Calculus Stewart ET 5e 0534393217;16. Vector Calculus; 16.9 The Divergence Theorem
1. The vectors that end near P are longer than the vectors that start near P , so the net flow is inward
1 1
near P and divF(P ) is negative. The vectors that end near P are shorter than the vectors that start
1 2 1 2
near P , so the net flow is outward near P and divF(P ) is positive.
2 2
2. (a) The vectors that end near P are shorter than the vectors that start near P , so the net flow is
1 1
outward and P is a source. The vectors that end near P are longer than the vectors that start near P ,
1 2 2
so the net flow is inward and P is a sink.
2
(b) F(x, y)= x, y
1
2
divF=
2
F=1+2y . The y value at P is positive, so divF=1+2y is positive,
1 2
thus P is a source. At P , y< 1 , so divF=1+2y is negative, and P is a sink.
1 1 1
3. divF=3+x+2x=3+3x , so
E
div FdV =
0 0 0
(3x+3)dxdydz=
9 (notice the triple integral is three 2
times the volume of the cube plus three times x ). To compute
S
F dS on S : n=i, F=3 i+y j+2z k, and
1 S
1
F dS=
S
1
3dS=3;
S : F=3x i+x j+2xz k, n= j and
2 S
2
F dS=
S
2
xdS=
1 ; 2
S : F=3x i+xy j+2x k, n=k and
3 S
3
F dS=
S
3
2xdS=1;
S : F=0,
4 S
4
F dS=0; S : F=3x i+2x k, n= j and
5 S
5
F dS=
S
5
0dS=0; 9 . 2
S : F=3x i+xy j, n= k and
6 S
6
F dS=
S
6
0dS=0. Thus
S
F dS=
4. divF=2x+x+1=3x+1 so
1
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.9 The Divergence Theorem
2
2 4 r 0 2
2
div FdV =
E E 22
(3x+1)dV =
0 0
(3rcos +1)rdzdr d
=
0 0 2 0
r(3rcos +1)(4 r )d dr
2 =2 =0 2
= r(4 r ) 3rsin +
2
dr 1 4 r 4
2 0
=2
0
(4r r )dr=2
3
2r
=2 (8 4)=8
On S : The surface is z=4 x y ,x +y
1
2
2 2
2
4 , with upward orientation, and F=x i+xy j+(4 x y ) k.
2
2
2
Then F dS =
S
1
[ (x )( 2x) (xy)( 2y)+(4 x y )]dA
D 2 2 2 2 2 2 2 2
2
2
2
=
D 2
2x(x +y )+4 (x +y ) dA=
0 0
(2rcos
2 0
r +4 r )rdr d
=
0
2 5 2 1 4 r cos +2r r 5 4 64 sin +4 5
2 0
r=2 r=0
d =
64 cos +4 d 5
=
=8
2
On S : The surface is z=0 with downward orientation, so F=x i+xy j, n= k and
2
F ndS=
S
2
0dS=0.
S
2
Thus
2
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.9 The Divergence Theorem
F dS=
S S
1
F dS+
S
2
F dS=8 .
5. div F=x+y+z , so
2 1 1 2 1
div FdV =
E 0 0 0 2
(rcos +rsin +z)rdzdr d =
0 0
r cos +r sin +
2
2
1 r dr d 2
=
0 1
1 1 1 cos + sin + 3 3 4
2
d =
1 (2 )= 4 2
3 1 2 1
Let S be the top of the cylinder, S the bottom, and S the vertical edge. On S , z=1 , n=k , and F=xyi+y j+x k, so
S
1
F dS=
S
1
F ndS=
S
1
xdS=
0 0
(rcos )rdr d = sin
2 0
1 3 r 3
1 0
=0 . On S ,
2
z=0, n= k, and F=xyi so
S
2
F dS=
S z
2
0dS=0. S is given by r( , z)=cos i+sin
3
j+z k ,
0
2 , 0 F dS =
z
1. Then r
2 1
r =cos i+sin (cos
2
j and
2
F (r
D 2 2
r )dA=
z 0 0
sin +zsin
)dzd 1 sin 2 2
2 0
S
3
=
0
cos
1 2 sin + sin 2 2 = 2 .
