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

43. I,. = foL foL foL k(y2 + z2)dz.dydx = k foL foL (Ly2 + ~L3) dydx = k foL ~L4 dx = ~kLs.
By symmetry, l x = 1 11 =I.= ~k£ 5 •
SECTION 15.7 TRIPLE INTEGRALS 0 275
45. 1::: = JJJE(x 2 + y 2 ) p(x, y, z) dV = JJ [J;• k(x 2 + y 2 ) dz] dA = JJ k(x 2 + y 2 )hdA
z2+y2:5a2
:z:2+y2:5a.2
(b) (x,y, z) wherex = "* f~ 1 !:2 f0 1 - 11
~ = ~ t 1
J:2 J;- v z Jx 2 + y 2 dzdydx.
x Jx 2 + y 2 dzdyd$, y = "* f 1 J:z f 0
1 -v y Jx 2 + y 2 dz dy dx, and
) f 1 r~ f"(1 ) d d d
3 ""
11
49. (a m = Jo Jo Jo + x + y + z z y x = 32 + 24
(b) {X, y, z) = ( m- 1 J; fo~ f~ x(1 + x + y + z) dzdydx,
m- 1 f 0
1
f0~ J~ y(1 + x + y + z) dzdydx,
m - 1 f 0
1
fo~ j~ z(1 + x + y +z)dzdydx)
. 28 307r + 128 457r + 208 )
= ( 9n + 44 ' 45n + 220 ' 135n + 660
t {~ {" . 68 + 1571'
(c) 1::: = fo f o fo (x2 + y2)(1 + x + y + z) dz dy dx = 240
51. (a) f(x, y, z) is a joint density function, so we know JJJR 3 f(x, y, z) dV = 1. Here we have
JJJR3 f(x,y,z) dV = f~oo f~oo }:" 00
f(x,y, z) dzdydx = f 0
2
J: J: Cxyzdzdydx
Then we must have 8G = 1 => C - 1 - 8'
= C J: xdx f 0
2
ydy f 0
2
zdz = C [~x 2 )~ [h 2 )~ [~z 2 )~ = 8C
'
1 rt d r1 d rl d 1 [ 1 2]1 [ 1 2] 1 [ 1 2] 1 1 ( 1 )3 1
= ii Jo x X Jo 11 Y Jo z z = ii 2x o 2Y o 2z o = 8 2 = 1M
(c) P(X + Y + Z::; 1) = P((X, Y, Z) E E) where E is the solid regio·n ·in the first octant bounded by the coordinate planes
and the plane x + y + z = 1. The plane x + y + z = 1 meets the xy-plane in the line x + y = 1, so we have
P(X.+ Y + Z::; 1) = fff Ef(x, y, z) dV = j~ 1 j~ -x j 0
1 -:z:-y kxyz dz dy dx
= k J; fot - :z: xy[~z2 J ;:~ - :z:-y dydx = fs fo1 fol -z xy(l - x - y)2 dydx
= 1~ f 0
1
f 0
1
- "'[(x
3 - 2x 2 + x)y + (2x 2 - 2x)y 2 + xy 3 ] dydx
= 110 fo1 [(x3 - 2x2 + xHy2 + (2x2- 2x)h3 + x(fy'')J::- :r: dx
- 1 f1( 4 2 6 3 . 4 4 + 5) d - 1 ( 1 ) - 1
- 192 Jo X- X + X . - X X X - 192 30 - 57iii)
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