Solução_Calculo_Stewart_6e

F.
SECTION TX.10 16.5 APPLICATIONS OF DOUBLE INTEGRALS ET SECTION 15.5 ¤ 237
17. I x = D y2 ρ(x, y)dA = 1
e
x
y 2 · ydydx= 1
1 y4 y=e x
dx = 1 1
0 0 0 4 y=0 4 0 e4x dx = 1 1 e4x 1
= 1
4 4 0 16 (e4 − 1),
I y = D x2 ρ(x, y) dA = 1
e
x
x 2 ydydx= 1
x2 1
y2 y=e x
dx = 1 1
0 0 0 2 y=0 2 0 x2 e 2x dx
= 1 1
2 2 x2 − 1 x + 1
2 4 e
2x 1
[integrate by parts twice] = 1 0
8 (e2 − 1),
and I 0 = I x + I y = 1 16 (e4 − 1) + 1 8 (e2 − 1) = 1 16 (e4 +2e 2 − 3).
19. As in Exercise 15, we place the vertex opposite the hypotenuse at (0, 0) and the equal sides along the positive axes.
I x = a a−x
y 2 k(x 2 + y 2 ) dy dx = k a a−x
(x 2 y 2 + y 4 ) dy dx = k a 1
0 0 0 0 0 3 x2 y 3 + 1 y5 y=a−x
5
dx
y=0
= k a 1
0 3 x2 (a − x) 3 + 1 (a − x)5 dx = k
1 1
5 3 3 a3 x 3 − 3 4 a2 x 4 + 3 5 ax5 − 1 x6 − 1 (a − x)6 a
= 7
6 30 0 180 ka6 ,
I y = a a−x
0
= k a
0
x 2 k(x 2 + y 2 ) dy dx = k a a−x
(x 4 + x 2 y 2 ) dy dx = k a 0 0 0 0 x 4 y + 1 3 x2 y 3 y=a−x
dx
y=0
x 4 (a − x)+ 1 3 x2 (a − x) 3 dx = k 1
5 ax5 − 1 6 x6 + 1 3
and I 0 = I x + I y = 7 90 ka6 .
1
3 a3 x 3 − 3 4 a2 x 4 + 3 5 ax5 − 1 6 x6 a
0 = 7
180 ka6 ,
21. Using a CAS, we find m = ρ(x, y) dA = π
sin x
xy dy dx = π2
D 0 0
8 .Then
x = 1
xρ(x, y) dA =
8 π
sin x
x 2 ydydx = 2π m D
π 2 0 0
3 − 1 π and
y = 1
yρ(x, y) dA = 8 π
sin x
xy 2 dy dx = 16
2π
m
π 2 9π ,so(x, y) = 3 − 1 π , 16
.
9π
D
0
0
The moments of inertia are I x = D y2 ρ(x, y) dA = π
sin x
xy 3 dy dx = 3π2
0 0
64 ,
I y = D x2 ρ(x, y) dA = π
sin x
x 3 ydydx= π2
0 0
16 (π2 − 3),andI 0 = I x + I y = π2
64 (4π2 − 9).
23. I x = D y2 ρ(x, y)dA = h b
0 0 ρy2 dx dy = ρ b
dx h
0 0 y2 dy = ρ x b 1 y3 h
= ρb 1
h3 = 1 0 3 0 3 3 ρbh3 ,
I y = D x2 ρ(x, y)dA = h b
0 0 ρx2 dx dy = ρ b
0 x2 dx h
dy = ρ 1
0 3 x3 b
0 [y]h 0 = 1 3 ρb3 h,
and m = ρ (area of rectangle) = ρbh since the lamina is homogeneous. Hence x 2 = Iy m = 1
3 ρb3 h
ρbh = b2 3
⇒ x = b √
3
and y 2 = Ix 1
m = 3 ρbh3
ρbh
= h2
3
⇒ y = h √
3
.
25. In polar coordinates, the region is D = (r, θ) | 0 ≤ r ≤ a, 0 ≤ θ ≤ π 2
,so
I x = D y2 ρdA= π/2 a ρ(r sin 0 0 θ)2 rdrdθ= ρ π/2
sin 2 dθ a
0
= ρ 1
θ − 1 sin 2θ π/2 1 r4 a
= ρ
π 1 a4 = 1 2 4 0 4 0 4 4 16 ρa4 π,
0 r3 dr
I y = D x2 ρdA= π/2 a ρ(r cos 0 0 θ)2 rdrdθ= ρ π/2
cos 2 dθ a
0
= ρ 1
θ + 1 sin 2θ π/2 1 r4 a
= ρ
π 1 a4 = 1 2 4 0 4 0 4 4 16 ρa4 π,
0 r3 dr
and m = ρ · A(D) =ρ · 1
4 πa2 since the lamina is homogeneous. Hence x 2 = y 2 =
1
16 ρa4 π
1
4 ρa2 π = a2
4
⇒ x = y = a 2 .