d =
1 1 3 cos + 3 4
=
2
Thus
S
F dS=0+0+
6. div F=1+1+1=3, so
E
div FdV =
E
3dV =3 (volume of ball) =3
4 3
=4 . To find
S
F dS
we use spherical coordinates. S is the unit sphere, represented by r( , )=sin cos i+sin sin j+cos k , 0 ,0 2 . Then r r =sin
2
cos i+sin
2
sin
2
j+sin
cos
k (see Example 17.6.10 [ET 16.6.10]) and k. Thus
F(r( , ))=sin F dS =
S D
cos i+sin r )dA=
sin
j+cos
3
F (r
(sin
0 0
cos
2
+sin
3
sin
2
+sin
cos
2
)d d
3
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.9 The Divergence Theorem
2
=
0
d
0 x
sin
d =(2 )(2)=4
7. div F= Theorem,
x
(e sin y)+
y
(e cos y)+
1 1 2
x
z
(yz )=e sin y e sin y+2yz=2yz , so by the Divergence
1 0 1 0 1 0
2
x
x
F dS =
S E
div FdV =
1 0
2yzdzdydx=2 dx ydy zdz 1 2 z 2
2 0
= 2 x
1 y 2
0 0 0 2 1 0 3
=2
4 3 3 3 3
8. div F= F dS =
S
x
(x z )+
2 3
y
(2xyz )+
1 2 3
z
3
(xz )=2xz +2xz +4xz =8xz , so by the Divergence Theorem,
1 1 2 2 3 3 3
div FdV =
E
8xz dzdydx=8 xdx dy z dz
1 2 3
= 8
2
1 2 x 2
2
1 1
y
2 2
1 4 z 4
3 3
=0
9. div F=3y +0+3z , so using cylindrical coordinates with y=rcos F dS =
S E 2
, z=rsin
, x=x we have
(3y +3z )dV =
1 0
2
2
2
1 2
(3r cos
0 0 1
2
2
+3r sin 9 2
2
2
)rdxdr d
= 3 d
0
r dr dx=3(2 )
1
3
2
1 4
(3)=
10. div F=3x y 2x y x y=0, so
S
2
2
2
F dS=
E
0dV =0.
11. div F=ysin z+0 ysin z=0, so by the Divergence Theorem,
S
F dS=
E
0dV =0.
12. divF=2xy+2xy+2xy=6xy, so F dS =
S E
6xydV
4
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.9 The Divergence Theorem
1 2 2y 2 x 2y
1 2 2y
=
0 0 1 2 2y 0
6xydzdxdy=
0 2 2 0
6xy(2 x 2y)dxdy
1 0
=
0 1 0 0
(12xy 2 6x y 12xy )dxdy=
3
6x y 2x y 6x y
1 0
2
3
2 2 x=2 2y x=0
dy
= y(2 2y) dy=
2 2 2
8 5 4 3 2 y +6y 8y +4y 5
=
2 5
13. div F=y +0+x =x +y so F dS =
S 2 E 2 0 3 2
(x +y )dV =
0 0 r 3 5
2
2
2
2 4
2
r rdzdr d =
0 0 4
2
2
2
r (4 r )dr d 32 3
3
2
= d
0
(4r r )dr=2
r
1 6 r 6
2 0
=
14. div F=4x +4xy so F dS =
S 2 E 1
4x(x +y )dV =
0 0 0
2
2
2
1 rcos +2
(4r cos )rdzdr d
2 0
3
=
0 0
(4r cos
2 2
5
2
+8r cos )dr d =
3
4
2 8 2 cos + cos 3 5
d =
2 3
15. div F=12x z+12y z+12z so F dS =
S E 2
12z(x +y +z )dV = sin
0
2
2
2
2
R
12( cos
0 0 0 R 5 0
)( ) sin 1 2 sin 2
2
2
d d d 1 6
6 R 0
=12 d
0
cos
d
d =12(2 )
0
=0
16. F dS =
S E /2
3(x +y +1)dV =
0 0 1
2
2
2
/2 2
3( sin d =2
2
2
+1) 93 5
2
sin
d d d 1 3 + cos 3
/2
=2
0
93 3 sin +7sin 5
cos
7cos
0
=
194 5
5
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.9 The Divergence Theorem
17.
S
F dS=
E
3 x dV =
1 1 0
2
1 1 2 x
4
y
4
3 x dzdydx=
2
341 60
2+
81 sin 20
1
3 3
18.
By the Divergence Theorem, the flux of F across the surface of the cube is
/2 /2 /2
F dS=
S 0 0 0
cos xcos y+3sin ycos ycos z+5sin zcos zcos x dzdydx=
2 2 2
2
2
4
4
6
19 64
2
.
