value(s) of a making div F
WO
W
A�ci�m
-
1032
19.3 THE DIVERGENCE OF A VECTOR FIELD
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
r
26. The vector field F (1"'') = /111"113 is not defined at the
origin. Nevertheless, we can attempt to use the flux defi­
nition to compute div F at the origin. What is the result?
27. The divergence of a magnetic vector field B must be zero
everywhere. Which of the following vector fields cannot
be a magnetic vector field?
32. Due to roadwork ahead, the traffic on a highway slows
linearly from 55 miles/hour to 15 miles/hour over a
2000-foot stretch of road, then crawls along at 15
miles/hour for 5000 feet, then speeds back up linearly to
55 miles/hour in the next 1000 feet, after which it moves
steadily at 55 miles/hour.
(a) Sketch a velocity vector field for the traffic flow.
(b) Write a formula for the velocity vector field
(miles/hour) as a function of the distance x feet from
the initial point of slowdown. (Take the direction of
motion to be i and consider the various sections of
the road separately.)
(c) Compute div at x = 1000, 5000, 7500, 10,000.
Be sure to include the proper units.
v
(a) B(x, y, z) =-yi+xJ+ (x+y)k
(b) B(x,y,z)
(c) B(x,y,z)
= -zi+yJ+xk
= (x2 - y 2 - x)i + (y - 2xy)J
28. Let F(x,y,z) = zk.
(a) Calculate div F .
(b) Sketch F. Does it appear to be diverging? Does this
agree with your answer to part (a)?
v
v
· 33. The velocity field in Problem 32 does not give a com­
plete description of the traffic flow, for it takes no account
of the spacing between vehicles. Let p be the density
(a) Calculate div F .
(cars/mile) of highway, where we assume that p depends
(b) Sketch F. Does it appear to be diverging? Does this
only on x.
agree with your answer to part (a)?
29. Let F (?) =
r /llr 11
3
(in 3-space),
r =I- 5.
(a) Using your highway experience, arrange in ascend­
ing order: p(O), p (lOOO), p(5000).
(b) What are the units and interpretation of the vector
field pv?
(c) Would you expect pv to be constant? Why? What
does this mean for div(pv )?
(d) Determine p(x) if p(O) = 75 cars/mile and pv is
constant.
(e) If the highway has two lanes, find the approximate
number of feet between cars at x = 0, 1000, and
5000.
Problems 30-31 involve electric fields. Electric charge pro­
duces a vector field E , called the electric field, which rep­
resents the force on a unit positive charge placed at the
point. Two positive or two negative charges repel one another,
whereas two charges of opposite sign attract one another. The
divergence of E is proportional to the density of the electric
charge (that is, the charge per unit volume), with a positive
constant of proportionality.
UIIH�II
30. A certain distribution of electric charge produces the
electric field shown in Figure 19.36. Where are the
charges that produced this electric field concentrated?
Which concentrations are positive and which are nega­
tive?
r
34. Let =xi+ Y] + zk and c = C1i + C2 ) + C3k, a
constant vector; let S be a sphere of radius R centered at
the origin. Find
(b) f (r x c) . dA
(a) div(r x c)
8
y
35. Show that if a is a constant vector and f(x, y, z) is a
function, then div(f = (grad !) .
a)
a.
36. Show that if g(x,y, z) is a scalar-valued function and
F (x, y, z) is a vector field, then
-4 -3 -2 -1
1
2
x
div(gF)=(gradg)·F +g div F.
37. If f(x, y,z) and g(x, y,z) are functions with continuous
second partial derivatives, show that
div(grad f x grad g) = 0.
Figure 19.36
31. The electric field at the point r as a result of a point
charge at the origin is E (?) =
/llf 11 3 .
kr
(a) Calculate div E for r =f. 0.
(b) Calculate the limit suggested by the geometric defi­
nition of div E at the point (0, 0, 0).
(c) Explain what your answers mean in terms of charge
density.
In Problems 38-40 use Problems 36 and 37 to find the diver­
gence of the vecto/ field. The vectors and b are constant.
a
-
38. F =
40.
1
l
llf lP
-
a
-
xr
c = (b . na x .,:
39.
B
41. Let F(x, y) = u(x,y)i + v(x,y)J be a 2--dimensional
vector field. Let F(x, y) be the magnitude of F and let
e(x, y) be the angle of F with the positive x-axis at the
point (x, y), so that u = F cos (J and v = F sine. Let
T be the unit vector in the direction of F, and let fJ be
the unit vector in the direction of k x F, perpendicular
to F . Show that
This problem shows that the divergence of the vector
field F is the sum of two terms. The first term, F(JfJ ,
is due to changes in the direction of F perpendicular to
the flow lines, so it reflects the extent to which flow lines
of F fail to be parallel. The second term, Fi' , is due to
changes in the magnitude of F along flow lines of F .
42. In Problem 64 on page 1015 it was shown that the rate
of heat loss from a volume Vin a region of non-uniform
temperature equals k f8(grad T) · dA, where k is a con­
stant, Sis the surface bounding V, and T(x, y, z) is the
temperature at the point (x, y, z) in space. By taking the
limit as V contracts to a point, show that, at that point,
�� = Bdiv gradT
where B 'is a constant with respect to x, y, z, but may
depend on time, t.
1033
43. A vector field, v, in the plane is a point source at the ori­
gin if its direction is away from the origin at every point,
its magnitude depends only on the distance from the ori­
gin, and its divergence. is zero away from the origin.
(a) Explain why a point source at the origin must be of
- for some
the form v- = [f(x 2 +y2)] (xi- + yj)
positive function f.
(b) Show that v =K(x 2 +y 2)-1 (xi + yJ) is a point
source at the origin if K > 0.
(c) What is the magnitude llv II of the source in part (b)
as a function of the distance from its center?
(d) Sketch the vector fieldv = (x 2 +y2)-1 (xi +yJ).
1
(e) Show that rp = tog(x2 + y 2 ) is a potential func­
tion for the source in part (b).
45. div(2xi) = 2i.
46. ForF(x,y,z) = (x 2+y)i+(2y+z)J-z2 k we have
divF = 2xi +2J - 2zk.
47. The divergence of f(x, y, z)
div f(x, y, z) = 2x + z + y.
x2 + yz is given by
In Problems 48-50, give an example of:
48. A vector field F (x, y, z) whose divergence is a nonzero
constant.
49. A nonzero vector field
zero.
F (x, y, z)
whose divergence is
v,
(a) Explain why a point sink at the origin must be of
the form v = [f(x 2 +y 2 )] (xi + yJ) for some
negative function f.
