Thomas Calculus 13th [Solutions]

Chapter 16 Additional and Advanced Exercises 1231
R 2 2 2 2 2
, R R
, , R , R R
, , R , R R
, , R , R R
, , R , R R
, ,
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
R 2 2
, R R
, , R , R R
,
2 2 2 2 2 2
7. Set up the coordinate system so that
2 2 2
( a, b, c) (0, R, 0) ( x, y, z) x ( y R)
z
2 2 2 2 2 2 2 2 2
x y z 2Ry R 2R 2 Ry;let f ( x, y, z) x y z R and p i
f 2 2 2
2 x f
i 2 y j 2 z k f 2 x y z 2 R d 2R
f i
dz dy 2x
dz dy
2
2R
2Ry
Mass = ( x, y, z) d 2R 2Ry R dz dy R dz dy
x
2 2 2
R y z
S R R
yz
yz
2
2 2
R y
2 2 2
2 2 2 1
4 R R R y R Ry R
4 2 2 sin z
R 0
dz dy R 2 2 2 R
R Ry dy
2 2
R y z R y
0
R 2 2 3/2
R
3
2 R 2R 2Ry dy 2 R 1 (2R 2 Ry)
16 R
R
3R
R 3
i j k
8. r( r, ) ( r cos ) i ( r sin ) j k, 0 r 1, 0 2 rr
r cos sin 0
r sin r cos 1
2 2 2 2 2 2 2
(sin ) i (cos ) j rk rr
r 1 r ; 2 x y 2 r cos r sin 2r
3/2
1
2 1 2 2
2 2
2
Mass ( x, y, z) d 2r 1 r dr d 1 r d 2 2 2 1 d
0 0 0 3 0 3
S
0
4 2 2 1
3
9.
M x 2
4 xy and N 6 y M 2 4 and N
b a
6 Flux (2 4 6)
x
x y y
= 0 0
x y dx dy
b a
2 2 2
4 ay 6 a dy a b 2 ab 6 ab . We want to minimize
0
2 2
f ( a, b) a b 2ab 6 ab ab( a 2b 6). Thus,
2
fa ( a , b ) 2 ab 2 b 6 b 0 and
2
fb ( a, b) a 4ab 6a 0 b(2a 2b 6) 0 b 0 or b a 3. Now
2
b 0 a 6a 0 a 0
2
or a 6 (0,0) and (6, 0) are critical points. On the other hand, b a 3 a 4 a( a 3) 6a
0
2
3a 6a 0 a 0 or a 2 (0, 3) and (2, 1) are also critical points. The flux at (0, 0) 0, the flux at
(6, 0) 0, the flux at (0, 3) 0 and the flux at (2, 1) 4. Therefore, the flux is minimized at (2, 1) with
value 4.
10. A plane through the origin has equation ax by cz 0. Consider first the case when c 0. Assume the plane
2 2 2
is given by z ax by and let f ( x, y, z) x y z 4. Let C denote the circle of intersection of the plane
with the sphere. By Stokes Theorem, F dr F n d , where n is a unit normal to the plane. Let
C
S
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