19. For S we have n= k , so F n=F ( k)= x z y = y (since z=0 on S ). So if D is the unit disk,
1 1
we get
S
1
F dS=
S
1
F ndS=
D
( y )dA=
0 0
2
2
1
r (sin
2
2
2
)rdr d =
1 4
. Now since S is closed, we can
2
use the Divergence Theorem. Since divF= use spherical coordinates to get
S
2
x
E
(z x)+
y
2 0
1 3 2 2 2 2 2 y +tan z + (x z+y )=z +y +x , we 3 z
/2 1 0 0 2 2
F dS= 1 4
div FdV = 13 . 20
sin
d d d =
2 . Finally 5
F dS=
S S
2
F dS
S
1
F dS=
2 5
=
20. As in the hint to Exercise 19, we create a closed surface S =S S , where S is the part of the
2 1
paraboloid x +y +z=2 that lies above the plane z=1 , and S is the disk x +y =1 on the plane z=1
1
2
2
2
2
oriented downward, and we then apply the Divergence Theorem. Since the disk S is oriented
1
downward, its unit normal vector is n= k and F ( k)= z= 1 on S . So
1
F dS=
S
1
F ndS=
S
1
( 1)dS= A(S )=
S 1
1
. Let E be the region bounded by S . Then
2
6
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.9 The Divergence Theorem
12 2 r S
2
12 0 0
F dS =
2
div FdV =
E E
1dV =
0 0 1
rdzd dr=
(r r )d dr
3
=(2 )
1 = 4 2 F dS=
S S
2
Thus the flux of F across S is
F dS
S
1
F dS=
2
(
)=
3 . 2
2 2
21. Since for
x | x|
3
=
x i+y j+z k (x +y +z ) y
2 2 2 3/2
and
x x
2
y x | x|
(x +y +z )
2 3
2
2
2 3/2 2 2
and
(x +y +z ) z (x +y +z )
2 2 3/2 2
2
2
2 3/2
=
(x +y +z ) 3x (x +y +z )
2 2 2 5/2
2
2
with similar expressions
z
2 2
, we have
div
=
3(x +y +z ) 3(x +y +z ) (x +y +z )
2 2 2 5/2
=0 , except at (0, 0, 0) where it is undefined.
22. We first need to find F so that
S
F ndS=
S
(2x+2y+z )dS, so F n=2x+2y+z .
2
2
But for S, n=
2
x i+y j+z k x +y +z
2 2 2
=x i+y j+z k. Thus F=2 i+2 j+z k and divF=1. If
2
B={(x, y, z)| x +y +z
2
2
1}, then
S
(2x+2y+z )dS=
B
dV =V (B)=
4 3 4 (1) = . 3 3
23.
S
a ndS=
E
div adV =0 since div a=0.
24.
1 3
F dS=
S
1 3
div FdV =
E
1 3
3dV =V (E)
E
25.
S
curl F dS=
E
div(curl F)dV =0 by Theorem 17.5.11 [ET 16.5.11].
26.
7
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.9 The Divergence Theorem
D f dS=
S n S
(
f n)dS=
E
div(
f )dV =
E
2
f dV
27.
S
(f
g) ndS=
E
div( f
g)dV =
E
(f
2
g+ g
f )dV by Exercise 17.5.25 [ET 16.5.25].
28.
S
(f g
g g f= f
f ) ndS=
E
(f (f
S
2
g+ g g g
f ) (g f ) ndS=
2
f+ g (f
E 2
f ) dV [by Exercise 27]. g g
2
But
g, so that
f )dV.
29. If c=c i+c j+c k is an arbitrary constant vector, we define F= f c= fc i+ fc j+ fc k . Then
1 2 3 1 2 3
divF=div f c= F ndS=
S
f f f c+ c+ c= x 1 y 2 z 3
f c and the Divergence Theorem says
S
F dS=
E
div FdV
f cdV. In particular, if c=i then
E S
f i ndS=
E
f idV fn dS=
S 2 E
fn dS=
S 1 E
f dV (where n=n i+n j+n k ). Similarly, if c= j we have 1 2 3 x fn dS=
S 3 E
f dV, y
and c=k gives f ndS =
S S
f dV. Then z fn dS
S 2
fn dS
1
i+ i+
j+
S
fn dS
3
k f dV z k
=
E
f dV x
E
f dV y
j+
E
=
E
f f f i+ j+ k dV = x y z
f dV
E
as desired.
30. By Exercise 29,
S
pndS=
E
pdV, so
8
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.9 The Divergence Theorem
F=
S
pndS=
E
pdV =
E
( gz)dV =
E
( g k)dV
=
g
E
dV
k=
gV (E) k density g= gV (E), thus F= W k as desired.
But the weight of the displaced liquid is volume
9