Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 6.1 AREAS BETWEEN CURVES ¤ 265
19. 2y 2 =4+y 2 ⇔ y 2 =4 ⇔ y = ±2,so
2
A = (4 + y 2 ) − 2y 2 dy
−2
=2
2
0
(4 − y 2 ) dy [by symmetry]
=2 4y − 1 y3 2
=2
3
8 − 8 0 3 =
32
3
21. The curves intersect when 1 − y 2 = y 2 − 1 ⇔ 2=2y 2 ⇔ y 2 =1 ⇔ y = ±1.
1
A = (1 − y 2 ) − (y 2 − 1) dy
−1
1
= 2(1 − y 2 ) dy
−1
=2· 2
1
0
(1 − y 2 ) dy
=4 y − 1 y3 1
=4
1 − 1 3 0 3 =
8
3
23. Notice that cos x =sin2x =2sinx cos x ⇔
2sinx cos x − cos x =0 ⇔ cos x (2 sin x − 1) = 0 ⇔
2sinx =1or cos x =0 ⇔ x = π 6 or π 2 .
A =
π/6
0
(cos x − sin 2x) dx +
= sin x + 1 2 cos 2x π/6
0
π/2
π/6
(sin 2x − cos x) dx
+ − 1 2 cos 2x − sin x π/2
π/6
= 1 2 + 1 2 · 1
2 − 0+ 1 2 · 1 + 1
2 − 1 − − 1 2 · 1
2 − 1 2
=
1
2
25. The curves intersect when x 2 = 2
x 2 +1
x 4 + x 2 =2 ⇔ x 4 + x 2 − 2=0 ⇔
(x 2 +2)(x 2 − 1) = 0 ⇔ x 2 =1 ⇔ x = ±1.
1
2
A =
−1 x 2 +1 − x2 dx =2
1
=2 2tan −1 x − 1 3 x3 0
1
0
⇔
2
x 2 +1 − x2 dx
=2 2 · π − 1
4 3 = π −
2
≈ 2.47
3