Solução_Calculo_Stewart_6e

F.
434 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10
35. Thedarkershadedregion(fromθ =0to θ =2π/3)represents 1 of the desired area plus 1 2 2
of the area of the inner loop.
From this area, we’ll subtract 1 2
of the area of the inner loop (the lighter shaded region from θ =2π/3 to θ = π), and then
double that difference to obtain the desired area.
2π/3
1
A =2
1
0 2 2 +cosθ2 dθ − π
1 1
2π/3 2 2 +cosθ2 dθ
= 2π/3
1 +cosθ 0 4 +cos2 θ dθ − π
1 +cosθ 2π/3 4 +cos2 θ dθ
= 2π/3
1 +cosθ + 1 (1 + cos 2θ) 0 4 2
dθ
θ
=
4 +sinθ + θ sin 2θ
+
2 4
π
= + √ 3
+ π − √
3
6 2 3 8
= π 4 + 3 4
√
3=
1
4
− π
1 +cosθ + 1 (1 + cos 2θ) 2π/3 4 2
dθ
2π/3 θ
−
4 +sinθ + θ sin 2θ
+
2 4
0
π +3
√
3
− π
4 + π 2
+
π
+ √ 3
+ π − √ 3
6 2 3 8
π
2π/3
37. The pole is a point of intersection.
1+sinθ =3sinθ ⇒ 1=2sinθ ⇒ sin θ = 1 2
⇒
θ = π or 5π .
6 6
The other two points of intersection are 3
, π
2 6 and 3 , 5π
2 6 .
39. 2sin2θ =1 ⇒ sin 2θ = 1 ⇒ 2θ = π , 5π , 13π 17π
,or .
2 6 6 6 6
By symmetry, the eight points of intersection are given by
(1,θ),whereθ = π 12 , 5π
, 13π
12 12
(−1,θ),whereθ = 7π
12 , 11π
, 19π
12 12
,and
17π
12 ,and
,and
23π
12 .
[There are many ways to describe these points.]
41. The pole is a point of intersection. sin θ =sin2θ =2sinθ cos θ ⇔
sin θ (1 − 2cosθ) =0 ⇔ sin θ =0or cos θ = 1 ⇒
2
√3
θ =0, π, π ,or− π 3 3
⇒ the other intersection points are , π 2 3
√3
and , 2π [by symmetry].
2 3