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

SECTION 15.5 APPLICATIONS OF DOUBLE INTEGRALS 0 261
·Y
?r/4 2
= r r 1rr/4 [
r 3 4 ] r=2
cos 8 sinfJdrd8= .!:.... cos8sin8 d()
Jo J 1 o 4 r = 1
151rr/ 4 . 15 [sin 2 ()] .,.; 4 15
= - sm8cos9d(J = - -- = -
4 0 4 2 0 16
• . 2 2
41. (a) We integrate by parts with u = x and dv =·xe-x dx. Then du = dx and v = - ~e-x , so
f "" x 2 e- "' 2 dx = lim rt x 2 e-"' 2 dx = lim ( - l xe- "' )t 2 + rt l e- "' 2 dx)
Jo t-+oo Jo t-+oo 2 0 Jo 2
= lim (_.!.te-t 2) + .!. roo e-x 2 dx = 0 + .!. roo e- "' 2 dx
t-+oo 2 2 Jo 2 Jo [by !'Hospital's Rule]
1 f "" _.,2 d
= 4 -oo e X
[by Exercise 40(c)]
2
[since e-"' is an even function]
(b) Let u = ..JX. Then u 2 = x .::? dx = 2udu .::?
00
f. 0
.../Xe- "'dx = lim I; .../Xe-"'dx = lim I 0
,/f.ue-" 2 00
2u du = 2I 0
u 2 e- u 2 du = 2 U.J7T)
· t-oo # t-oo
[by part(a)] = ~ft.
15.5 Applications of Double Integrals
1. Q = IIv O"(x, y) dA = I~ I: (2x + 4y) dydx =I~ [2xy + 2y 2 ] ~=~ dx
= I~ (lOx + 50- 4x- 8) dx = J~ (6x + 42) dx = [3x 2 + 42x]~ = 75 + 210 = 285 C
3. m = IIv p(x, y) dA = I 1
3
I 1
4
ky
2
dydx = k I 1
3
dx .{ 1
4
y
2
dy = . k [xJ ~ [h 3 ]~ = k(2)(21) = 42k,
x = ~ IIv xp(x,y) dA = 4 ~k I : I 1
4
kxy
2
dydx = 4
1
2 I 1
3
xdx I 1
4 y
2
dy = f2 [ tx 2 ]~ [k y 3 ]~ = 4\(4)(21) = 2,
- 1 ff ( )dA - 1 f a f 4k a d dx- 1 f sdx r4 ad _ 1 [ ]3 [1 4] 4 - 1 ( 2 )( 2 11:;) 85
Y = m D yp x , Y - 42k 1 . 1 Y y - 42 1 Jt y y - 42 X 1 4Y 1 - 42 -4- = 28
Hence m = 42k, (x, Y) = (2, ~) .
r2f3-x( ) r2 [ 1 2]v=3-:z: dx r2 [ ( a ) 1 ( 3 )2 1 2]
5. m= Jo .,12 x+y dydx = Jo xy+2Y v=x/ 2 =J~ x 3- 2x + 2 - x - 8x dx
= J 0
2
(-~x 2 + ~) dx = [- ~(tx 3 ) + ~x ] ~ = 5,
. r2J·3-x( 2 )d dx r2 ( 2 1 2]11=8- :z: dx r2 (9 9 3) d 9
lvfv = Jo :r:/ 2 x + xy y = Jo x y + 2xy 1l=x/2 = Jo 2x - gX x = 2•
lvfx = Jo
f2J3-v(
.,
2)d d r2.[1 2 1 3)11=3-x d r 2 (g 9 )dx 12 xy + y y x = Jo 2xy + 3Y
9
11 =, 12 x = Jo - 2x = .
--) (lvfy Mx) (3 3)
Hencem = 6, ( x_, y = m '--;:; = 4' 2 .
7. m = J~ 1 J~ -"' 2 kydydx = k f~ 1 [ ~y 2 ]~=~-"' 2 dx = ~k f~ 1 (1 - x 2 ? dx = ~ k f~ 1 (1 - 2x2 + x 4 ) dx
= ~k [x- ix 3 + %x 5 ] ~ 1 = ~k (1 - i + -k + 1- % + %) = 1
85 k,
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