Thomas Calculus 13th [Solutions]

1186 Chapter 16 Integrals and Vector Fields
2 2
z x y and
2
2 2 2 2 2 2 2 2 2 2
x y z 9 x y x y 9 2 x y 9 2r
9
r
3 0 r 3 .
2 2
2
6. In cylindrical coordinates, r( r, ) ( r cos ) i ( r sin ) j 4 r k (see Exercise 5 above with
2 2 2
x y z 4, instead of
2 2
z x y and
2 2 2
x y z 9 ). For the first octant, let
0 . For the domain of r:
2
2
2 2 2 2 2 2 2 2 2 2
x y z 4 x y x y 4 2 x y 4 2r 4 r 2.
Thus, let 2 r 2 (to get the portion of the sphere between the cone and the xy -plane ).
7. In spherical coordinates,
z 3 cos for the sphere;
1 2
2 3
2 2 2 2
x sin cos , y sin sin , x y z 3 3
z
3 1
3 3
z
2 2 3 2 2
3 cos cos ; 3 cos
cos . Then r( , ) 3 sin cos i 3 sin sin j 3 cos k,
2
3 3
and 0 2 .
8. In spherical coordinates,
2 2 2 2
x sin cos , y sin sin , x y z 8 8 2 2
x 2 2 sin cos , y 2 2 sin sin , and z 2 2 cos . Thus let
r( , ) 2 2 sin cos i 2 2 sin sin j 2 2 cos k; z 2 2 2 2 cos
1
2
3
4
cos ; z 2 2 2 2 2 2 cos cos 1 0. Thus 0
3
and
0 2 .
4
9. Since
2
z 4 y , we can let r be a function of x and y
r( x, y) xi yj 4 y k . Then z 0
2
2
0 4 y y 2. Thus, let 2 y 2 and 0 x 2.
10. Since
y x 2 , we can let r be a function of x and
2
x 2 x 2. Thus, let 2 x 2 and 0 z 3.
11. When x 0 , let
2
z r( x, z) xi x j zk . Then y 2
2 2
y z 9 be the circular section in the yz -plane . Use polar coordinates in the yz-plane
y 3 cos and z 3 sin . Thus let x u and v r( u, v) ui (3 cos v) j (3 sin v)
k where
0 u 3, and 0 v 2 .
12. When y 0, let
2 2
x z 4 be the circular section in the xz -plane. Use polar coordinates in the xz-plane
x 2 cos and z 2sin . Thus let y u and v r( u, v) (2 cos v) i uj (3 sin v)
k where
2 u 2, and 0 v (since we want the portion above the xy -plane ).
13. (a) x y z 1 z 1 x y . In cylindrical coordinates, let x r cos and y r sin
z 1 r cos r sin r( r, ) ( r cos ) i ( r sin ) j (1 r cos r sin ) k,0 2
and 0 r 3.
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