Thomas Calculus 13th [Solutions]

Section 16.5 Surfaces and Area 1189
i j k
and ry j rv ry 10 sin v 0 10 cos v 10 cos v i 10 sin v k | rv ry| 10
0 1 0
A
2 1 2 1 2
10 du dv 10u dv
0 1 0 1 0
2 10 dv 4 10
23.
2 2
z 2 x y and
2 2 2 2
z x y z 2 z z z 2 0 z 2 or z 1. Since
2 2
z x y
we get z 1 where the cone intersects the paraboloid. When x 0 and y 0, z 2 the vertex of the
paraboloid is (0, 0, 2). Therefore, z ranges from 1 to 2 on the cap
0 (when x 0 and y 0 at the vertex). Let x r cos , y r sin , and
2
r ranges from 1 (when
2
z 2 r . Then
r( r, ) ( r cos ) i ( r sin ) j 2 r k , 0 r 1, 0 2 r (cos ) i (sin ) j 2rk and
r ( r sin ) i ( r cos ) j r r cos sin 2r
r
i j k
r sin r cos 0
2 2 4 2 4 2 2 2
i j k rr
r
2r cos 2r sin r | | 4r cos 4r sin r r 4r
1
3/2
1
2 1 2 2
1 2 2 5 5 1
A r 4r 1 dr d 4r 1 d d 5 5 1
0 0 0 12 0 12 6
0
r
2 2
0,
x y 1 ) to
24. Let x r cos , y r sin and
z x
2 y
2 r
2 . Then
2
r( r, ) ( r cos ) i ( r sin ) j r k , 1 r 2,
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
3/2
2
4 2 4 2 2 2 2 2 2 2
1 2
4r cos 4r sin r r 4r 1 A r 4r 1 dr d 4r 1 d
0 1 0 12
1
2 17 17 5 5
d 17 17 5 5
0 12 6
25. Let x sin cos , y sin sin , and
2 2 2
x y z 2 and
2 2 2
z cos x y z 2 on the sphere. Next,
2 2 2 2 2
z x y z z 2 z 1 z 1 since
portion of the sphere cut by the cone, we get
. Then
r( , ) 2 sin cos i 2 sin sin j 2 cos k,
4
, 0 2
z 0 . For the lower
4
r 2 sin cos i 2 cos sin j 2 sin k and r 2 sin sin i 2 sin cos j
i j k
2 2
r r 2 cos cos 2 cos sin 2 sin 2sin cos i 2sin sin j 2sin cos k
2 sin sin 2 sin cos 0
Copyright
2014 Pearson Education, Inc.