(b) Show thatv =K(x 2 +y2)-1 (xi +yJ) is a point
sink at the origin if K < 0.
(c) Determine the magnitude llv II of the sink in part (b)
as a function of the distance from its center.
1
2
(d) Sketch v = -(x 2 +y )- (xi + yJ).
2
(e) Show that rp = log(x + y2) is a potential func­
tion for the sink in part (b).
1
55. A constant vector field
divergence.
F
ai + bJ
+
ck
56. If a vector field F in 3-space has zero divergence then
F = ai +bJ +ck where a, band care constants.
57. If F is a vector field in 3-space, and f is a scalar func­
tion, then div(! F) = fdivF.
58. If F_js a vector field in 3-space, and F = grad f, then
divF = 0.
59. If F is a vector field in 3-space, then grad(div F) =
O.
Are the statements in Problems 51-63 true or false? Give rea­
sons for your answer.
61. If f(x, y, z) is any given continuous scalar function, then
there is at least one vector field F such that divF = f.
·G) =F(div G)+(div F)G
53. div F is a scalar whose value can vary from point to
point.
I�
has zero
60. The field F ( r) =
51. div(F + G) =div F + div G
.,
54. If F is a vector field in 3-space, then divF is also a vec­
tor field.
50. A vector field that is not divergence free.
52. grad(F
,,.
,,,
44. A vector field,
in the plane is a point sink at the ori­
gin if its direction is toward the origin at every point, its
magnitude depends only on the distance from the origin,
and its divergence is zero away from the origin.
Strengthen Your Understanding
In Problems 45-47, explain what is wrong with the statement.
JI
r is divergence free.
62. If F and G are vector fields satisfying divF = divG
then F = G.
63. There exist a scalar function f and a vector field F sat­
isfying div(grad !) =grad(div F ).
I�
1034
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
19.4 THE DIVERGENCE THEOREM
64. For r =xi+ yJ + zk, an arbitrary function J(x, y, z)
and an arbitrary vector field, F (x, y, z), which of the
following is a vector field, and which is a constant vector
field?
19.4
(a) gradf
(b)
(d) (div i)F
(e) grad(div F)
(divF)i
(c)
(divr)i
Flux out of vV
Flux out of W
= (
lw
div F dV.
We have calculated the flux in two ways, as a flux integral and as a volume integral. Therefore,
these two integrals must be equal. This result holds even if vV is not a rectangular solid. Thus, we
have the following result.3
The Divergence Theorem is a multivariable analogue of the Fundamental Theorem of Calculus;
it says that the integral of the flux density over a solid region equals the flux integral through the
boundary of the region.
Theorem 19.1: The Divergence Theorem
The Boundary of a Solid Region
A solid region is an open region in 3-space. The boundary of a solid region may be thought of as the
slcin between the interior of the region and the space around it. For example, the boundary of a solid
ball is a spherical surface, the boundary of a solid cube is its six faces, and the boundary of a solid
cylinder is a tube sealed at both ends by disks. (See Figure 19.37). A surface which is the boundary
of a solid region is called a closed s111face.
If HI is a solid region whose boundary Sis a piecewise smooth surface, and if f is a smooth
vector field on an open region containing HI and S, then
r
ls
Hf = Solid cube
W =Solid Cylinder
W=Ball
S = Sphere
f . dA
=
where S is given the outward orientation.
S = 6 square faces
S = Tube and two disks
r
lw
div f dV,
Example1
Use the Divergence Theorem to calculate the flux of the vector field f (r) = r through the sphere
of radius a centered at the origin.
Solution
In Example 5 on page 1010 we computed the flux using the definition of a flux integral, giving
dA = 41ra 3
r
Now we use div F
Figure 19.37: Several solid regions and their boundaries
ls r
( r · dA = (
Consider a solid reoion HI in 3-space whose boundary is the closed surface S. There are two ways
to find the total flux of a vector field F out of Hf. One is to calculate the flux of F through S:
·
= div(xi + yJ + zk) = 3 and the Divergence Theorem:
ls
Calculating the Flux from the Flux Density
b
L Flux out of small boxes � L div f !:::,. V.
We have approximated the flux by a Riemann sum. As the subdivision gets finer, the sum approaches
an integral, so
THE DIVERGENCE THEOREM
HIIM
=
1035
lw
div F' dV
= (
lw
3 dV
= 3 · i 1ra3 = 41ra3..
3
-
-
Example2
Flux out of W = ls F · dA.
f
= x2
_ (x, y, z) (
through the cube in Figure 19.40.
Another way is to use div f, which gives the flux density at any point in vV. We subdivide VV
into small boxes, as shown in Figure 19.38. Then, for a small box of volume!:::,. V,
Flux out of box � Flux density · Volume
Use the Divergence Theorem to calculate the flux of the vector field
= div F !:::,. V.
+ y2 )i + (y2 + z2 )J + (x2 + z2 )k
z
What happens when we add the fluxes out of all the boxes? Consider two adjacent boxes, as
shown in Figure 19.39. The flux through the shared wall is counted twice, once out of the box on
each side. When we add the fluxes, these two contributions cancel, so we get the flux out of the solid
region formed by joining the two boxes. Continuing in this way, we find that
Fluxes through
inner wall cancel
x
Figure 19.40
Solution
Figure 19.38: Subdivision of
region into small boxes
Figure 19.39: Adding the flux out
of adjacent boxes
The divergence of F is div F
2x + 2y + 2z. Since div f is positive everywhere in the first
quadrant, the flux through Sis positive. By the Divergence Theorem,
3 A proof of the Divergence Theorem using the coordinate definition of the divergence. can be found in the online supple­
ment at www.wiley.com/college/hughes-hallett.
1036
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
1 F.
8
dA
=
111
1
1
0
1 2(x+y+z)dxdydz = 1
19.4 THE DIVERGENCE THEOREM
1
1
2
l\x +2x(y+z))\ dydz
Example4
O
-
},
\ +2(y+ z)dydr [(y+:': 2yz{ dz
� !,'
= i\2 + 2z) dz= (2z + z
2
{ = 3.
Solution
Calculate
0=
3
'
Is F . dA, using the Divergence Theorem if possible, for the following surface s:
and thus
1
defined.
Notice that the flux is not zero, although div F is zero eveEy where it is
re in T,V,
whe
very
(b) Suppose Hi is the solid region enclosed by S2 . Since div F = 0 e
the Divergence Theorem in this case, giving
r F.
ls2
dA
=
r
lw
div F dV
r F.
ls2
dA.
div ff dV
= o.
