Solução_Calculo_Stewart_6e

F.
276 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
TX.10
59. The cross-section of the base corresponding to the coordinate x has length
y =1− x. The corresponding square with side s has area
A(x) =s 2 =(1− x) 2 =1− 2x + x 2 . Therefore,
V =
Or:
1
0
A(x) dx =
1
0
(1 − 2x + x 2 ) dx
= x − x 2 + 1 x3 1
=
1 − 1+ 1 3 0 3 − 0=
1
3
1
0
(1 − x) 2 dx =
0
1
u 2 (−du) [u =1− x] = 1
3 u3 1
0 = 1 3
61. The cross-section of the base b corresponding to the coordinate x has length 1 − x 2 . The height h also has length 1 − x 2 ,
so the corresponding isosceles triangle has area A(x) = 1 2 bh = 1 2 (1 − x2 ) 2 . Therefore,
V =
1
1
(1 − 2 x2 ) 2 dx
−1
1
=2· 1
2
0
(1 − 2x 2 + x 4 ) dx [by symmetry]
= x − 2 3 x3 + 1 x5 1
= 1 − 2 + 1
5 0 3 5 − 0=
8
15
63. (a) The torus is obtainedbyrotatingthecircle(x − R) 2 + y 2 = r 2 about
the y-axis. Solving for x, we see that the right half of the circle is given by
x = R + r 2 − y 2 = f(y) and the left half by x = R − r 2 − y 2 = g(y). So
V = π r
−r [f(y)] 2 − [g(y)] 2 dy
=2π r
0
R 2 +2R r 2 − y 2 + r 2 − y 2
−
=2π r
0 4R r 2 − y 2 dy =8πR r
0
r2 − y 2 dy
R 2 − 2R r 2 − y 2 + r 2 − y 2
dy
(b) Observe that the integral represents a quarter of the area of a circle with radius r, so
8πR r
r2 − y
0
2 dy =8πR · 1
4 πr2 =2π 2 r 2 R.
65. (a) Volume(S 1 )= h
0 A(z) dz = Volume(S 2) since the cross-sectional area A(z) at height z is the same for both solids.
(b) By Cavalieri’s Principle, the volume of the cylinder in the figure is the same as that of a right circular cylinder with radius r
and height h,thatis,πr 2 h.