Thomas Calculus 13th [Solutions]

(b) In a fashion similar to cylindrical coordinates, but working in the
y u cos v,
z u sin v where
Section 16.5 Surfaces and Area 1187
yz -plane instead of the xy -plane, let
2 2
u y z and v is the angle formed by ( x, y, z ), ( x ,0,0), and ( x, y,0)
with ( x ,0,0) as vertex. Since x y z 1 x 1 y z x 1 u cos v u sin v , then r is a function
of u and v r( u, v) (1 u cos v u sin v) i ( u cos v) j ( u sin v) k , 0 u 3 and 0 v 2 .
14. (a) In a fashion similar to cylindrical coordinates, but working in the xz -plane instead of the xy -plane, let
x u cos v,
z u sin v where
2 2
u x z and v is the angle formed by ( x, y, z ), ( y , 0, 0), and
( x, y , 0) with vertex ( y ,0,0). Since x y 2z 2 y x 2z 2, then
r( u, v) ( u cos v) i ( u cos v 2u sin v 2) j ( u sin v) k , 0 u 3 and 0 v 2 .
(b) In a fashion similar to cylindrical coordinates, but working in the
y u cos v,
z u sin v where
yz -plane instead of the xy -plane, let
2 2
u y z and v is the angle formed by ( x, y, z ), ( x , 0, 0), and
( x, y , 0) with vertex ( x ,0,0). Since x y 2z 2 x y 2z 2, then
r( u, v) ( u cos v 2u sin v 2) i ( u cos v) j ( u sin v) k , 0 u 2 and 0 v 2 .
15. Let x w cos v and z w sin v . Then
2 2 2 2
( x 2) z 4 x 4x z 0
2 2 2 2 2
w cos v 4wcos v w sin v 0 w 4w cos v 0 w 0 or w 4cos v 0 w 0 or
w 4 cos v . Now w 0 x 0 and y 0, which is a line not a cylinder. Therefore, let
2
w 4 cos v x (4 cos v)(cos v) 4cos v and z 4 cos v sin v . Finally, let y u . Then
2
r( u, v) 4 cos v i uj (4cos v sin v) k , v and 0 u 3.
2 2
16. Let y wcos
v and z wsin v . Then
2 2 2 2
y ( z 5) 25 y z 10z
0
2 2 2 2 2
w cos v w sin v 10wsin v 0 w 10wsin v 0 w( w 10sin v) 0 w 0 or w 10sin v.
Now w 0 y 0 and z 0, which is a line not a cylinder. Therefore, let w 10sin v y 10sin v cos v
and
2
z 10sin v . Finally, let x u . Then
0 v .
2
r( u, v) ui (10sin v cos v) j 10sin v k , 0 u 10 and
17. Let x r cos and y r sin . Then r( r, ) ( r cos ) i ( r sin ) j
2 r sin
k , 0 r 1 and 0 2
2
r (cos ) (sin ) sin
r i j k and ( r sin ) ( r cos ) r cos
2
2
rr
r
cos
i j k
r sin
sin
r cos
sin
2
r cos
2
r i j k
(sin )( r cos ) 2 2
2 2
r sin cos i r sin r cos j r cos r sin k r j rk
2 2 2 2 2
2 5r 2 1 5r 2 5r
2 5
| r | r
r r r A dr d d d
4 2 0 0 2 0 4 0 2
0
2 2 1
18. Let x r cos and y r sin z x r cos , 0 r 2 and 0 2 . Then
r( r, ) ( r cos ) i ( r sin ) j ( r cos ) k rr
(cos ) i (sin ) j (cos ) k and
r ( r sin ) i ( r cos ) j ( r sin ) k
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