{ F . dA = { F . dA -
h
r
1�
F . dA
=
r
ls,
1�{ F . dA,
F . dA.
In Example 3 we showed that Is, F · dA = 41r, so Is2 F · dA = 41r also. Note that it would have
been more difficult to compute the integral over the ellipsoid directly.
Electric Fields
The electric field produced by a positive point charge q placed at the origin is
a
�
� s1
s,
=
lw
ls2
defined every where inside
(a) We cannot use th e Divergence Theorem directly Eecause F is not
ery where on S1, the flux
ev
ard
w
the sphere (it is not defined at the origin). Since F points out
out of S1 is positive. On S1 ,
a
dA,
F · dA = IIF lldA = 3
a
so
2
dA = 2._(Area of S1) = \41ra = 41r.
F · dA = 2_
1
dA
fi-0,
3
'
16,
But S consists of S2 oriented outward and S1 oriented inward, so
(a) S1 is the sphere of radius a centered at the origin.
(b) S2 is the sphere of radius a centered at the point (2a, 0, 0).
Solution
llfll
f ft . dA = f
f
llfll
f
=
The ellipsoid contains the sphere ; let Hi be the solid region between the m. Since W does not contain
the origin, div F is defined and e qual to zero e very where in VV. Thus, if Sis the boundary of 1'V,
then
ls
In Example 3 on page 1028 we saw that the following vector field is divergence free :
F (f ) =
r F.
ls,
An important application of the Diverge nce Theorem is the study of divergence-free vector fields.
-
F (f ) =
show that
The Divergence Theorem and Divergence-Free Vector Fields
Example3
Le t S1 be the sphere of radius 1 centered at the origin and let S2 be the ellipsoid x2 +y2 +4z2
both oriented outward. For
1037
=
r
lw
OdV
we
can use
Using Example 3, w e see that the flux of the electric field through any sphere centered at the origin
is 41rq. In fact, using the idea of Example 4, we can show that the flux of E through any closed
surface containing the origin is 41rq. See Problems 36 and 37 on page 1039. This is a special case of
Gauss's Law, which state s that the flux of an electric field through any closed surface is proportional
to the total charge enclosed by the surface . Carl Friedrich Gauss (1777-1855) also discovered the
Divergence Theorem, which is sometimes called Gauss's Theorem.
= o.
y S, even in case s
The Diveroence Theorem applie s to any solid re gion VV and its boundar
region ?et ween
solid
e
th
is
Hi
if
,
e
exampl
For
e s.
where the boundary consists of two or more surfac
pomt, then
ame
s
e
h
t
at
red
e
nt
e
c
h
bot
2,
the sphere 31 of radius 1 and the sphere S2 of radius
.
outward
he
t
requues
m
e
eor
h
T
e
nc
e
rg
e
Div
the boundary of Hi consists of both S and S2 . The
(See
center.
e
th
ard
w
to
points
S
on
and
1
orientation, which on 32 points away from the center
1
Figure 19.41.)
Exercises and Problems for Section 19.4
Exercises
J�
For Exercises 1-5, compute the flux integral F ·dA in two
ways, ifpossible, directly and using the Divergence Theorem.
In each case, S is dosed and oriented outward.
1. F (f) = f and S is the cube enclosing the volume
0 :::; x :::; 2, 0 :::; y :::; 2, and O :::; z :::; 2.
Figure 19.41: Cut-away view of the region W between two spheres,
showing orientation vectors
2. F = Y] and Sis a closed vertical cylinder of height 2,
with its base a circle ofradius 1 on the xy-plane, centered
at the origin.
3. F =x 2 i+2y2 J +3z2 k andSis the surface ofthe box
with faces x = 1, x = 2, y = 0, y = 1, z = 0, z = 1.
4. F = (z2 + x)i + (x 2 + y)J + (y 2 + z)k andSis the
closed cylinder x2 + z2 = 1, with O :::; y :::; 1, oriented
outward.
5. F = -zi + xk and Sis a square pyramid with height
3 and base on the xy-plane ofside length 1.
Find the flux of the vector fields in Exercises 6-12 out of the
closed box O :::; x :::; 2, O :::; y :::; 3, O :::; z :::; 4.
6· F
7. G
= 4i +7J -k
= yi + zk
1038
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
8. fi = xyi + zJ + yk
9. J = xy 2J +xk
10. N = ez i + sin(xy)k
19.4 THE DIVERGENCE THEOREM
11. 111 = ((3x + 4y)i + ( 4y + 5z).7 + (5z + 3x)k
12. 111 where div M = xy + 5
Problems
(c) Use your answers to parts (a) and (b) to calculate the
flux out of a box of side 10 centered at the origin and
with sides parallel to the coordinate planes. (The box
is also oriented outward.)
13. Find the flux of F = zi + y.7 + xk out of a sphere of
radius 3 centered at the origin.
14. Find the flux of F = xyi + yz.7 + zxk out of a sphere
of radius 1 centered at the origin.
15. Find the flux of F = x 3 i + y 3. 7 + z3 k through the
closed surface bounding the solid region x 2 + y 2 S 4,
OSzS5, oriented outward.
2
2
2
16. The region W lies between the spheres x + y + z = 4
2
2
2
and x2+y 2 + z = 9 and within the cone z = x + Y
S, ori> O; its boundary is the closed7surface,
with z 33 -:'
3
ented outward. Find the flux of F = x i + y J + z k
out of S.
17. Find the flux of F through the closed cylinder of ra­
dius 2, centered on the z-axis, with 3 S z S 7, if
F = (x + 3e yz )i + (ln(x2 z2 + 1) + y).7 + zk.
e Y z i + (tan( 0.00l x 2 z2 ) +
2
2 2
y )J + (ln(l + x y ) + z2 )k out of the closed box
OSxS5, 0 S y S 4, 0SzS3.
2 2
(b) Can you use the Divergence Theorem to compute
2
the flux of /I \r 11 out of a closed cylinder of ra­
dius 1, length 2, centered at the origin, and with its
axis along the z-axis?
(c) Compute the flux of /I \r 112 out of the cylinder in
part (b).
2
(d) Find the flux of i;' /I \r 11 out of a closed cylinder of
radius 2, length 2, centered at the origin, and with its
axis along the z-axis.
r
r
26. Let S be the cube in the first quadrant with side 2, one
corner at the origin and edges parallel to the axes. Let
2
2 -:'
2 k­
2
2
F- 1 = (xy + 3xz )i- + (3x y + 2yz )J + 3zy -
19. Find the flux of F = x i + z.7 + yk out of the closed
cone x = y 2 + z2 , withOSx S 1.
