Thomas Calculus 13th [Solutions]

Section 16.6 Surface Integrals 1197
4 4 2 4 4 2 4 2 2 2
| r r | a sin cos a sin sin a sin cos a sin ; z a cos
2 2 2 /2 2 2 2 2 4
G( x, y, z) a cos G( x, y, z) d a cos a sin d d a
0 0
3
S
i j k
5. Let the parametrization be r( x, y) xi yj (4 x y) k rx
i k and ry j k rx ry
1 0 1
0 1 1
1 1 1
y
i j k | rx
ry| 3 F( x, y, z) d (4 x y) 3 dy dx 3 4y xy dx
0 0 0
2
S
0
1
7 7
0 3 x dx
2 3 x x
2 2
3 3
0
2 1
2 1
6. Let the parametrization be r( r, ) ( r cos ) i ( r sin ) j rk , 0 r 1 (since 0 z 1 ) and 0 2
i j k
rr
(cos ) i (sin ) j k and r ( r sin ) i ( r cos ) j rr
r cos sin 1
r sin r cos 0
2 2 2
( r cos ) i ( r sin ) j rk | rr
r | ( r cos ) ( r sin ) r r 2; z r and x r cos
2 1
F( x, y z) r r cos F( x, y, z) d ( r r cos ) r 2 dr d
0 0
S
2 1 2 2 2
2 (1 cos ) r dr d
0 0
3
7. Let the parametrization be r( r, ) ( r cos ) i ( r sin ) j (1
2
r ) k , 0 r 1 (since 0 z 1 ) and
0 2 rr
(cos ) i (sin ) j 2rk and r ( r sin ) i ( r cos ) j
i j k
rr
r cos sin 2r 2
2r cos i
2
2r sin j rk | rr
r |
r sin r cos 0
2
2
2
2
2 2 2
2r cos 2r sin r r 1 4 r ; z 1 r and
2 2 2
x r cos H ( x, y z) r cos 1 4r
S
2 1 2 2 2 2 2 1 3 2 2
H ( x, y, z) d r cos 1 4r r 1 4r dr d r 1 4r cos dr d 11
0 0 0 0
12
8. Let the parametrization be r( , ) (2sin cos ) i (2sin sin ) j (2cos ) k (spherical coordinates with
2 2 2
2 on the sphere), 0 ; x y z 4 and
4
2 2 2 2 2
z x y z z 4 z 2 z 2 (since
z
2
0 ) 2cos 2 cos , 0
2 4
2 ; r (2cos cos ) i (2cos sin ) j (2sin ) k
i j k
and r ( 2sin sin ) i (2sin cos ) j r r 2cos cos 2cos sin 2sin
2sin sin 2sin cos 0
2 2
4sin cos i 4sin sin j 4sin cos k
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