Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 8.2 AREA OF A SURFACE OF REVOLUTION ¤ 361
27. Since a>0,thecurve3ay 2 = x(a − x) 2 only has points with x ≥ 0.
[3ay 2 ≥ 0 ⇒ x(a − x) 2 ≥ 0 ⇒ x ≥ 0.]
The curve is symmetric about the x-axis (since the equation is unchanged
when y is replaced by −y). y =0when x =0or a, so the curve’s loop
extends from x =0to x = a.
d
dx (3ay2 )= d
dx [x(a − x)2 ] ⇒ 6ay dy
dx = x · 2(a − x)(−1) + (a − x)2 ⇒ dy
dx
2 dy
= (a − x)2 (a − 3x) 2
= (a −
x)2 (a − 3x) 2 3a the last fraction
·
=
dx
36a 2 y 2 36a 2 x(a − x) 2 is 1/y 2
=
(a − x)[−2x + a − x]
6ay
(a − 3x)2
12ax
⇒
⇒
2 dy
1+ =1+ a2 − 6ax +9x 2
dx
12ax
= 12ax
12ax + a2 − 6ax +9x 2
12ax
= a2 +6ax +9x 2
12ax
=
(a +3x)2
12ax
for x 6= 0.
a
a
(a) S = 2πy ds =2π
x=0
0
a
√
x (a − x)
√
3a
· a +3x √
12ax
dx =2π
a
0
(a − x)(a +3x)
6a
= π (a 2 +2ax − 3x 2 ) dx = π
a 2 x + ax 2 − x 3 a
3a 0
3a
= π 0
3a (a3 + a 3 − a 3 )= π 3a · a3 = πa2
3 .
Note that we have rotated the top half of the loop about the x-axis. This generates the full surface.
(b) We must rotate the full loop about the y-axis, so we get double the area obtained by rotating the top half of the loop:
a
a
S =2· 2π xds=4π x √ a +3x dx =
4π a
x=0
0 12ax 2 √ x 1/2 (a +3x) dx =
2π a
√ (ax 1/2 +3x 3/2 ) dx
3a 0
3a 0
= √ 2π 2
3a 3 ax3/2 + 6 a
5 x5/2 = 2π √
3 2
0
3 √ a 3 a5/2 + 6
5 a5/2 = 2π √
3 2
3 3 + 6 a
5
2 = 2π √
3 28
a 2
3 15
= 56π √ 3 a 2
45
29. (a) x2
a + y2
y (dy/dx)
=1 ⇒ = − x ⇒ dy
2 b2 b 2 a 2 dx = − b2 x
⇒
a 2 y
2 dy
1+ =1+ b4 x 2
dx a 4 y = b4 x 2 + a 4 y 2
= b4 x 2 + a 4 b 2 1 − x 2 /a 2
2 a 4 y 2 a 4 b 2 (1 − x 2 /a 2 )
= a4 + b 2 x 2 − a 2 x 2
= a4 − a 2 − b 2 x 2
a 4 − a 2 x 2 a 2 (a 2 − x 2 )
dx
= a4 b 2 + b 4 x 2 − a 2 b 2 x 2
a 4 b 2 − a 2 b 2 x 2
The ellipsoid’s surface area is twice the area generated by rotating the first-quadrant portion of the ellipse about the x-axis.
Thus,
S =2
a
0
2πy
= 4πb
a 2 a
√
=
0
1+
a 2 −b 2
2 dy
a
b a4 − (a
dx =4π a2 − x 2 − b 2 )x 2
dx
0 a
2 a √ dx = 4πb
a 2 − x 2 a 2
a4 − u 2
du
√
a2 − b 2 u = √ a 2 − b 2 x 30
=
√ √
4πb a a2 − b 2
a 2√ a4 − a
a 2 − b 2 2
2 (a 2 − b 2 )+ a4 a2 − b 2
2 sin−1 a
a
4πb u
a 2√ a4 − u
a 2 − b 2 2
2 + a4
⎡
=2π⎢
⎣ b2 +
0
a4 − (a 2 − b 2 )x 2 dx
2 sin−1 u
a 2 a
a 2 b sin −1 √
a2 − b 2
a √
a2 − b 2
0
√
a 2 −b 2
⎤
⎥
⎦