Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 8.2 AREA OF A SURFACE OF REVOLUTION ¤ 359
8.2 Area of a Surface of Revolution
1. y = x 4 ⇒ dy/dx =4x 3 ⇒ ds = 1+(dy/dx) 2 dx = √ 1+16x 6 dx
(a) By (7), an integral for the area of the surface obtained by rotating the curve about the x-axis is
S = 2πy ds = 1
0 2πx4 √ 1+16x 6 dx.
(b) By (8), an integral for the area of the surface obtained by rotating the curve about the y-axis is
S = 2πx ds = 1
0 2πx √ 1+16x 6 dx.
3. y =tan −1 x ⇒ dy
dx = 1
1+x 2 ⇒ ds =
(a) By (7), S = 2πy ds = 1
0 2π tan−1 x
1+
1+
2
dy
1
dx = 1+
dx
(1 + x 2 ) dx. 2
1
(1 + x 2 ) 2 dx.
(b) By (8), S = 2πx ds = 1
2πx 1
1+
0
(1 + x 2 ) 2 dx.
5. y = x 3 ⇒ y 0 =3x 2 .So
S = 2
2πy 1+(y
0 0 ) 2 dx =2π 2
√
0 x3 1+9x 4 dx [u =1+9x 4 , du =36x 3 dx]
= 2π
36
145 √
1 udu=
π
18
7. y = √ 1+4x ⇒ y 0 = 1 2 (1 + 4x)−1/2 (4) =
145 √
2
3 u3/2 = π 27 145 145 − 1
1
2
√ 1+4x
⇒ 1+(y 0 ) 2 =
S = 5
2πy 1+(y
1 0 ) 2 dx =2π
5 √ 5+4x
1 1+4x
1+4x dx =2π 5 √
1 4x +5dx
=2π 25 √
9 u 1 du 4
u =4x +5,
du =4dx
= 2π 4
9. y =sinπx ⇒ y 0 = π cos πx ⇒ 1+(y 0 ) 2 =1+π 2 cos 2 (πx). So
S = 1
2πy 1+(y
0 0 ) 2 dx =2π 1
sin πx
1+π
0 2 cos 2 (πx) dx
1+ 4
1+4x =
5+4x
1+4x .So
25
2
3 u3/2 = π 3 (253/2 − 9 3/2 )= π (125 − 27) = 98 π
3 3
9
u = π cos πx,
du = −π 2 sin πx dx
−π
=2π 1+u
2
− 1
π
π du = 2 π
1+u2 du
2 π −π
= 4 π
1+u2 du 21 = 4 u
1+u2 + 1 2
π 0
π 2 ln u + π
1+u 2 0
= 4 π √
1+π2 + 1 2
π 2
ln π + √ 1+π 2 − 0
=2 √ 1+π 2 + 2 π ln π + √ 1+π 2
11. x = 1 3 (y2 +2) 3/2 ⇒ dx/dy = 1 2 (y2 +2) 1/2 (2y) =y y 2 +2 ⇒ 1+(dx/dy) 2 =1+y 2 (y 2 +2)=(y 2 +1) 2 .
So S =2π 2
1 y(y2 +1)dy =2π 1
4 y4 + 1 y2 2
=2π 4+2− 1 − 1
2 1 4 2 =
21π
. 2
13. y = 3√ x ⇒ x = y 3 ⇒ 1+(dx/dy) 2 =1+9y 4 .So
S =2π 2
x 1+(dx/dy)
1 2 dy =2π 2
y3
1+9y
1 4 dy = 2π 2
36 1
√ √
145 145 − 10 10
= π 27
1+9y4 36y 3 dy = π 18
2
3
1+9y
4 3/2 2
1