Solução_Calculo_Stewart_6e

F.
436 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10
51. The curve r =sin
θ
2 is completely traced with 0 ≤ θ ≤ 4π.
L =
4π
0
sin 2 θ
2
+
1
4 cos2 θ
2
dθ ≈ 9.6884
r 2 + dr 2
dθ =sin
2 θ
2 + 1 cos
θ 2
2 2
⇒
53. The curve r =cos 4 (θ/4) is completely traced with 0 ≤ θ ≤ 4π.
r 2 +(dr/dθ) 2 =[cos 4 (θ/4)] 2 + 4cos 3 (θ/4) · (− sin(θ/4)) · 1
4
L = 4π
0
=cos 8 (θ/4) + cos 6 (θ/4) sin 2 (θ/4)
=cos 6 (θ/4)[cos 2 (θ/4) + sin 2 (θ/4)] = cos 6 (θ/4)
cos6 (θ/4) dθ = 4π
cos 3 (θ/4) dθ
0
=2 2π
0
cos 3 (θ/4) dθ [since cos 3 (θ/4) ≥ 0 for 0 ≤ θ ≤ 2π] =8 π/2
0
cos 3 udu u = 1 4 θ
68
=8 1
(2 + 3 cos2 u)sinu π/2
= 8 [(2 · 1) − (3 · 0)] = 16 0 3 3
2
55. (a) From (10.2.7),
S = b
a 2πy (dx/dθ) 2 +(dy/dθ) 2 dθ
= b
a 2πy r 2 +(dr/dθ) 2 dθ [from the derivation of Equation 10.4.5]
= b
a 2πr sin θ r 2 +(dr/dθ) 2 dθ
(b) The curve r 2 =cos2θ goes through the pole when cos 2θ =0
2θ = π ⇒ θ = π . We’ll rotate the curve from θ =0to θ = π and double
2 4 4
this value to obtain the total surface area generated.
r 2 =cos2θ ⇒ 2r dr
2 dr
dθ = −2sin2θ ⇒ = sin2 2θ
dθ r 2
S =2
=4π
π/4
0
π/4
0
2π √
cos 2θ sin θ cos 2θ + sin 2 2θ /cos 2θdθ=4π
√
cos 2θ sin θ
1
√
cos 2θ
dθ =4π
π/4
0
⇒
= sin2 2θ
cos 2θ .
π/4
sin θdθ=4π − cos θ π/4
0
0
√
cos 2θ sin θ
cos 2 2θ +sin 2 2θ
cos 2θ
dθ
√2
= −4π − 1 =2π 2 − √
2
2