F 2 = (xy 2+5e yz)i +(yz2 +7sin(xz)).J +(x2 z+cos (xy))k
z3 y 3 -:'
2
x3 7
2
2
F 3 = (xz + 3 ) i + (yz + 3 ) J + (zy + 3 ) k .
2
J
20. Suppose F is a vector field with div F = 10. Find the
flux of F out of a cyUnder of height a and radius a, cen­
tered on the z-axis and with base in the xy-plane.
21. A cone has its tip at the point (0, 0, 5) and its base the
disk D, x2 + y 2 S 1, in the xy -plane. The surface of the
cone is the curved and slanted face, S, oriented upward,
and the flat base, D, oriented downward. The flux of the
constant vector field F = ai + b. 7 + ck through S is
given by
1
_
Arrange the flux integrals of F1, F2 , F3 out of Sin
increasing order.
27. LetdivF =2(6-x) andOSa,b,cSlO.
(a) Find the flux of F out of the rectangular solidO S
xSa, 0Sy Sb, 0SzSc.
(b) For what values of a, b, c is the flux largest? What is
that largest flux?
f . dA = 3.22.
Is it possible to calculate ID F · dA? If so, give the
answer. If not, explain what additional information you
would need to be able to make this calculation.
28. Assumer #
r·
23. A vector field F satisfies div F = 0 everywhere. Show
that Is F · dA = 0 for every closed smface S .
r
3
l\r 11
F
be the vector field
.
2
(a) Calculate the flux of F out of the unit sphere x +
2
2
y + z = 1 oriented outward.
(b) Calculatediv F. Show your work and simplify your
answer completely.
6. Let F = r /1\r 11 3 .
(a) Calculate the flux of F out of the unit sphere x 2 +
y 2 + z2 = 1 oriented outward.
(b) Calculate div F. Simplify your answer completely.
(c) Use your answer to parts (a) and (b) to calculate the
flux out of a box of side LO centered at the origin and
with sides parallel to the coordinate planes. (The box
is also oriented outward.)
22. If V is a volume surrounded by a closed surface S, show
dA = V.
that � Is
24. Let = xi + yj + zk and let
given by
1
2
25. (a) Finddiv(r/l\rll ) wherer = xi+y.7forr#O.
J
18. .Find the .flux of F
31. (a) Let div F = x2 +y 2+z2+3. Calculate Is F ·dA
1
where S1 is the sphere of radius I centered at the ori­
gin.
(b) Let 82 be the sphere of ra dius 2 centered at the ori­
gin; let 83 be the sphere of radius 3 centered at the
origin; let 84 be the box of side 6 centered at the
origin with edges pai:allel to the axes. Without cal­
culating them, arrange the following integrals in in­
creasing order:
r
In Problems 29-30, find the flux of F = /l\f 11 3 through
the surface. [Hint: Use the method of Problem 28.]
29.
S is the ellipsoid x2 + 2y 2 + 3z2
=
6.
30. Sis the closed cylinder y 2 + z2 = 4, -2 SxS2.
s.
f · dA.
32. Suppose div F = xyz 2 .
(a) Find div F at the point (1, 2, 1). [Note: You are
given div F, not F .]
(b) Using your answer to part (a), but no other informa­
tion about the vector field F, estimate the flux out of
a small box of side 0.2 centered at the point (1, 2, 1)
and with edges parallel to the axes.
(c) Without computing the vector field F, calculate the
exact flux out of the box.
33. Supposediv F = x +y +3. Find a surface S such that
Is F · dA is negative, or explain why no such surface
exists.
2
35. If a surface S is submerged in an incompressible fluid, a
force F is exerted on one side of the surface by the pres­
sure in the fluid. If the z-axis is vertical, with the positive
direction upward and the fluid level at z = 0, then the
component of force in the direction of a unit vector i1 is
given by the following:
F . 11
(a) What is the value of div F? Include units.
(b) Assume the heat flows outward symmetrically. Ver­
ify that F = ai;' , where = xi + y.7 + zk and a
is a suitable constant, satisfies the given conditions.
Find a:.
(c) Let T(x, y, z) denote the temperature inside the
earth. Heat flows according to the equation F =
-k grad T, where k is a constant. Explain why this
makes sense physically.
(d) If T is in °C, then k = 30,000 watts/km 0 C. Assum­
ing the earth is a sphere with radius 6400 km and
surface temperature 20° C, what is the temperature
at the center?
r
=-
1
zo9a . dA,
where 8 is the density of the fluid (mass/�olume), g is the
acceleration due to gravity, and the smface is oriented
away from the side on which the force is exerted . In this
problem we consider a totally submerged closed surface
enclosing a volume V. We are interested in the force of
the liquid on the external smface, so Sis oriented inward.
Use the Divergence Theorem to show that:
(a) The force in the i and J directions is zero.
(b) The force in the k direction is 8g V, the weight of
the volume of fluid with the same volume as V. This
is Archimedes' Principle.
36. According to Coulomb's Law, the electrostatic field E at
the point due to a charge q at the origin is given by
r
- (r) =
2
34. As a result of radioactive decay, heat is generated uni­
formly throughout the interior of the earth at a rate of 30
watts per cubic kilometer. (A watt is a rate of heat pro­
duction.) The heat then flows to the earth's surface where
it is lost to space. Let F(x, y, z) denote the rate of flow
of heat measured in watts per square kilometer. By def­
inition, the flux of F across a surface is the quantity of
heat flowing through the surface per unit of time.
1039
E
q
r
l\r ll3·
(a) Compute div E .
(b) Let Sa be the sphere of radius a centered at the ori­
gin and oriented outward. Show that the flux of E
tlu·ough Sa is 41rq.
(c) Could you have used the Divergence Theorem in
part (b)? Explain why or why not.
(d) Let S be an arbitrary, closed, outward-oriented sur­
face surrounding the origin. Show that the ·flux of
E through S is again 41rq. [Hint: Apply the Diver­
gence Theorem to the solid region lying between a
small sphere Sa and the surface S.]
37. According to Coulomb's Law, the electric field E at the
point due to a charge q at the point O is given by
r
r
E- (r)
=q
(f-fo)
.
l\r -r o ll3
Suppose S is a closed, outward-oriented surface and that
r o does not lie on S. Use Problem 36 to show that
1E
s
,
. dA = { 41rq if q lies inside S
O
if q lies outside S.
