Solução_Calculo_Stewart_6e

F.
83. tan ψ =tan(φ − θ) =
=
dy
dθ − dx
dθ tan θ
dx
dθ + dy =
dθ tan θ
dy
dx − tan θ
1+ dy =
dx tan θ
dr
dr
dθ sin θ + r cos θ − tan θ
dr
dr
dθ cos θ − r sin θ +tanθ
tan φ − tan θ
1+tanφ tan θ =
= r cos2 θ + r sin 2 θ
dr
dθ cos2 θ + dr = r
dr/dθ
dθ sin2 θ
TX.10SECTION 10.4 AREAS AND LENGTHS IN POLAR COORDINATES ¤ 431
dy/dθ
dx/dθ − tan θ
1+ dy/dθ
dx/dθ tan θ
dθ cos θ − r sin θ
dθ sin θ + r cos θ =
r cos θ + r · sin2 θ
cos θ
dr dr
cos θ +
dθ dθ · sin2 θ
cos θ
10.4 Areas and Lengths in Polar Coordinates
1. r = θ 2 , 0 ≤ θ ≤ π 4 . A = π/4
3. r =sinθ, π 3 ≤ θ ≤ 2π 3 .
2π/3
1
A =
2 sin2 θdθ= 1 4
π/3
= 1 4
2π
3 − 1 2
− √ 3
2
0
π/4
π/4
1
2 r2 1
dθ =
2 (θ2 ) 2 1
dθ =
2 θ4 dθ = 1
θ5 π/4
= 1 π
5
= 1
10 0 10 4 10,240 π5
0
0
2π/3
− π 3 + 1 2
π/3
√3
(1 − cos 2θ) dθ = 1 4
2
= 1 4
θ −
1
sin 2θ 2π/3
= 1 2π − 1 sin 4π − π + 1 sin
2π
2 π/3 4 3 2 3 3 2 3
π
+ √
3
= π + √ 3
3 2 12 8
5. r = √ 2π
2π √ 2
2π
1
θ, 0 ≤ θ ≤ 2π. A =
2 r2 1
dθ =
2 θ dθ =
1
θdθ= 1
θ2 2π
= π 2
2 4 0
0
0
0
7. r =4+3sinθ, − π ≤ θ ≤ π .
2 2
π/2
1
A = ((4 + 3 sin 2 θ)2 dθ = 1 2
−π/2
= 1 2 · 2 π/2
=
π/2
0
0
π/2
−π/2
(16 + 24 sin θ +9sin 2 θ) dθ = 1 2
16 + 9 · 1
2 (1 − cos 2θ) dθ [by Theorem 5.5.7(a)]
π/2
−π/2
41
− 9 cos 2θ dθ = 41
θ − 9 sin 2θ π/2
= 41π
− 0 − (0 − 0) = 41π
2 2 2 4 0 4 4
9. The area above the polar axis is bounded by r =3cosθ for θ =0
to θ = π/2 [not π]. By symmetry,
A =2 π/2
0
=9 π/2
0
1
2 r2 dθ = π/2
(3 cos θ) 2 dθ =3 2 π/2
0
1
2 (1 + cos 2θ) dθ = 9 2
cos 2 θdθ
0
θ +
1
sin 2θ π/2
= 9 π +0 − (0 + 0) = 9π 2 0 2 2 4
Also, note that this is a circle with radius 3 2 ,soitsareaisπ 3
2
2
= 9π 4 .
11. The curve goes through the pole when θ = π/4,sowe’llfindtheareafor
0 ≤ θ ≤ π/4 and multiply it by 4.
A =4 π/4
0
1
2 r2 dθ =2 π/4
(4 cos 2θ) dθ
0
=8 π/4
0
cos 2θdθ=4 sin 2θ π/4
0
=4
(16 + 9 sin 2 θ) dθ [by Theorem 5.5.7(b)]