Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

280 0 CHAPTER 15 MULTIPLE INTEGRALS
15.9 Triple Integrals in Spherical Coordinates
1. (a)
X
(6.¥. f)
If
I I
1 I
I I
'1r I I
7i /6 :
I
I
I
I
I
y
From Equat!ons 1, x = psin¢cos9 = 6 sin 2J cos i = 6 · ~ · ~ = !.
y = p sin rP sin 9 = 6 sin ~ sin ~ == 6 · ~ · :1/- = ~, and
z = pcos¢' = 6cos 2J = 6 · :1/- = 3VJ, so the point is(~ . ~ ,3../3) in
rectangular coordinates.
(b)
x = 3 sin 3 _; cos ~ = 3 · ~ · 0 = 0,
y = 3 sin 3 ;
sin ~ = 3 · ~ · 1 = ¥, and
X
z= 3cos 3 ; = 3 (- ~) = ·~¥. so thepointis (o.¥,-¥) in
·rectangular coordinates.
. . z 0 71'
3. (a) From Equations 1 and 2, p = .jx2 + y2 + z2 = ../0 2 + (-2)2 + 0 2 = 2, cos¢ = p = '2 = 0 => ¢' = 2'' and
cos 9 = p s: rP = 2
sin~ 71' / 2
) = 0 => 9 = 3 ; [since y < 0). Thus spherical coordinates are ( 2, 3 ; , -i).
z - J2 371'
(b) p =. v'1 + 1 + 2 = 2, cos¢ = P = - 2
- ::::} =
4 , and
X - 1
cos 9 = -- = ::-:--:-::---:-:7
• psin¢ 2sin(311'/4)
371' 37!')
are (
2, 4 , 4
.
- 1 1 371'
-,..-::::--:- - -- => 9 = - 4
[since y > 0). Thus spherical coordinates
2 (J2/2) - J2
5. Since ¢ = ~, the surface is the top half of the right circular cone with vertex at the origin and axis the positive z-axis.
x 2 + (y - ~) 2 + z 2 = ~ . Therefore, the surface is a sphere of radius ~ centered at ( 0, ~, 0).
9. (a) x = psin¢>cos8, y = psin ¢>sinO, and z = pcos ¢,so the equation z 2 = x 2 + y 2 becomes
(p cos¢ ) 2 = (p sin cos 8) 2 + (p sin ¢> sin 8) 2 or p 2 cos 2 ¢J = p 2 sin 2 ¢'. If p ;f 0, this becomes co~ 2 = sin 2 rp. (p = 0
corresponds to the origin which is included in the surface.) There are many equivalent equations in spherical coordinates,
such as tan 2 ¢ = 1, 2 cos 2 ¢' = 1, cos 2¢ = 0, or even rP = ~ . ¢' = :~ 4 ,..
(b) x 2 + z 2 = 9