Thomas Calculus 13th [Solutions]

1192 Chapter 16 Integrals and Vector Fields
32. (a) The point ( x, y, z ) is on the surface for fixed x f ( u ) when
y g( u)sin
v and
2
z g( u) cos v x f ( u), y g( u) cos v , and z g( u)sin v r( u, v)
2
f ( u) i g( u)cos v j g( u)sin v k , 0 v 2 , a u b
(b) Let u y and
0 v 2 , 0 u
2 2
x u f ( u)
u and
2
g( u) u r( u, v) u i ( u cos v) j ( u sin v) k,
2
2
2
33. (a) Let w z 1 where w cos and z
2
sin x y
cos x cos cos
2
2 2
c
c
a b
a
x a cos cos , y b sin cos , and z csin
r( , ) ( a cos cos ) i ( bsin cos ) j ( csin ) k
(b) r ( asin cos ) i ( bcos cos ) j and r ( a cos sin ) i ( bsin sin ) j ( c cos ) k
r
r
i j k
a sin cos b cos cos 0
a cos sin b sin sin c cos
2 2
bc cos cos i acsin cos j absin cos k
2
y
and cos sin
b
2 2 2 2 4 2 2 2 4 2 2 2 2
| r r | b c cos cos a c sin cos a b sin cos , and the result follows.
2 2 2 2 2 2 2 2 2 2 4 2 2 2 4
1/2
A | r r | d d a b sin cos b c cos cos a c sin cos d d
0 0 0 0
34. (a) r( , u) (cosh u cos ) i (cosh u sin ) j (sinh u)
k
(b) r( , u) ( a cosh u cos ) i ( bcosh u sin ) j ( csinh u)
k
35. r( , u) (5cosh u cos ) i (5cosh u sin ) j (5sinh u) k r ( 5cosh u sin ) i (5cosh u cos ) j and
i j k
ru
(5sinh u cos ) i (5sinh u sin ) j (5cosh u) k r ru
5cosh u sin 5cosh u cos 0
5sinh u cos 5sinh u sin 5cosh u
2 2
25cosh u cos i 25cosh u sin j (25cosh u sinh u) k . At the point ( x0 , y 0, 0), where
we have 5sinh u 0 u 0 and x0 25cos , y 0 25sin the tangent plane is
2 2
5 x0i y0j ( x x0 ) i ( y y0 ) j zk
0 x0x x0 y0 y y0 0 x0 x y0
y 25
2
2
2
36. Let z w 1 where z cosh u and
2
c
c
w 2
sinh u w x y x cos
2 2
a b a
w and y
w
b
sin
x a sinh u cos , y bsinh u sin , and z c cosh u
r( , u) ( a sinh u cos ) i ( bsinh u sin ) j ( c cosh u) k,
0 2 , u
2
2 2
x0 y0 25
37.
2 2 2 2 2
p k, f 2xi 2 yj k | f | (2 x) (2 y) ( 1) 4x 4y
1 and | f p| 1;
2 2
z 2 x y 2; thus
| f | 2 2 2 2 2 2
S dA 4x 4y 1 dx dy 4r cos 4r sin 1r dr d
| f p|
R R R
3/2
2
2 2 2 2 2 2
4r 1 r dr d 1 4r 1 d 13 d 13
0 0 0 12 0 6 3
0
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