[Section 8.2] Area of a surface of revolution
1. y = x4 , 0 ≤ x ≤ 1 ⇒ dydx = 4x3 (a) about x-axis:
S = Z
2πy ds = Z 1
0
2πx4p
1 + 16x6dx (b) about y-axis:
S = Z
2πx ds = Z 1
0
2πxp
1 + 16x6dx
8. y = cos(2x) , 0 ≤ x ≤ π6 ⇒ dydx = −2 sin(2x)
S = Rπ/6
0 2π cos(2x) q
1 + 4 sin2(2x) dx
= R
√3/2
0 π√
1 + 4u2du by u = sin(2x)
= Rπ/3 0 π√
1 + tan2θsec22θdθ by u = tan θ2
= π2Rπ/3
0 sec3(θ) dθ
= π2[12(sec θ tan θ + ln | sec θ + tan θ|)]π/30
= π4(2√
3 + ln(2 +√
3))
15. x =p
a2− y2, 0 ≤ y ≤ a2 ⇒ dxdy = √−y
a2−y2
S = Ra/2 0 2πp
a2− y2 r
(√−y
a2−y2)2+ 1 dy
= Ra/2 0 2πp
a2− y2q
y2
a2−y2 + 1 dy
= Ra/2 0 2πp
a2− y2√ a
a2−y2dy
= Ra/2 0 2πa dy
= πa2
25. S =R∞
1 2πyp1 + (y0)2dx , y = 1x
S = 2πR∞ 1
1 x
q
1 +x14dx
= 2πR∞ 1
√x4+1 x3 dx
> 2πR∞ 1
√ x4
x3 dx ∀x > 0
= 2πR∞ 1
1
xdx div.
Hence, S diverge.
1
29. (a) xa22 +yb22 = 1 , a > b ⇒ dydx = −2xa2/2yb2 = −ab22xy
1 + (dy
dx)2= 1 + b4x2
a4y2 = 1 + b2x2
a2(a2− x2)= a4− (a2− b2)x2 a2(a2− x2)
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 = 2Ra
0 2πyp1 + [dy/dx]2dx
= 4πRa 0
b a
√a2− x2
√
a4−(a2−b2)x2 a√
a2−x2 dx
= 4πba2
Ra
0 pa4− (a2− b2)x2dx
= 4πba2
R
√a2−b2/a 0
√a4− a4u2√a2
a2−b2 du let a2u =√
a2− b2x
= √4πa2b
a2−b2
R
√a2−b2/a 0
√1 − u2du
= √4πa2b
a2−b2
Rsin−1(√ a2−b2/a)
0 cos2θ dθ let u = sin θ
= √4πa2b
a2−b2
Rsin−1(√ a2−b2/a) 0
1+cos(2θ)
2 dθ
= √4πa2b
a2−b2[θ2+sin(2θ)4 ]|sin−1 (
√
a2 −b2 /a)
0 (remind sin(2θ) = 2 sin θ cos θ)
= √2πa2b
a2−b2[sin−1u + u cos(sin−1u)]|
√
a2 −b2 /a
0 by θ = sin−1u
= √2πa2b
a2−b2[sin−1u + u√ 1 − u2]|
√
a2 −b2 /a 0
= √2πaa22b
−b2[sin−1(
√ a2−b2
a ) +
√ a2−b2
a
q
1 − a2a−b2 2]
= 2π[√a2b
a2−b2 sin−1(
√a2−b2
a ) + b2]
(b) Similarly, change a and b, we get
S = 2π[ ab2
√b2− a2sin−1(
√b2− a2
b ) + a2]
31. S =Rb
a 2π[c − f (x)]p1 + [f0(x)]2dx
36. g(x) = f (x) + c ⇒ g0(x) = f0(x)
Sg = Rb
a2πg(x)p1 + [g0(x)]2dx
= Rb
a2π[f (x) + c]p1 + [f0(x)]2dx
= Rb
a2πf (x)p1 + [f0(x)]2dx + 2πcRb
ap1 + [f0(x)]2dx
= Sf+ 2πcL
2
