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

278 D CHAPTER 15 MULTIPLE INTEGRALS
17. In cylindrical coordinates, E is given by {(r, B, z} I 0 ~ B ~ 21r, 0 ~ r ~ 4, - 5 ~ z ~ 4}. So
19. The paraboloid z = 4 - x 2 - y 2 = 4 - r 2 intersects the xy-plane in the circle x 2 + y 2 = 4 or r 2 = 4 =? r = 2, so in
cylindrical co~rd inate s, E is given by { (1·, B, z} I 0 ~ (} ~ 1r / 2, 0 ~ r ~ 2, 0 ~ z ~ 4- r 2 }. Thus ·
JJJE (x + y + z) dV = Io" 12 I; I 0
4
-r
2
(rcos{J + r sin(}+ z) r dzdr d(} = I 0 " 12 I; [r 2 (cos 0 + sin O) z + ~7·z 2 ]~=~-r 2 dr d{J
= Io"/ 2 J; [(4r 2 - r 4 )(cos 0 + sin 9) + ~r(4- r· 2?] dr d{J
= f '1r / 2 [(1r 3 - l r 5 ) (cos(} +sin 0) - .1.. (4 - r 2 ) 3 ] r = Z d9
.Jo 3 5 12 r=O
= I 0 " 12 [~~(cosO + sin 0) + Jf] dO= [ ~(sin 0 - cos B)+ JfB] ; 12
= lfi(1 - 0) + ¥. ~ - ~(0 - 1)- 0 = ~7r + \ 2 5 8
21 . In cylindrical coordinates, E is bounded by the cylinder,. = 1, the plane z = 0, and the cone z = 2r. So
E = {(r, 0, z) I 0 ~ 8 ~ 27r, 0 ~ r ~ 1, 0 ~ z ~ 2r} and
I I IE x 2 dV = J;'T( f 0
1
J;r r 2 cos 2 9 r dz dr dO = I 0
21f J 0
1
[r 3 cos 2 0 z ] ::~·· dr d9 = I;" Io 1 2r 4 cos 2 8 dr dO
= I 02
'T( [%r 5 cos 2 B]~=~ dO= % I 0
2 " cos
2
0 d(} = ~I;"~ (1 +cos 28) dO,;, t [B + t sin 28] ~1f = 2 ;
23. In cylindrical coordinates, E is bounded below by the cone z = r and above by the sphere r 2 + z 2 = 2 or z = v'2 - r 2 . The
con~ and the sphere·intersect when 2r 2 = 2 =? r = 1, so E = { (r, 8, z) I 0 ~ 0 ~ 21r, 0 ~ r ~ 1, r ~ z ~ v'2 - r 2 }
and the volume is
IIIE dV = J;" I0 1 .fr~ rdzdrdO =I;" I; [rzJ~::P drdO = I 0
2
" J 0
1
(r~- r 2 ) drd(}
= J 0
21f dO Io 1 (rv'2 - r 2 - r 2 ) dr = 21r [-~(2- r 2 ) 3 1 2 - ~r 3 J:
= 27r (- ~) (1 + 1-2 3 1 2 ) = - j7r (2- 2V2) = ~7r (V2 - 1)
25. (a) The paraboloids intersect when x 2 + y 2 = 36 - 3x 2 - 3y 2 =? x 2 + y 2 = 9, so the region of integration
is D = {(x,y) I x 2 + y 2 ::; 9}. Then, in cylindrical coordinates,
. E = {( r, B, z) I r 2 ~ z ~ 36 - 3r 2 , 0 ~ r ~ 3, 0 ~ (} ~ 27r} and
V = g" I; fr 3 26
-
3
r 2 r dz dr d(} = J;" I; (36r - 4r 3 ) dr dB = J;" [18r 2 - r 4 ] ::~dB= J 0
2
" 81 dO= 16271".
(b) For constant density K , m = J(V = 1627rJ( from part (a). Since the region is homogeneous and symmetric,
Nfyz = Mxz = 0 and
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