Solução_Calculo_Stewart_6e

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F. 25. To findtheareainsidetheleminiscater 2 =8cos2θ and outside the circle r =2, we first note that the two curves intersect when r 2 =8cos2θ and r =2, that is, when cos 2θ = 1 2 .For−π
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
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