Thomas Calculus 13th [Solutions]

Section 16.5 Surfaces and Area 1193
38.
2 2
p k, f 2xi 2 yj k | f | 4x 4y
1 and | f p| 1;
2 2
| f |
2 x y 6 S dA
| f p|
R
3/2
6
2 2 2 2 6 2 2
1 2
4x 4y 1dx dy 4r 1r dr d 4r 1r dr d 4r 1 d
0 2 0 12
R
R
2
2 49 d 49
0 6 3
2
39. p k, f i 2j 2 k | f | 3 and | f p | 2; x y and
| f | 3
1 2 y
3
1 2
S dA dx dy
| | 2
2 dx dy
f
1 y 2
1 3 3 y dy
p
4
R
R
2
2
x 2 y intersect at (1,1) and (1, 1)
40.
p k, 2 2
| f |
f 2xi 2 k | f | 4x 4 2 x 1 and | f p| 2 S dA
| f p|
R
3
2
2 x 1 3 x 2 3 2 2
3/2
1 1 3/2
dx dy x 1dy dx x x 1dx x 1 (4) 1 7
2 0 0 0
3 3 3 3
R
0
41.
2 2 2 2 2
p k, f 2xi 2j 2 k | f | (2 x) ( 2) ( 2) 4x 8 2 x 2 and | f p| 2
2
| f |
2
2 3 2 2 2 2
3/2
2 x 2
x
S dA dx dy x 2 dy dx 3x x 2 dx x 2 6 6 2 2
| f p| 2 0 0 0
R
R
0
42.
2 2 2
p k, f 2xi 2y 2 zk | f | 4x 4y 4z
8 2 2 and | f p| 2 z;
2 2 2 2
| f | 2 2
z x y x y 1; thus, S dA dA 2 1 dA
| f p| 2z z
R R R
1
2 1 r dr d 2
2 dA 2 2 1 2 d 2 2 2
2 2
0 0 2
2 x y
2 r 0
R
2 2 2
x y z 2 and
43.
2 2
| f | 2
p k, f ci k | f | c 1 and | f p| 1 S dA c 1 dx dy
| f p|
R
R
2 1 2 2
2 1
2
0 0 1 c
c r dr d d c
0 2
1
44.
p k, f 2xi 2 zj | f |
2
(2 x) 2
(2 z) 2 and | f p | 2z
for the upper surface, z 0
S
| f | 1/2 1/2 1/2
dA 2 dA 1 dx dy 2
1 dy dx 1 dx
| f p| 2z
2 1/2 0 2 1/2 2
1 x 1 x 1 x
R R R
sin
1
1/2
x
1/2 6 6 3
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