Thomas Calculus 13th [Solutions]

1230 Chapter 16 Integrals and Vector Fields
xy-plane n k (outward normal) F n 1 Flux across the base
R xy
2
1 dx dy a . Therefore, the total flux across the closed surface is
2 3 2 3
a 2 a a 2 a .
S
F n d
CHAPTER 16 ADDITIONAL AND ADVANCED EXERCISES
1. dx ( 2sin t 2 sin 2 t) dt and dy (2cos t 2 cos 2 t) dt; Area 1 x dy y dx
2 C
1
2
(2cos t cos 2 t)(2cos t 2cos 2 t) (2sin t sin 2 t)( 2sin t 2sin 2 t)
dt
2 0
1
2
1
2
6 (6cos t cos 2t 6sin t sin 2 t) dt (6 6cos t) dt 6
2 0 2 0
2. dx ( 2sin t 2sin 2 t) dt and dy (2cos t 2cos 2 t) dt;Area
1 x dy y dx
2 C
1
2
(2cos t cos 2 t)(2cos t 2cos 2 t) (2sin t sin 2 t)( 2sin t 2sin 2 t)
dt
2 0
2 2
2
1 2 2(cos t cos 2t sin t sin 2 t) dt 1 (2 2cos 3 t) dt 1 2t 2 sin 3t
2
2 0 2 0 2 3 0
3. dx cos 2 t dt and dy cos t dt ;Area 1 x dy y dx 1 1
2 C
2 0 2
sin 2 t cos t sin t cos 2 t dt
1 2 2 1 2 1 1 3
sin t cos t (sin t) 2cos t 1 dt sin t cos t sin t dt cos t cos t 1 1 2
2 0 2 0 2 3 0 3 3
4. dx ( 2a sin t 2a cos 2 t) dt and dy ( b cos t) dt; Area= 1 x dy y dx
2 C
1
2 2 2
2ab cos t ab cos t sin 2t 2ab sin t 2ab sin t cos 2t dt
2 0
1
2 2 2 1
2
2
2ab 2ab cos t sin t 2 ab(sin t) 2cos t 1 dt 2ab 2ab cos t sin t 2ab sin t dt
2 0 2 0
3
2
1 2abt 2 ab cos t 2ab cos t 2 ab
2 3 0
5. (a) F( x, y, z,) zi xj yk is 0 only at the point (0, 0, 0), and curl F( x, y, z) i j k is never 0.
(b) F( x, y, z) zi yk is 0 only on the line x t, y 0, z 0 and curl F( x, y, z) i j is never 0.
(c) F( x, y, z) zi is 0 only when z 0 (the xy-plane) and curl F( x, y, z) j is never 0.
6.
2 2 xi yj zk xi yj zk
2 2 cy
F yz i xz j 2xyzk and n , so F is parallel to n when yz cx , xz ,
2 2 2
x y z
R R R
2
yz 2
2 2 2 2
and 2xyz cz xz 2xy y x y x and z c 2x z 2 x. Also,
R x y R
2 2 2 2 2 2 2 2 2 2 2R
x y z R x x 2x R 4 x R x R . Thus the points are: R , R , ,
2 2 2 2
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