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

SECTION 15.5 APPLICATIONS OF DOUBLE INTEGRALS D 263
15. Placing the vertex opposite the hypotenuse at (0, 0), p(x, y) = k(x 2 + y 2 ). Then
m = J; foa-"' k (x 2 + y 2 ) dy dx = k foa [ax 2 - x 3 + 4 (a - x) 3 ] dx = k[~ax 3 - tx 4 -
By symmetry,
1
12 (a - x) 4 ]~ = ~ ka 4 •
Hence (x, y) = (~a, ~a).
17. I :r: = JJ 0
y 2 p(x, y)dA = .{~ 1 J; -x 2 y 2 · ky dy dx = k j~ 1 [h 4 J~:~ -x 2 dx = tk f~ 1 (1 - :t2 ) 4 dx ·
- l kf 1 (x 8 -4x 6 +6x 4 - 4x 2 + l)dx = l k [lx 9 - ix 7 + !!x 5 - :!x 3 + x] 1 = .M..k
- 4 -1 4 9 7 5 3. - 1 315 '
Iy = fio x 2 p(x, y) dA = }~ 1 j~ -"' 2 kx~ydydx = k I~ 1 [ ~x 2 y 2 J ~:~-"' 2 dx = ~k t 1
x 2 (1- x 2 ? dx
1 k It < 2
2
4 6) d 1 k [ 1 3 2 s 1 1] 1 _ 8 k
= 2 • - 1 X - X +X X= 2 aX - s X + "'fX - 1 - 105 '
and Io = I, + Iv = 36~k + t~sk = 38t8sk.
19. As in Exercise 15, we place the vertex opposite the hypotenuse at (0, 0) and the equal sides along the positive axes.
Ix = I~' I; -xy2k(x2 +y2)dydx = k.fo" .fo"- "'(x2y2 +y4)dydx = k.fo"[~x2y3 + h5J~:~-x
dx
= k f'a [lx 2 (a- x) 3 + l(a - x) 5 ] dx = k [l (la 3 x 3 - 1a 2 x 4 + !l.ax 5 - lx 6 ) - l (a- x) 6 ] a = - 1 -ka 6
. 0 3 5 3 3 4 5 6 30 0 180 '
Iv = I; I;-x x 2 k(x 2 +y 2 ) dydx = k J 0
a.f;-"'(x 4 +x 2 y 2 )dydx = k .f 0
a[x 4 y + ~x 2 y 3 J::~ - x
dx
- k j ·a [x 4 (a - x) + l x 2 (a - x) 3 ) dx = k [lax 5 - lx 0 + l (l a 3 x 3 - !l.a 2 x 4 + !!ax 5 - lx 0 ) ) a - - 1 - ka 6
- 0 3 5 6. 3 3 4 5 6 0 - 180 '
21. I,. = IIo y 2 p(x, y)dA = Io" I~ py 2 dx dy =pI: dx I~' y 2 dy = p[ x ]~ [ ~y 3 ]~ = pb(th 3 ) = tpbh 3 ,
Iv = IIo x 2 p(x, y)dA = Io" I~ px 2 dxdy =pI: x 2 dx I~' dy = P[ tx 3 ]~ (y) ~ = ipb 3 h,
l l b 3 h b 2
and m = p (area of rectangle) = pbh since the lamina is homogeneous. Hence ¥ 2 = ...JL = L_ = -
m pbh 3
= 2 I :r: tpbh 3 h 2
andy = - = --= -
m pbh 3
=>
23. In polar coordinates, the region is D = { (r, 8) I 0 ~ r ~ a, 0 ~ 8 ~ ~},so
! 11 = .{[ 0
x 2 pdA = f 0
"" 12 J; p(r cos 8) 2 r dr d/J = p I 0
"" 12 cos 2 dB .f 0
o. r 3 dr
= p(t8 + t sin 28)~ 12 (tr 4 )~ =~Of) (ta 4 ) = ftPa 4 rr,
.l...pa4rr
and m = p · A(D) = p · trra 2 since the lamina is homogeneous. Hence ¥ 2 = y 2 = ~ = - => x = y = ~
2 .
4pa 1r 4
a2
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