Strengthen Your Understanding
In Problems 38-39, explain what is wrong with the statement.
38. The flux integral Is F · dA can be evaluated using the
Divergence Theorem, where F = 2xi - 3.7 and S
is the triangular surface with corners (1, 0, 0), ( 0, 1, 0),
(0, 0, 1) oriented away from the origin.
39. If Sis the boundary of a solid region W, where Sis ori­
ented outward, and F is a vector field, then
1
s
div F
1
dA =
f dV.
'
w
'"
1040
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
In Problems 40--41, give an example of:
of a solid
region
such
40. A surface
- is the boundary
- S that
that fs F · dA = 0 if F (x, y, z) = yi +xz.i + y2 k".
41. A vector field F such that the flux of
of radius 1 centered at the origin is 3.
F out of a sphere
Are the statements in Problems 42-46 true or false? The
smooth vector field F is defined everywhere in 3-space and
has constant divergence equal to 4.
42. The field F has a net inflow per unit volume at the point
(-3,4,0).
43. The vector field F could be the field
(y-5x)k.
F = xi +(3y)J +
44. The vector field F could be a constant field.
45. The flux of F through a circle of radius 5 lying anywhere
on the xy-plane and oriented upward is 4(1r52 ).
46. The flux of F through a closed cylinder of radius 1 cen­
tered along the y-axis, 0 S yS 3 and oriented outward
is 4(31r).
Are the statements in Problems 47-55 true or false? Give rea­
sons for your answer.
47. fs F · dA = div F.
48. If F is a vector field in 3-space, and W is a solid re­
gion with boundary surface S , then fs div F · dA =
fw F d\!.
vector field in 3-space,
and
49. If F is a diveroence-free
0
- S is
a closed surface oriented inward, then fs F · dA = 0.
50. If F is a vector field in 3-space satisfying div F = 1,
and S is a closed surface oriented outward, then fs F
dA is equal to the volume enclosed by S.
51. Let vV be the solid region between the sphere S1 of ra­
dius 1 and S2 of radius 2, both centered at the origin.
If F is a vector field in 3-space, then fw div F d\! =
F . dA - j'S1 F · dA, where both S1 and S2 are
S2
oriented outward.
I
52. Let S1 be the squareOS x S 1,0 S yS 1,z = 0 ori­
ented downward and let S2 be the squareO S x S 1,0S
y S 1, z:: 1 oriented ipwar�. If F is� vect�r field, then
F · dA +
F · dA , where W
r div F d\! =
Jw
S1
S2
is the solid cubeOS x S 1, 0 S yS 1,0 S z S 1.
J
• Calculating flux integrals over surfaces
Graphs: dA = (-fx i - jy ) +k) dx dy
Cylinders: dA = (cos Bi +sin BJ )R dz dB
Spheres:
dA = (sin¢ cos Bi +sin¢ sin BJ + cos ¢k)R2 d¢ dB
(3i + 4]) · dA a vector or a scalar? Calculate it.
yi - xJ
2. Is div ( 2
a vector or a scalar? Calculate it.
x +y2 )
J-=r
55. Let Sh be the surface consisting of a cylinder of height
2
2
h, closed at the top. The curved sides are x + y = 1,
2
2
forOS z S h, and the top x +y S 1, for z = _h, ort
ented outward. If F is divergence free, then fs F · dA
is independent of the height h.
h
56. Let F = (5x + 7y)i + (7y + 9z)J + (9z +llx)k,
and let Q.i be the flux of F tlU"ough the surfaces S;. for
i =1-4. Arrange Q; in ascending order, where
(a) S1 is the sphere of radius 2 centered at the origin
(b) S2 is the cube of side 2 centered at the origin and
with sides parallel to the axes
(c) S 3 is the sphere of radius 1 centered at the origin
(d) S4 is a pyramid with all four corners lying on S3
• Divergence
Geometric and coordinate definition of divergence, cal­
culating divergence, interpretation in terms of outflow
per unit volume.
• The Divergence Theorem
Stiitement of the theorem, divergence-free fields, har­
monic functions.
3. Let F (x, y, z)
i + 2] + k. Each of the surfaces
in (a)-(e) are squares of side length 4, with orientation
given by the normal vector shown. Compute the flux of
F through these surfaces.
y
x
I
x
)-
21.
y
In Exercises 4-7, are the flux integrals positive, negative, or
zero? Let S be a disk of radius 3 in the plane x = 7, centered
on the x-axis and oriented toward the origin.
4.1 (xJ +yk). dA
6.
1
(yi
5. 1 (xi
+ yk) . dA
+ xk) . dA
1.1((x - lO)i + (x + 10)]). dA
In Exercises 8-23, find the flux of the vector field through the
surface S.
8.
9.
10.
11.
12.
13.
14.
15.
19.
20.
z
(e)
0, then div F = 0 at all points inside S.
18.
y
I
54. If Sis a sphere of radius 1, centered at the origin, oriented
outward, and F is a vector field satisfying fs F · dA =
Exercises
1
x
17.
z
(d)
1041
16. F = (x2 + y2 )i + xyJ and S is the square in the
xy-plane with corners at (1,1,0), (-1,1, 0), (1, -1,O),
(-1, -1,0), and oriented upward.
y
y
J
_ _______
REVIEW EXERCISES AND PROBLEMS FOR CHAPTER NIN_ET_EEN
1. Let S be the disk of radius 5 perpendicular to the y-axis,
centered at (0, 7,0) and oriented toward the origin. Is
(c)
z
(b)
z
J
CHAPTER SUMMARY (see also Ready Reference at the end of the book)
• Flux Integrals
Oriented surfaces, definition of flux through a smface,
definition of flux integral as a limit of a Riemann sum
x
53. Let S1 be the square OS x S 1,0 S yS 1, z = 0 ori­
ented downward and let S2 be the squareO S x S 1, 0 S
y S 1, z:: 1 oriented �pwar�. If F =_ cos(:i_yz)k, then
r div F d\! = o F · dA + S1 F · dA , where W
Jw
S
is the solid cubeO S-x S 1,0 Sy S 1,0 S zS 1.
J
tl "GJ
z
(a)
REVIEW EXERCISES AND PROBLEMS FOR CHAPTER NINETEEN
F = 2i + 3] through a disk of radius 5 in the plane
y = 2 oriented in the direction of increasing y.
F = i + 2] through a sphere of radius 3 at the origin.
F = yJ through the square of side 4 in the plane y = 5.
The square is centered on the y-axis, has sides parallel to
the axes, and is oriented in the positive y-direction.
