Thomas Calculus 13th [Solutions]

1216 Chapter 16 Integrals and Vector Fields
S
/2 /2 a 2
/2 /2
sin
f n d dV
d d d a sin d d
2 2 2
0 0 0
2
x y z
0 0
D
/2 /2 /2
a cos d a d a
0 0 0
2
21. The integrals value never exceeds the surface area of S. Since | F | 1, we have | F n| | F| | n | (1)(1) 1 and
D
F d F n d
[Divergence Theorem]
S
S
F n d
[A property of integrals]
(1) d F n 1
S
Area of S.
22.
F
a b 1
a b
2x 4 y 6z 12 Flux ( 2x 4y 6z 12) dz dy dx ( 2x 4 y 9) dy dx
0 0 0 0 0
a
2 2 2 f
2
2xb 2b 9b dx a b 2ab 9 ab ab( a 2b 9) f ( a, b); 2ab 2b 9b and
0
a
f 2
f
f
a 4ab 9a so that 0 and 0 b( 2a 2b 9) 0 and a( a 4b 9) 0 b 0 or
b
a
b
2a 2b 9 0, and a 0 or a 4b 9 0. Now b 0 or a 0 Flux 0; 2a 2b 9 0 and
a 4b 9 0 3a 9 0 a 3 b 3 so that f 3, 3 27 is the maximum flux.
2 2 2
2 xy
23. By the Divergence Theorem, the net outward flux of the field F xyi (sin xz y ) j ( e x)
k over the
surface S will be equal to the integral of F y 2 y 3y
over the region D bounded by S. We will
2
integrate using the area in the zx-plane bounded by z 0 and z 1 x as the base. The y height at any point
( x, z ) will be 2 z . Thus the integral of div F over D is
1 2
1 1 2
2 1 x 2 1 2
x
2 z
z
3
1 x
3
3 y dy dz dx y dz dx (2 z)
2 dz dx
0 0
1
2 0
1 0 2
1 0
1 2
1 x 1
1
3 2 3 3 5 7
(2 z) dx 4 ( x 1) dx x x x x
0 1
1
1
1 1 7 1 3 1 184
2 2 2 3 10 14 35
2
2 2 2
3/2
24. The field F ( xi yj zk ) / x y z is discussed in Example 5 in Section 16.8, where we show that
the flux of F across any closed surface enclosing the origin is 4 . Note that the divergence of F is not
defined at the origin, so we need an argument like that shown in Example 5.
M g N g P g
x y z x x y y z z
25. (a) div( gF) gF
( gM ) ( gN ) ( gP)
g M g N g P
g g g
M N P g
M N P
g F g
x y z x y z
F
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