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

294 0 CHAPTER 15 MULTIPLE INTEGRALS
41.
f 3
rJ 9 2
- x
(x 3 + xy 2 )dydx = { 3 {~ x(x 2 +y 2 )dydx
1o 1 -J9-x2 1o 1 -~
= J::~~2 f~ (r cos O)(r2 ) r dr dO
43. From the graph, it appears that 1 - x 2 = e"' at x ~ - 0.71 and at
x = 0, with 1 - x 2 > e"' on ( -0.71, 0). So the desired integral is
= J::~~2 cos 0 dO J; r 4 dr
= (sino]:::/2 (ir5] ~ = 2 ·. i (243) = 4 : 6 = 97.2
JJ"n y2dA ~ f~o.n J.1.,-x2 y2 dydx
= l Jo [(1 - x2)3 - e3"'] dx
3 -0.71
=-o.25
45. {a) f(x, y) is a joint density function, so we know that JJR2 f(x, y) dA = 1. Since f(x, y) = 0 outside the rectangle
(0, B) x (0, 2], we can say
Then 15C = 1 :::?
JJR2 f (x,y) dA = f~oo f~oo f(x,y) dydx = J: J: C(x + y) dydx
C = ft.
= C .r; [xy + h 2 J ~:~ dx = C J0 3 (2x + 2) dx = C[x 2 + 2x)~ = 15C
(c) P(X + Y ~ 1) = P ((X, Y) E D) where Dis the triangular regiol} shown in
the figure. Thus
y
P(X +Y ~ 1) = JJ0 f(x,y) dA = f 01 J;-x -ft (x+y)dydx
=· ..1.. rl [ + 1 2]u= 1 -"' dx
15 Jo xy 2Y u=O
= fs f; [x(1 - x) + ~ ( 1 - x?] dx
- 1 f1(1 .2)d - 1 [ 1 3]1 1
- 30 Jo - X X - 30 X- 3X 0 = 45
0 I X
47.
. z
J_ l 1
j'l .,2 j '1- IJ ( ) J.l r.1-=Jv"ii ! ( ) d d d
0
fx,y ,zdzdydx= 0
. 0
- v"ii x,y,z xyz
X
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