F = xi through a square of side 3 in the plane x = -5.
The square is centered on the x-axis, has sides parallel to
the axes, and is oriented in the positive x-direction.
F = (y+3)] through a square of side 2 in the xz-plane,
oriented in the negative y-direction.
22.
F = zi + yJ + 2xk and S is the rectangle z = 4,
O S x S 2, 0 S y S 3, oriented in the positive z­
direction.
= (x + cos z)i +yJ + 2xk and S is the rectangle
y S 3, 0S z S 4, oriented in the positive
x-direction.
F
x
= 2, 0 S
= x2 i + (x + e Y )J - k, and S is the rectangle
y = -1, 0 S xS 2, 0 S zS 4, oriented in the negative
y-direction.
F
F = (5 +xy)i +zJ +yzk and Sis the 2 x 2 square
plate in the yz-plane centered at the origin, oriented in
the positive x-direction.
F = xi + Y.1 and S is the surface of a closed cylinder
of radius 2 and height 3 centered on the z-axis with its
base in the xy-plane.
F = -yi + xJ + zk and S is the surface of a closed
cylinder of radius 1 centered on the z-axis with base in
the plane z = -1 and top in the plane z = 1.
23. F = x i + y j +zk and S is the cone z = V�
x2 + y 2 ,
oriented upward with x2 +y2 S 1, x 2 0, y 2 0.
-
2-
2-
-
In Problems 24-27, give conditions on one or more of the con­
stants a, b, c to ensure that the flux integral fs F · dA has the
given sign.
24. Positive for F = ai +b] +ck arid Sis a disk perpen­
dicular to the y-axis, through (0, 5, 0) and oriented away
from the origin.
25. Negative for F = ai + b] + ck and S is the upper
half of the unit sphere centered at the origin and oriented
downward.
26. Positive for F = axi +ay] + azk and S is a sphere
centered at the origin.
27. Positive for F = ai + b] + ck and S is the triangle
cut off by x + y +z = 1 in the first quadrant, oriented
upward.
28. Calculate the flux of F = xyi + yzJ + zxk out of the
closed boxO S x S 1, 0 S y S 1,0 S zS 1
(a) Directly (b) Using the Divergence Theorem.
29. Compute the flux integral fs(x3 i +2y] + 3k) . dA,
where Sis the 2 x 2 x 2 rectangular surface centered at
the origin, oriented outward. Do this in two ways:
(a) Directly (b) Using the Divergence Theorem
F = xk through the square O S x S 3, 0 S yS 3 in
the xy-plane, with sides parallel to the axes, and oriented
upward.
In Exercises 30-35, use the Divergence Theorem to calculate
the flux of the vector field out of the surface.
= 6i + x2 ] - k, through a square of side 2 in the
plane z = 3, oriented upward.
31.
F = i - J - k through a cube of side 2 with sides
parallel to the axes.
F
30.
F = -xi +2y] +(3+2z)k out of the sphere of radius
2 centered at (1,2,3).
F = (2x + y)i +(3y + z)J + (4z +x)k and Sis the
sphere of radius 5 centered at the origin.
1042
32.
33.
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
F = x2i + y2J + e"'Y k
and S is the cube of side 3
in the first octant with one corner at the origin, and sides
parallel to the axes.
F = ( e 2 + x) i + (2y + sin x)J
+ (e Sis the sphere of radius 1 centered at (2, 1, 0).
"'2
z) k
and
34.
35.
F = x3 i + y3J + (x2 + y2 ) k
cylinder x2
not included) x2
outward.
(a)
(c)
J
• 8 1 is a horizontal square of side 1 with one corner
at (0, 0, 2), above the first quadrant of the xy-plane,
oriented upward.
• 8 2 is a horizontal square of side 1 with one corner at
(0,0, 3), above the third quadrant of the xy-plane,
oriented upward.
• 83 is. a square of side J2 in the xz-plane with one
corner at the origin, one edge along the positive x­
axis, one along the negative z-axis, oriented in the
negative y- direction.
• 84 is a square of side J2 with one corner at the ori­
gin, one edge along the positive y-axis, one corner
at (1, 0, 1), oriented upward.
37. Let f(x, y, z)
= xy + exyz . Find
(a) grad f
(b) Ic grad f · di", where C i.s the line from the point
(1,1,1) to (2, 3,4).
(c) Is grad f · d.A, where S is the quarter disk in the
xy-plane x2 + y2 ::::; 4,x 2 0,y 2 0, oriented up­
ward.
38. The flux of the constant vector field ai +bJ +ck through
the square of side 2 in the plane x = 5, oriented in the
positive x-direction, is 24. Which of the constants a, b, c
can be determined from the information given? Give the
vaJue(s).
39. (a) Let F = (x2 + 4)i + yJ. Which of the following
flux integrals is the largest? Explain.
Isi F · d.A, where 81 is the disk of radius I in
the plane x = 2 centered on the x-axis and oriented
away from the origin.
Is2 F · d.A, where 82 is the disk of radius 1 in
the plane y = 4 centered on the y-axis and oriented
away from the origin.
Is3 F · d.A, where 83 is the disk of radius 1
in the plane x = -3 centered on the x-axis and ori­
ented toward the origin.
(b) Calculate the integral you chose in part (a).
40. Figure 19.42 shows a cross-section of the earth's mag­
netic field. Assume that the earth's magnetic and geo­
graphic poles coincide. Is the magnetic flux through a
horizontal plate, oriented skyward, positive, negative, or
zero if the plate is
+ y2 = 4 with O ::::;
and S is the closed
z ::::; 5.
F = xi + yJ + zk and Sis the open cylinder (ends
Problems
36. Arrange the following flux integrals, Is; F · d.A, with
i = 1, 2, 3,4, in ascending order if F = -i - + k
and if the Si are the following surfaces:
REVIEW EXERCISES AND PROBLEMS FOR CHAPTER NINETEEN
+ y2
::::; 1, with O ::::; z ::::; 1, oriented
At the north pole?
On the equator?
(b)
.�
..
North pole
.�
�
i�
..
�
South pole
At the south pole?
•
.
Figure 19.42
41. (a)
Let div(F) = x 2 + y 2 - z2 . Estimate the flux out
of a small sphere of radius 0.1 centered at each of
the following points
(ii) (0,0,1)
(i) (2,1,1)
(b) What do the signs of your answers to part (a) tell you
about the vector field near each of these two points?
For Problems 42--44,
(a) Find the flux of the given vector field out of a cube in the
first octant with edge length c, one corner at the origin
and edges along the axes.
