Solução_Calculo_Stewart_6e

F.
432 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10
13. One-sixth of the area lies above the polar axis and is bounded by the curve
r =2cos3θ for θ =0to θ = π/6.
15. A= 2π
0
= 1 2
= 1 2
= 1 2
= 1 2
A =6 π/6
0
= 12 2
1
(2 cos 2 3θ)2 dθ =12 π/6
cos 2 3θdθ
0
π/6
0
(1 + cos 6θ) dθ
=6 θ + 1 6 sin 6θ π/6
0
=6 π
6
= π
1
(1 + 2 sin
2 6θ)2 dθ = 1 2π
2
(1 + 4 sin 6θ 0 +4sin2 6θ) dθ
2π
0 1+4sin6θ +4· 1
(1 − cos 12θ) 2
dθ
2π
(3 + 4 sin 6θ − 2 cos 12θ) dθ
0
3θ −
2
cos 6θ − 1 sin 12θ 2π
3 6 0
(6π −
2
− 0) − 0 − 2 − 0 =3π
3 3
17. Theshadedloopistracedoutfromθ =0to θ = π/2.
A = π/2
0
= 1 π/2
2 0
= 1 4
π
2
1
2 r2 dθ = 1 2
π/2
0
sin 2 2θdθ
1
2 (1 − cos 4θ) dθ = 1 4
=
π
8
θ −
1
4 sin 4θ π/2
0
19. r =0 ⇒ 3cos5θ =0 ⇒ 5θ = π ⇒ θ = π .
2 10
A = π/10 1
(3 cos −π/10 2 5θ)2 dθ = π/10
9cos 2 5θdθ= 9 π/10
(1 + cos 10θ) dθ = 9 0 2 0 2 θ +
1
sin 10θ π/10
= 9π
10 0 20
21. This is a limaçon, with inner loop traced
out between θ = 7π 6
solving r =0].
11π
and [found by
6
3π/2
1
A =2 (1 + 2 sin 2 θ)2 dθ =
7π/6
3π/2
7π/6
= θ − 4cosθ +2θ − sin 2θ 3π/2
= 9π
7π/6 2
1+4sinθ +4sin 2 θ dθ =
3π/2
7π/6
−
7π
+2√ 3 − √
3
= π − 3√ 3
2 2
2
1+4sinθ +4· 1
2 (1 − cos 2θ) dθ
23. 2cosθ =1 ⇒ cos θ = 1 2
⇒ θ = π 3 or 5π 3 .
A =2 π/3
0
= π/3
0
1
[(2 cos 2 θ)2 − 1 2 ] dθ = π/3
(4 cos 2 θ − 1) dθ
0
4
1 (1 + cos 2θ) − 1 dθ = π/3
(1 + 2 cos 2θ) dθ
2 0
= θ +sin2θ π/3
0
= π 3 + √ 3
2