Solução_Calculo_Stewart_6e

F.
236 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
9. Note that sin(πx/L) ≥ 0 for 0 ≤ x ≤ L.
m = L
0
sin(πx/L)
ydydx= L 1
0 0 2 sin2 (πx/L) dx = 1 2
M y = L sin(πx/L)
x · ydydx= 1 L
0 0 2 0
= 1 · x 1
x − L sin(2πx/L) L
− 1 L
2 2 4π 0 2 0
= 1 4 L2 − 1 2
L
1
4 x2 + cos(2πx/L) L2
= 1 4π 2 4 L2 − 1 2
0
1 x − L sin(2πx/L) L
= 1 L,
2 4π 0 4
x sin2 (πx/L) dx
1
2 x − L 4π sin(2πx/L) dx
integrate by parts with
u = x, dv =sin 2 (πx/L) dx
1
4 L2 + L2
4π 2 − L2
4π 2
= 1 8 L2 ,
M x = L sin(πx/L)
y · ydydx= L
1
0 0 0 3 sin3 (πx/L) dx = 1 L
3 0 1 − cos 2 (πx/L) sin(πx/L) dx
[substitute u =cos(πx/L)] ⇒ du = − π sin(πx/L)]
L
= 1 3 −
L
π cos(πx/L) −
1
3 cos3 (πx/L) L
= − L
0 3π −1+
1
− 1+ 1
3 3 =
4
L. 9π
Hence m = L L 2
4 , (x, y) = /8
L/4 , 4L/(9π)
=
L/4
L
2 , 16
.
9π
11. ρ(x, y) =ky = kr sin θ, m = π/2 1
0 0 kr2 sin θdrdθ = 1 k π/2
3
sin θdθ = 1 k 0 3
− cos θ π/2
= 1 k,
0 3
M y = π/2 1
0 0 kr3 sin θ cos θdrdθ= 1 k π/2
4
sin θ cos θdθ= 1 k 0 8
− cos 2θ π/2
= 1 k,
0 8
M x = π/2
0
1
0 kr3 sin 2 θdrdθ= 1 k π/2
sin 2 θdθ= 1 k θ +sin2θ π/2
= π k.
4 0 8 0 16
Hence (x, y) = 3
8 , 3π
16
.
13. ρ(x, y) =k x 2 + y 2 = kr,
m = ρ(x, y)dA = π 2
kr · rdrdθ
D 0 1
= k π
0 dθ 2
1 r2 dr = k(π) 1
3 r3 2
1 = 7 3 πk,
M y = xρ(x, y)dA = π 2 (r cos θ)(kr) rdrdθ= k π
cos θdθ 2
D 0 1 0 1 r3 dr
= k sin θ π 1 r4 2
= k(0)
15
0 4 1 4 =0
[this is to be expected as the region and density
function are symmetric about the y-axis]
Hence (x, y) =
M x = yρ(x, y)dA = π 2 (r sin θ)(kr) rdrdθ= k π
sin θdθ 2
D 0 1 0
= k − cos θ π 1 r4 2
= k(1 + 1)
15
0 4 1 4 =
15
k. 2
0, 15k/2
7πk/3
= 0, 45
14π
.
15. Placing the vertex opposite the hypotenuse at (0, 0), ρ(x, y) =k(x 2 + y 2 ).Then
m = a
0
a − x
By symmetry,
k x 2 + y 2 dy dx = k a
0 0
Hence (x, y) = 2
5 a, 2 5 a .
1 r3 dr
ax 2 − x 3 + 1 3 (a − x)3 dx = k 1
3 ax3 − 1 4 x4 − 1 12 (a − x)4 a
0 = 1 6 ka4 .
M y = M x = a a − x
ky(x 2 + y 2 ) dy dx = k a
1 (a − 0 0 0 2 x)2 x 2 + 1 (a − x)4 dx
4
= k 1
6 a2 x 3 − 1 4 ax4 + 1
10 x5 − 1 20 (a − x)5 a
0 = 1 15 ka5