(b) Use your answer to part (a) to find div F at the origin
using the geometric definition.
(c) ComputedivF at the origin using partial derivatives.
42.
43.
F = xi
F = 2i + Y] + 3k
In Problems 45--49, calculate the flux of F through the cylin­
der x 2 + y2 = 2, -3 ::::; z ::::; 3 and its base, oriented outward.
The cylinder is open at the top.
+ x2J + 5k
45.
F
47.
F = zi +xJ +yk
F = y 2i + x 2J + 7zk
ft = x3i + y3J + f
46. F
48.
49.
= y2i + z2J
52. (a) Let F be a smooth vector field defined throughout
3-space. What must be true of F if the flux of F
through any closed surface is zero?
(b) What value of a ensures that the flux of F =
a(e"' +y2 -x)i +12y(l-e"' )J through any closed
surface is zero?
a = a1i + a2J + a3k be a constant vector and let
r = xi + yJ + zk .
53. Let
(a) Calculatediv(r x a).
(b) Calculate the flux of i" x a out of a cube of side 5
centered at the origin with edges parallel to the axes.
54. Find the flux of F out of the closed surface S given by
x2 + y2 + z2 = 100. You are given that F is contin­
uous, with div F = 3 inside the cube -2 ::::; x ::::; 2,
-2 ::::; y ::::; 2, -2 ::::; z ::::; 2 and div F = 5 outside the
cube.
55. The closed surface S consists of 81 , the cone x
Jy 2 + z2 for O ::::; x ::::; 2, and a disk 82. Let F
3xi
+ (x + y )k
2
2
+ 4yJ + 5zk.
(a) In what plane does the disk 82 lie? How is it ori­
ented?
(b) Find the flux of F through
(ii) 81
(i) 82
56. Find the flux integral, using i"
(a)
r r.
= xi + yJ + zk.
cl.A' where 81 is the dis)< x
Js 1
oriented away from the origin.
= 5, y2 + z 2
r.
::::; 7,
(b) r
d.A' where 82 is the closed cylinder y2
Js2
z2 = 7, 0 ::::; x ::::; 5.
(c)
44. ft= xi +yJ
= z2i
50. Find the constant vector field F parallel to i + k and
giving a flux of --7 through S, the cone z = r, with
0 ::::; z ::::; 4, oriented upward.
2
51. (a) Finddiv(r/llf 11 ) wherer = xi +yJ forr =!= 6.
(b) Where in 3-space isdiv(f' /I ff 11 2 ) undefined?
(c) Find the flux of i" /I Ir fl 2 through the closed cylin­
der of radius 2, length 3, centered at (5, 0, 0), with
its axis parallel to the z-axis.
(d) Find the flux of i" /I Ii" 11 2 through the curved sides of
the open cylinder in part (c).
r
+
i" . cl.A, where 83 is the curved side of the open
Js3
cylinder y2 + z 2 = 7, 0 ::::; x ::::; 5, oriented away
from the x-axis.
57. Let F (x, y, z) = Ji (x, y, z)i + h(x, y, z)J + k be
a vector field with the property that div F = 5 every­
where. Let S be the hemisphere z = -.J9 - x2 - y2 ,
with its boundary in the xy-plane and oriented downward. Find 1 F
58. Let F
·
= r/1'17113 .
d.A.
(a) Calculate div F . Where is div F undefined?
1043
(b) Find ]"5 F · d.A where S is a sphere of radius IO
centered at the origin.
(c) Find 8 F · d.A where B1 is a box of side 1 cen­
1
tered at the point (3, 0,0) with sides parallel to the
axes.
(d) Find 8 F · d.A where B2 is a box of side l cen­
2
tered at the origin with sides parallel to the axes.
(e) Using your results to parts (c) and (d), explain how,
with no further calculation, you can find the flux of
this vector field through any closed surface, provided
the origin does not lie on the surface.
I
I
59. The gravitational field,
origin is given by
F , of a planet of mass m at the
F = -Gm
fc:
lfr 113
.
Use the Divergence Theorem to show that the flux of the
gravitational field through the sphere of radius a is inde­
pendent of a. [Hint: Consider the region bounded by two
concentric spheres.]
60. A basic property of the electric field E is that its diver­
gence is zero at points where there is no charge. Suppose
that the only charge is along the z-axis, and that th� elec­
tric field E points radially out from the z-axis and. its
magnitude depends only on the distance r from the z­
axis. Use the Divergence Theorem to show that the mag­
nitude of the field is proportional to 1/r. [Hint: Consider
a solid region consisting of a cylinder of finite length
whose axis is the z-axis, and with a smaller concentric
cylinder removed.]
61. A fluid is flowing along a cylindrical pipe of radius a in
the i direction. The velocity of the fluid at a radial dis­
tance r from the center of the pipe is = u(l-r2 /a 2 )i.
v
(a) What is the significance of the constant u?
(b) What is the velocity of the fluid at the wall of the
pipe?
(c) Find the flux through a circular cross-section of the
pipe.
62. A closed surface S encloses a volume 1,v. The function
p(x, y, z) gives the electrical charge density at points in
space. The vector field J (x, y, z) gives the electric cur­
rent density at any point in space and is defined so that
the current through a small area d.A is given by
Current through small area � J · d.A.
(a) What do the following integrals represent, in terms
of electricity?
(i)
!v p dV
(ii)
1
J . d.A
(b) Using the fact that an electric current through a
surface is the rate at which electric charge passes
through the surface per unit time, explain why
( J · d.A �
ls
-�
at
(1
w
p dV)
.
1
1044
PROJECTS FOR CHAPTER NINETEEN
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
if its direction is awa.y from the origin at every point, its
magnitude depends only on the distance from the origin,
and its divergence is zero except at the origin. (Such a
vector field might be used to model the photon flow out
of a star or the neutrino flow out of a supernova.)
(a) Show that v = K(x2 +y2 +z 2 )-312(xi +yJ +zk)
63. (a) A river flows across the xy-plane in the positive
x-direction and aronnd a circular rock of radius 1
centered at the origin. The velocity of the river can
be modeled using the potential function ¢ = x +
(x/(x2 + y2 )). Compute the velocity vector field,
v = grad¢.
(b) Show that div v = 0.
(c) Show that the flow of v is tangent to the circle
x2 + y2 = 1. This means that no water crosses the
circle. The water on the outside must therefore all
flow around the circle.
(d) Use a computer to sketch the vector field v in the
region outside the unit circle.
64. A vector field is a point source at the origin in 3-space
is a point source at the origin if I( > 0.
