Thomas Calculus 13th [Solutions]

Chapter 16 Practice Exercises 1223
which is Wyoming so then S is part of the surface
2 2 2
1/2
z R x y . Let R xy be the projection of S onto
2 2
the xy -plane . The surface area of Wyoming is 1 d 1 z z dA
x y
S Rxy
2 2
1/2
x
y
2 Rsin 49 2 2
1 dA R dA R R r r dr d (where
2 2 2 2 2 2 1/2
R x y R x y 2 2 2
R x y
1 Rsin 45
1 and
Rxy
Rxy
2 are the radian equivalent to 104 3 and 111 3 , respectively) 1/2
Rsin 49
1 2 2
R R r d
1
Rsin 45
2 2 2
1/2
2 2 2
1/2
2
R R R sin 45 R R R sin 49 d ( 2 1) R cos 45 cos 49
2
1
7 2 7 2
R cos 45 cos 49 (3959) cos 45 cos 49 97,751 sq. mi.
180 180
19. A possible parametrization is r( , ) (6sin cos ) i (6sin sin ) j (6cos ) k (spherical coordinates); now
6 and z 3 3 6cos cos 1 2 and z 3 3 3 3 6cos
2 3
3
cos 2 ; also 0 2
2 6 6 3
20. A possible parametrization is r( r, ) ( r cos ) i ( r sin ) j r
2
k (cylindrical coordinates); now
r
2 2
2
2
x y z r
2
and 2 z 0 2 r 0 4 r
2
2
0 0 r 2 since r 0; also
0 2
21. A possible parametrization is r( r, ) ( r cos ) i ( r sin ) j (1 r)
k (cylindrical coordinates);
2 2
now r x y z 1 r and 1 z 3 1 1 r 3 0 r 2; also 0 2
2
22. A possible parametrization is
y
r( x, y) xi yj 3 x k for 0 x 2 and 0 y 2
2
23. Let x u cos v and z u sin v , where
2
r( u, v) ( u cos v) i 2 u j ( u sin v)
k is a possible parametrization;
2 2
u x z and v is the angle in the xz -plane with the x-axis
0 u 1 since u 0; also, for just the upper half of the paraboloid, 0 v
2 2
0 y 2 2u 2 u 1
24. A possible parametrization is
10 sin cos i 10 sin sin j 10 cos k , 0 and
2
0
2
i j k
25. ru i j, rv i j k ru rv 1 1 0 i j 2 k | ru rv
| 6
1 1 1
Surface Area
R
uv
1 1
| ru
rv
| du dv 6 du dv 6
0 0
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