Thomas Calculus 13th [Solutions]

1208 Chapter 16 Integrals and Vector Fields
i j k
2 2
8. curl F F 2 j; f ( x, y, z) 4x y z f 8xi j 2zk
x y z
1 1
z tan y x 1
2 x
4 z
f
| f |
n and p j | f p | 1 d dA | f | dA; F n 1 2j
f 2
| f | | f p| | f | | f |
F n d 2 dA F n d 2 dA 2 (Area of R) 2( 1 2) 4 , where R is the
elliptic region in the
S
xz -plane enclosed by
R
2
4x
2
z 4.
9. Flux of F F n d F dr , so let C be parametrized by r ( a cos t) i ( a sin t) j, 0 t 2
C
S
dr
2 2 2 2 2
( a sin t) i ( a cos t) j F dr
ay sin t ax cost a sin t a cos t a
dt
dt
2 2 2
Flux of F F dr
a dt 2 a
C 0
10.
i j k
( y i ) x y z
k ; n
|
f 2xi 2 yj 2zk
1
f |
i y j z k ( i ) n z;
d
2 2 2
2 x y z
z
y 0 0
(Section 16.6, Example 6, with a 1 ) ( yi) n d ( z) 1 dA
2
dA , where R is the disk
S R R
2
x
2
y 1 in the xy-plane.
11. For the upper hemisphere with z 0, the boundary C is the unit circle of radius 1 centered at the origin in the
xy-plane. An outward normal on the upper hemisphere corresponds to counterclockwise circulation around the
boundary, so the boundary can be parametrized as r( ) (cos ) i (sin ) j 0 k , with 0 2 . Thus
dr ( sin d ) i (cos d ) j . For the field A ( y z ) i e j (cos xz) k,
the flux of F A across the
upper hemisphere is, by Stokes Theorem, equal to the circulation of A on the boundary. Since z 0 and
y sin on the boundary, the field A on the boundary is (sin ) i j k . The circulation of A on C is
2 2
A dr ((sin ) i j k) (( sin d ) i (cos d ) j) cos sin d
0
C C
2
cos 1 cos2 1 d
0
2
12. Since the outward normal on the bottom hemisphere corresponds to clockwise circulation on the boundary, the
flux of F through the bottom hemisphere will be and the total flux through the sphere will be 0.
xyz
i j k
13. F
x y z
5i 2j 3 k; rr
(cos ) i (sin ) j 2rk and r ( r sin ) i ( r cos ) j
2z 3x 5y
i j k
2 2
rr
r
rr
r cos sin 2r 2r cos i 2r sin j rk; n and d | r |
| r |
r r dr d
r r
r sin r cos 0
Copyright
2014 Pearson Education, Inc.