(b) Determine the magnitude llv II of the source in part
(a) as a function of the distance from its center.
(c) Compute the flux of v through a sphere of radius r
centered at the origin.
(d) Compute the flux of v through a closed surface that
does not contain the origin.
CAS Challenge Problems
65. Let S be the part of the ellipsoid x 2 + y2 + 2z2 = 1 ly­
ing above the rectangle -1/2 S x S 1/2, -1/2 S y S
1/2, oriented upward. For each vector field (a)-(c), say
whether you expect Is F · dA to be positive, negative,
or zero. Then evaluate the integral exactly using a com­
puter algebra system and find numerical approximations
for..your answers. Describe and explain what you notice.
(a)
Cb)
Cc)
66. Let F = (z + 4)k, and let S be the surface with normal
pointing in the direction of the negative y-axis parame­
terized, for O S s. S 2, 0 S t S 21r, by
i"(s,t)=s2 costi +sJ +s2 sintk.
(a) Sketch S and, without evaluating the integral, say
whether Is F · dA is positive, negative, or zero.
Give a geometric explanation for your answer.
(b) Evaluate the flux integral. Does it agree with your
answer to part (a)?
F = xi
ff = (x + l)i
ff = yJ
PROJECTS FOR CHAPTER NINETEEN
-
=
xi+ yJ + zk
(x2 + y2 + z2)3/2
and let oriented surfaces S1 and S2 be as in Figure 19.43. The surface S2 is on the unit sphere
centered at the origin. Lines from the origin to the boundary of S1 intersect the unit sphere at
the boundary of S2 . Both surfaces are oriented away from the origin. Show that
(a) The divergence of F is zero.
Cb) Is, ft · dA = Is ft . dA
Is, F
· di"
the origin.
=
A vector field is spherically symmetric about the origin if, on every sphere centered at the
origin, it has constant magnitude and points either away from or toward the origin. A vector
field that is spherically :rymmetric about the origin can be written in terms of the spherical
coordinate p = 117"'11 as F = f(p)e P where f is a function of the distance p from the origin,
f (0) = 0, and Pis a unit vector pointing away from the origin.
(a) Show that
e
div ft
= ;2 :
P
(p2 J(p)),
Pi= o.
(b) Use part (a) to show that if F is a spherically symmetric vector field such that div F
away from the origin then, for some constant k,
=
O
(c) Use part (a) to confirm the Divergence Theorem for the flux of a spherically symmetric
vector field through a sphere centered at the origin.
(d) A form of Gauss's Law for an electric field E states that div E (r) = c5(r), where c5 is a
scalar-valued function giving the density of electric charge at every point. Use Gauss's Law
and part (a) to find the electric field when the charge density is
oW)
= { �o
for some nonnegative a. Assume that
where.
E
llrll :=::: a
llrll > a,
is spherically symmetric and continuous every­
Gauss's Law states that the flux of an electric field through a closed surface; S, is proportional
to the quantity of charge, q, enclosed within S. That is,
F (x, y)
(c)
2. Divergence of Spherically Symmetric Vector Fields
3. Gauss's Law Applied to a Charged Wire and a Charged Sheet
1. Solid Angle
Let
1045
2
D, the area of S2 . The value Dis called the solid angle subtended by S1 at
Figure 19.43
hi. dA =
kq.
In this project we use Gauss's Law to calculate the electric field of a uniformly charged wire in
part (a) and a flat sheet of charge in part (b).
(a) Consider the-electric field due to an infinitely long, straight, uniformly charged wire. (There
is no current running through the wire-all charges are fixed.) Assuming that the wire is
infinitely long means that we can assume that the electric field is perpendicular to any
cylinder that has the wire as an axis and that the magnitude of the field is constant on any
such cylinder. Denote by Er the magnitude of the electric field due to the wire on a cylinder
of radius r. (See Figure 19.44.)
Imagine a closed surface S made up of two cylinders, one of radius a and one of
larger radius b, both coaxial with the wire, and the two washers that cap the ends. (See
Figure 19.45.) The outward orientation of S means that a normal on the outer cylinder
points away from the wire and a normal on the inner cylinder points toward the wire.
(i) Explain why the flux of E, the electric field, through the washers is 0.
(ii) Explain why Gauss's Law implies that the flux through the inner cylinder is the same
as the flux through the outer cylinder. [Hint: The charge on the wire is not inside the
surface SJ.
(iii) Use part (ii) to show that Eb/ Ea = a/b.
(iv) Explain why part (iii) shows that the strength of the fi�ld due to an infinitely long
uniformly charged wire is proportional to l/r.
1046
Chapter Nineteen FLUX INTEGRALS AND DIVERGENCE
t
Wire
Figure 19.44
]ia
b
Wire
Figure 19.45
Figure 19.46
(b) Now consider an infinite flat sheet uniformly covered with charge. As in part (a), symmetry
shows that the electric field E is perpendicular to the sheet and has the same magnitude at
all points that are the same distance from the sheet. Use Gauss's Law to explain why, on any
one side of the sheet, the electric field is the same at all points in space off the sheet. [Hint:
Consider the flux through the box with sides parallel to the sheet shown in Figure 19.46.]
4. Flux Across a Cylinder: Obtaining Gauss's Law from Coulomb's Law
An electric charge q is placed at the origin in 3-space. The induced electric field
point with position vector f is given by Coulomb's Law, which says
-
E (f) = q
r
3
llf 11 '
E (r) at the
r =1 o.
In this project, Gauss's Law is obtained for a cylinder enclosing a point charge by direct calcu­
lation from Coulomb's Law.
(a) Let S be the open cylinder of height 2H and radius R given by x 2 + y 2 = R 2 ,
H, oriented outward.
(i) Show that the flux of E the electric field, through S is given by
,
1s
(ii) What are the limits of the flux
-
E · dA = 47rq
-H
:::; z :::;
H
)H2
+ R2
J� E · dA if
• H ---+ 0 or H ---+ oo when R is fixed?
• R ---+ 0 or R ---+ oo when His fixed?
(b) Let T be the outward-oriented, closed cylinder of height 2H and radius R whose curved
side is given by x2 +y2 = R 2 , -H :::; z :::; H, whose top is given by z = H, x 2 +y 2 :::; R2 ,
and bottom by z = -H, x2 + y 2 :::; R 2 . Use part (a) to show that the flux of the electric
field, E through T is given by
,
lE .
dA = 47rq.
Notice that this is Gauss's Law. In particular, the flux is independent of both the height, H,
and radius, R, of the cylinder.