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

262 D CHAPTER 15 MULTIPLE INTEGRALS
Mv ·= .f~ 1 ,{0 1 -:r 2 kxydydx = kf~ 1 axy 2 ]~=~- :r 2 dx = ~kf~ 1 x(1 - x 2 ) 2 dx = ~ kf~ 1 (x- 2x 3 + x 5 )dx
- lk(lx 2 - lx 4 + lx 6 ] 1 - l k (1- l + l - l + l _ l) - 0
- 2 2 2 6 -1 - 2 2 . 2 6 2 2 6 - '
M x = J 1 r 1 -"'~ ky 2 dydx = kf 1 (1y 3 ] v=l-x2
dx = lkf 1 (1 - x 2 ) 3 dx = lk { 1 (1'- 3x 2 + 3x 4 - x 6 )dx
- 1 Jo -1 3 v=o s -1 3 . - 1
H _ 8 k (- -) _ (o 32k/105) _ (o 4)
encem - 15 , x , y - ,· Bkf l ~ - •7 ·
9. Note thatsin(7Tx/ L) ~ 0 for 0 ::::; x::::; L.
m = J:' J;in(rrxfL) ydydx = g t sin 2 (7Tx/L) dx =tax- 4 ~ s in (27rx /L) j~ = ~ L ,
M. _ j 'Lj·sin(7rx/L) . d dx _ .!
y - o o
rL · 2( jL) dx
X Y Y - 2 Jo X SID 11'X
[
integrate by parts with ]
u = x , dv = sin 2 (7rx/ L) dx
= t · x(tx - 4r;, sin (27Tx/L)) ] ~ - ~ J 0
L [tx - 4 ~ sin(27Tx/ L)] dx
= l£2- l [.!x2 + L2 cos(211'xj L)]L = .!£2 - l (.!£2 + L2 - L2) = l£2
4 2 4 41f2' 0 4 2 4 41f2' 41f2' 8 '
M:z: = J 0
L J;in("'xf L ) y · y dy dx = f 0
L ~ sin 3 (7Tx/ L ) dx = ~ f 0
L [1 - .cos 2 (7Tx/ L )] sin(7Tx/ L ) dx
[substituteu=cos(7Tx/L)) ::} du= -fsin(7Tx/L))
= t(- ~)[cos(7Tx / L) - 1 cos 3 (7Tx/L)]~ = - 3~ (- 1 + ~ -1 + i) = 9~L .
L _ _ (L 2 /8 4£/ (911')) (L 16)
Hencem = 4' (x,y) = L/ 4 ' L /4 = 2 ' 971' .
11 . p(x, y) = ky = kr sin O, m = )~,. 12 J; kr 2 sinO dr dB= ik )~.,. 12 sinO dtl = ik[- cosB]~ 12 = ftk,
Mv = J 0
,. 12 J; kr 3 sinBcos BdrdB = i k f 0
" 12 sinB cosB dB= ~k [- cos 28] ~ 12 = lk,
M x = J 0
,. 12 J 0
1
kr 3 sin 2 B dr dB = ik f 0
"' 12 sin 2 B dB = lk[B + sin 28] ~ 1 2 = fak.
Hence (x, y) = (i, ~~) .
13.
y
p(x, y) = k jx2 +y 2 = kr,
m = ffo p(x, y)dA = f 0
"' f 1
2
kr · r dr dB
= k f 0 "' dB f 1
2
r 2 dr = k(11') [ ir 3 ]~ = ~11'k,
= k [sinO]"' [.!r 4]2 = k(O) (.!&) = O . [this istobe expectedastheregionanddensity
0
4
1
4
function are symmetric about they-axis]
M:r = ffo yp(x, y)dA = fo,. t