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

264 D CHAPTER 15 MULTIPLE INTEGRALS
25. The right loop of the curve is given by D = {(r, 0) I 0 ~ r :::; cos 20, - r./4 ~ () ~ 71'/ 4}. Using a CAS, we
find m = f fo p(x,y) dA = ff 0 (x 2 + y 2 ) dA = r:~;4 f oco" 20 r 2 rdrd9 = ::. Then
1 i'!r 64 j "'/4 [ os20 64 ~ Tr/4 1cos20 16384J2
·x= - xp(x,y)dA= - (rcos9)r 2 rdrd8= - r 4 cos9dr d8= and
m o 3 7r -Tr/4 o 371' - Tr/4 0 1039571'
1 /lr 64 ~ rr/4 1 cos28 64 j"'/4 cos20
y =- yp(x, y) dA = - 3
. (rsin 9) r 2 rdrd(J = - 3
r 4 sin 9drd0 = 0, so
m D 7r -rr/4 o 7r -Tr/4 1o
(- _) ( 16384v'2 o)
x, y = 1039571' ' .
The moments of inertia are
I _ [J 2 ( )dA -'J "/4 rcos20( . (})2 2 d dB - J"/4 j ·cos20 s . 2(J d d(J- 571' 4
:r -. D y p x, y - -Tr/ 4 Jo r sm r r r - - 7r/4 o r sm r - 384 - 105,
ly = JJ 0 x 2 p(x, y) dA = J~~;4 J; 0 "
571'
Io = I, + Iy = 192
28 (Hos0)
2 r
2 r dr dO = J~~~ 4 J 0 c"" 20 r 5 cos 2 0 drdO = :~ +
1 ~ 5 , and
27. (a) f(x, y) is a joint density function, so we know JfR2 j(x, y) dA = 1. Since f(x, Y).= 0 outside the
rectangle [0, 1) x [0, 2], we can say
Then 2C = 1 => C = t.
J JR2 f(x,y)dA = f~oo f~oo f(x,y)dy 'dx.= f 0
1
J; Cx(1 + y ) dy dx
= C f 1 x [y + ly 2 ]v= 2 dx = c' f 1 4xdx ~ C (2x 2 ] 1 = 2C
Jo 2 v=;O Jo o
1
(b) P (X :::; 1, Y ~ 1) = f~oo f~oo f(x, y) dy dx = J 0
J; tx(1 + y) dy dx
r l 1 [ 1 2] Y = 1 d r1 1 ( 3) dx 3 [ 1 2 ] 1 3
0 375·
= Jo 2x Y + 2Y v=O x = Jo 2x 2 = 4 2x o = 8 or ·
(c) P (X+ Y :::; 1) = P ((X , Y) ED) where Dis the triangular region shown in
the figure. Thus
· If 1 12: 1
P(X+Y :s; 1)=
0 f(x, y)dA = f 0 f 0 - 2x(1+y)dydx
Y
= t 1 lx[y + ly 2 ] v=l -:~: dx = f 1 lx(lx 2 - 2x + ~) dx
Jo 2 2 11 = o Jo 2 2 2 D
3 2 [ •I 3 2] 1
= l 1 (x - 4x + 3x) dx = l L - 4L + 3L
40 44 3 2()
= 4~ ~ 0. 1042
0
1 X
. 29. (a) f(x, y) ;;::: 0, so f is a joint density function if Jf'it2 f(x, y) dA = 1. Here, f(x, y) = 0 outside the fi rst quadrant, so
JJR2 f(x,y)dA= f ooo fo":'0.1e-(o.ox+?·211) dydx = 0.1 J~oc f ooo e-0.5:re-0.2y dy dx = O.l fooo e-o.sx dx f ooo e-0.2y dy
= 0. 1 lim r t e- 0·5"' dx lim rt e- 0·2 Y dy = O.l. lim (- 2e- 0·5"'] t lim [- se- 0·211 ] t
L--+oo Jo t-oo Jo t.--+eo 0 t --+oo 0
= 0.1lim [- 2(e- 0 ·5t
- 1)] lim [- 5(e- 0·2 t - 1)] = (0.1). (-2)(0 - 1) · (-5)(0 - 1) = 1
t --+oo
t --+oo
Thus f(x, y) is a joint density function.
. '
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