Solução_Calculo_Stewart_6e

F.
SECTION 16.4 TX.10 DOUBLE INTEGRALS IN POLAR COORDINATES ET SECTION 15.4 ¤ 233
15. One loop is given by the region
D = {(r, θ) |−π/6 ≤ θ ≤ π/6, 0 ≤ r ≤ cos 3θ },sotheareais
π/6
cos 3θ
π/6
r=cos 3θ
1
dA =
rdrdθ=
D
−π/6 0
−π/6 2 r2 dθ
π/6
1
=
−π/6 2 cos2 3θdθ=2
θ + 1 π/6
6 sin 6θ
π/6
0
1
2
r=0
1+cos6θ
2
dθ
= 1 2
17. By symmetry,
0
= π 12
A =2 π/4
0
sin θ
rdrdθ=2 π/4
0 0
1 r2 r=sin θ
dθ
2
r=0
= π/4
sin 2 θdθ= π/4 1
(1 − cos 2θ) dθ
0 0 2
= 1 2 θ −
1
sin 2θ π/4
2 0
= 1 π − 1 sin π − 0+ 1 sin 0 = 1 (π − 2)
2 4 2 2 2 8
19. V = x 2 + y 2 ≤4
x2 + y 2 dA = 2π 2
√
r2 rdrdθ= 2π
dθ 2
0 0 0 0 r2 dr = θ 2π 1 r3 2
=2π
8
0 3 0 3 =
16
π 3
21. The hyperboloid of two sheets −x 2 − y 2 + z 2 =1intersects the plane z =2when −x 2 − y 2 +4=1or x 2 + y 2 =3.Sothe
solid region lies above the surface z = 1+x 2 + y 2 and below the plane z =2for x 2 + y 2 ≤ 3, and its volume is
V =
x 2 + y 2 ≤ 3
2 − 1+x 2 + y 2
dA =
2π
√ 3
0
0
(2 − 1+r 2 ) rdrdθ
= 2π
dθ √ 3
(2r − r √ 1+r
0 0 2 )dr = θ
2π
r √ 2 − 1 (1 + 0
3 r2 ) 3/2 3
0
=2π 3 − 8 − 0+ 1
3 3 =
4
π 3
23. By symmetry,
2π
a 2π
V =2
a2 − x 2 − y 2 dA =2 a2 − r rdrdθ=2 2 dθ
0 0
0
x 2 + y 2 ≤ a 2
=2 θ
2π
a
− 1 0 3 (a2 − r 2 ) 3/2 =2(2π) 0+ 1 a3 = 4π 3 3 a3
0
a
0
r a 2 − r 2 dr
25. The cone z = x 2 + y 2 intersects the sphere x 2 + y 2 + z 2 =1when x 2 + y 2 +
x2 + y 2 2
=1or x 2 + y 2 = 1 2 .So
V =
x 2 + y 2 ≤ 1/2
= 2π
dθ 1/ √ 2
0 0
1 − x2 − y 2 − x 2 + y 2
dA =
2π
√
1/ 2
0
0
1 − r2 − r
rdrdθ
√
r 1 − r2 − r 2 dr = θ
√
2π
1/ 2
− 1 (1 − 0 3 r2 ) 3/2 − 1 3 r3 =2π √
− 1 √2
1
− 1 = π 3 3 2 − 2
0