Solução_Calculo_Stewart_6e

F.
308 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
TX.10
35. Area of 4POQ = 1 2 (r cos θ)(r sin θ) = 1 2 r2 sin θ cos θ. AreaofregionPQR = r
r cos θ
√
r2 − x 2 dx.
Let x = r cos u ⇒ dx = −r sin udufor θ ≤ u ≤ π . Then we obtain
2
√
r2 − x 2 dx = r sin u (−r sin u) du = −r 2 sin 2 udu= − 1 2 r2 (u − sin u cos u)+C
so area of region PQR= 1 2
= − 1 2 r2 cos −1 (x/r)+ 1 2 x √ r 2 − x 2 + C
−r 2 cos −1 (x/r)+x √ r 2 − x 2 r
r cos θ
= 1 2
0 − (−r 2 θ + r cos θrsin θ) = 1 2 r2 θ − 1 2 r2 sin θ cos θ
and thus, (area of sector POR)=(area of 4POQ)+(area of region PQR)= 1 2 r2 θ.
37. From the graph, it appears that the curve y = x 2 √ 4 − x 2 and the
line y =2− x intersectataboutx = a ≈ 0.81 and x =2, with
x √ 2 4 − x 2 > 2 − x on (a, 2). So the area bounded by the curve and the line is
A ≈ 2
√
a x
2
4 − x 2 − (2 − x) dx = 2
√
a x2 4 − x 2 dx − 2x − 1 x2 2
. 2 a
To evaluate the integral, we put x =2sinθ,where− π 2 ≤ θ ≤ π 2 .Then
dx =2cosθdθ, x =2 ⇒ θ =sin −1 1= π ,andx = a ⇒ θ = α 2 =sin−1 (a/2) ≈ 0.416.So
2 x2√ 4 − x
a 2 dx ≈ π/2
4sin 2 θ (2 cos θ)(2 cos θdθ)=4 π/2
sin 2 2θdθ=4 π/2 1
(1 − cos 4θ) dθ
α α α 2
=2 θ − 1 4 sin 4θ π/2
α
Thus, A ≈ 2.81 − 2 · 2 − 1 2 · 22 − 2a − 1 2 a2 ≈ 2.10.
=2 π
2 − 0 − α − 1 4 (0.996) ≈ 2.81
39. (a) Let t = a sin θ, dt = a cos θdθ, t =0 ⇒ θ =0and t = x ⇒ θ =sin −1 (x/a). Then
x
0
sin
−1 (x/a)
a2 − t 2 dt = a cos θ (a cos θdθ)
0
= a 2 sin −1 (x/a)
0
cos 2 θdθ= a2
2
=
θ a2
sin
−1 (x/a)
+ 1 2
2
sin 2θ = a2
0
2
√
a2 − x 2
= a2
2
sin −1 x
a
+ x a ·
= 1 2 a2 sin −1 (x/a)+ 1 2 x √ a 2 − x 2
a
sin −1 (x/a)
0
(1 + cos 2θ) dθ
sin
−1 (x/a)
θ +sinθ cos θ
0
− 0
(b) The integral x
√
a2 − t
0
2 dt represents the area under the curve y = √ a 2 − t 2 between the vertical lines t =0and t = x.
The figure shows that this area consists of a triangular region and a sector of the circle t 2 + y 2 = a 2 . The triangular region
has base x and height √ a 2 − x 2 , so its area is 1 2 x √ a 2 − x 2 . The sector has area 1 2 a2 θ = 1 2 a2 sin −1 (x/a).