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

274 0 CHAPTER 15 MULTIPLE INTEGRALS
Thus we also have
E = {(x, y, z) I 0 ::::; x::::; 1, 0 ::::; y ::::; X, 0 ::::; z::::; y} = {(x, y, z) I 0 ::::; y::::; 1, 0::::; z.::::; y, y ::::; x ::::; 1}
Then
= {(x,y,z) I 0 ::::; z::::; 1, z::::; y::::; 1, y::::; x ::::; 1} = {(x,y,z) I 0 ::::; x $ 1,0::::; z ::::; x, z::::; y ::::; x}
= {(x,y, z) I 0::::; z::::; 1, z::::; x::::; 1, z::::; y::::; x}.
J; fv 1 f~ f(x,y, z) dzdx_dy = f 0
1
J; f~ f(x,y,z) dz dydz = f 0
1
g 1: f(x,y, z ) dxdz dy
{=]; J; I: f(x,y,z) dxdyd~ = J; J; f:'" f(x,y, z) dydzdx
"~--
= f 0
1
f~ 1 [;"' f(x, y, z) dy dx dz
37. The region Cis the solid bounded by a circular cylinder of radius 2 with axis the z -axis for -2 ::::; z $ 2. We can write
fff 0
(4 + 5x 2 yz 2 ) dV = fffc 4dV + fffc 5x 2 yz 2 dV, but f(x,y,z) = 5x 2 yz 2 is an odd function with
. respect toy. Since Cis symmetrical about the xz-plane, we have fffc 5x 2 yz 2 dV = 0. Thus
39. m = JJJE p(x, y,z) dV = f 0
1
fofi fol+x+v 2dz dydx = J: J 0
fi 2(1 + x + y) dydx.
= rl (2y + 2xy + y2] y=fi dx = rl (2 v'x + 2x3/ 2 + x) dx = [ix3/ 2 + ix5/2 + lx2] 1 = 79
Jo y=D Jo 3 s 2
'
My: = fffE xp(x, y, z ) dV = f 0
1
fofi J~ +o:+v 2x dz dy dx = J; fofi 2x(1 + x + y) dydx
Mx~
0
so
, = r1 [2xy + 2 x2y + xy2]v=fi dx = r1(2xs12 + 2xsf2 + x2) dx = [ixo/2 + ix7/2 + J.x3] 1 = 179
J o v=D J o . s 1 3
= JJJE yp(x,y,z)dV = f 0
1
fofi J;+x+v 2ydzdydx = J 0
1
f 0
fi2y(1 +x+y)dydx
= r1 [y2 + xy2 + 1y3] v=fi dx = r1 (x + x2 + 1x3/2) dx = [lx2 + lx3 + ~x:> /2) 1 = 11
Jo 3 v=O Jo 3 2 3 15
. fff ( ) dV r l rfi r1+:r+v 2 d d .d J·l J·./i. [ 2] ~=l+ :r+v d. d r1 r./i. (1 )2 d d
lvlxv = E zp x, y, z = Jo Jo Jo z z y x = 0 0 z ==D y x = Jo Jo + x + y y x
= J 0
1
J/"(1 + 2x + 2y + 2xy + x 2 + y 2 ) dydx = J; [y + 2xy + y 2 + xy 2 + x 2 y.+ ty 3 J::f dx
_ r1 ( 'x + 1x3/2 + x + x2 + x5/2) dx _ ['lx3/2 + llx5/ 2 + l x2 + lx3 + 'lx7/2) 1 _
- Jo
571
v"' 3 - 3 1s 2 · 3 7
0 - 2 10
. 79 . (- __ ) (My= M:r= Mxv ) (358 33 571)
Thus the mass IS 30 and the Center Of mass IS X, Y, Z = --:;;:;:--, --:;;:;:--, m = 553 ' 79 > 553 ·
0
0
10
105
Mv: = J; f 0
a f 0
a [x 3 + x(y 2 + z 2 )] dxdydz = J; J; [ta 4 + ~a 2 (y 2 + z 2 ) ] dydz
_ ru (.1 s 1 s 1 a 2) d 1 6 1 6 1 6 M M b fE d ( )
- Jo 4a + 0 a + 2a Z Z = 4a + 3a = na = :z:z = xy y symmetry 0 an p X, y, Z
H ence (---) x,y,z = (1 7 ' 7)
12
a, 12 a, 12
a.
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