Solução_Calculo_Stewart_6e

F.
TX.10SECTION 10.4 AREAS AND LENGTHS IN POLAR COORDINATES ¤ 435
43.
From the first graph, we see that the pole is one point of intersection. By zooming in or using the cursor, we find the θ-values
of the intersection points to be α ≈ 0.88786 ≈ 0.89 and π − α ≈ 2.25. (Thefirst of these values may be more easily
estimated by plotting y =1+sinx and y =2x in rectangular coordinates; see the second graph.) By symmetry, the total
area contained is twice the area contained in the first quadrant, that is,
α
π/2
1
A =2
2 (2θ)2 1
dθ +2 (1 + sin 2 θ)2 dθ =
0
α
45. L =
α
0
4θ 2 dθ +
π/2
= 4
θ3 α
+
3
θ − 2cosθ + 1
θ − 1 sin 2θ π/2
0 2 4
= 4 α 3 α3 + π
+ π 2 4
=3
b
a
π/3
0
α
1+2sinθ +
1
2 (1 − cos 2θ) dθ
−
α − 2cosα +
1
2 α − 1 4 sin 2α ≈ 3.4645
π/3
π/3
r2 +(dr/dθ) 2 dθ = (3 sin θ)2 +(3cosθ) 2 dθ = 9(sin 2 θ +cos 2 θ) dθ
0
dθ =3 θ π/3
0
=3 π
3
= π.
As a check, note that the circumference of a circle with radius 3 is 2π
3
2 2 =3π, and since θ =0to π =
π
traces out 1 of the
3 3
circle (from θ =0to θ = π), 1 (3π) =π.
3
47. L =
=
b
a
2π
0
r2 +(dr/dθ) 2 dθ =
2π
θ 2 (θ 2 +4)dθ =
0
2π
0
(θ 2 ) 2 +(2θ) 2 dθ =
θ θ 2 +4dθ
2π
0
0
θ 4 +4θ 2 dθ
Now let u = θ 2 +4,sothatdu =2θdθ
θdθ= 1 2 du and
2π
0
4π
2 +4
√
θ θ 2 1
+4dθ =
2 udu=
1
u · 2 3/2 4(π 2 +1)
= 1 2 3
3 [43/2 (π 2 +1) 3/2 − 4 3/2 ]= 8 3 [(π2 +1) 3/2 − 1]
4
4
49. The curve r =3sin2θ is completely traced with 0 ≤ θ ≤ 2π. r 2 +
dr 2
dθ =(3sin2θ) 2 +(6cos2θ) 2 ⇒
L =
2π
0 9sin 2 2θ +36cos 2 2θdθ≈ 29.0653