Solução_Calculo_Stewart_6e

F.
362 ¤ CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION
TX.10
(b) x2
a + y2
x (dx/dy)
=1 ⇒ = − y ⇒ dx
2 b2 a 2 b 2 dy = y
−a2 b 2 x
⇒
2 dx
1+ =1+ a4 y 2
dy b 4 x = b4 x 2 + a 4 y 2
= b4 a 2 (1 − y 2 /b 2 )+a 4 y 2
2 b 4 x 2 b 4 a 2 (1 − y 2 /b 2 )
= b4 − b 2 y 2 + a 2 y 2
= b4 − (b 2 − a 2 )y 2
b 4 − b 2 y 2 b 2 (b 2 − y 2 )
= a2 b 4 − a 2 b 2 y 2 + a 4 y 2
a 2 b 4 − a 2 b 2 y 2
The oblate spheroid’s surface area is twice the area generated by rotating the first-quadrant portion of the ellipse about the
y-axis. Thus,
S =2
b
0
= 4πa
b 2
30
=
=
2πx
b
0
1+
2 dx
b
dy =4π
dy
0
b4 − (b 2 − a 2 )y 2 dy = 4πa
b 2
a b4 − (b
b2 − y 2 − a 2 )y 2
b
2 b dy
b 2 − y 2
b
√
b 2 −a 2
4πa u
b √ √
b4 − u 2 b 2 − a 2 2
2 + b4 u
b
2 sin−1 b 2
0
0
√
b4 − u 2
b 2 −a 2
√ √
4πa b
b √ b2 − a 2
b4 − b 2 b 2 − a 2 2
2 (b 2 − a 2 )+ b4 b2 − a 2
2 sin−1 b
du
√
u
b2 − a 2 = √
b 2 − a 2 y
⎡
=2π⎢
⎣ a2 +
Notice that this result can be obtained from the answer in part (a) by interchanging a and b.
31. The analogue of f(x ∗ i ) in the derivation of (4) is now c − f(x ∗ i ), so
S = lim
n
n→∞ i=1
2π[c − f(x ∗ i )] 1+[f 0 (x ∗ i )]2 ∆x = b
a 2π[c − f(x)] 1+[f 0 (x)] 2 dx.
33. For the upper semicircle, f(x) = √ r 2 − x 2 , f 0 (x) =−x/ √ r 2 − x 2 . The surface area generated is
S 1 =
r
=4π
−r
r
2π
r −
r 2 − x 2 1+ x2
0
r 2
√
r2 − x − r dx
2
For the lower semicircle, f(x) =− √ r 2 − x 2 and f 0 (x) =
Thus, the total area is S = S 1 + S 2 =8π
r
0
r r
r 2 − x dx =4π −
r 2 − x 2 2
x
√
r2 − x 2 ,soS 2 =4π
r 2
√ dx =8π
r2 − x 2
0
r
0
r 2 sin −1 x
r
r
0
ab 2 sin −1 √
b2 − a 2
b √
b2 − a 2
r
√
r2 − x 2 dx
r 2
√
r2 − x + r dx.
2
=8πr 2 π
=4π 2 r 2 .
2
yi−1 + yi
35. In the derivation of (4), we computed a typical contribution to the surface area to be 2π |P i−1P i|,
2
the area of a frustum of a cone. When f(x) is not necessarily positive, the approximations y i = f(x i ) ≈ f(x ∗ i ) and
y i−1 = f(x i−1 ) ≈ f(x ∗ i ) must be replaced by y i = |f(x i )| ≈ |f(x ∗ i )| and y i−1 = |f(x i−1 )| ≈ |f(x ∗ i )|. Thus,
2π y i−1 + y i
2
|P i−1 P i | ≈ 2π |f(x ∗ i )| 1+[f 0 (x ∗ i )]2 ∆x. Continuing with the rest of the derivation as before,
we obtain S = b
a 2π |f(x)| 1+[f 0 (x)] 2 dx.
⎤
⎥
⎦