Section 9.4: Area of a Surface of Revolution
Consider a continuous function f on the interval [a, b]. Revolving the curve y = f (x),
a ≤ x ≤ b about the x- or y-axis produces a surface known as a surface of revolution. A
general formula for the area of such a surface is
Z
SA = 2πrdL,
where L denotes the arc length function and r is the distance from the curve to the axis of
revolution (the radius).
There are two cases to consider.
1. Revolving about the x-axis.
(a) If the curve y = f (x), a ≤ x ≤ b is revolved about the x-axis, then the area of
the resulting surface is given by
Z b
p
SA = 2π
f (x) 1 + [f 0 (x)]2 dx.
a
(b) If the curve x = g(y), c ≤ y ≤ d is revolved about the x-axis, then the area of the
resulting surface is given by
Z d p
SA = 2π
y 1 + [g 0 (y)]2 dy.
c
(c) If the curve defined by x = x(t), y = y(t), α ≤ t ≤ β is revolved about the x-axis,
then the area of the resulting surface is given by
s 2
Z β
2
dx
dy
SA = 2π
y(t)
+
dt.
dt
dt
α
2. Revolving about the y-axis.
(a) If the curve y = f (x), a ≤ x ≤ b is revolved about the y-axis, then the area of
the resulting surface is given by
Z b p
SA = 2π
x 1 + [f 0 (x)]2 dx.
a
(b) If the curve x = g(y), c ≤ y ≤ d is revolved about the y-axis, then the area of the
resulting surface is given by
Z d
p
SA = 2π
g(y) 1 + [g 0 (y)]2 dy.
c
(c) If the curve defined by x = x(t), y = y(t), α ≤ t ≤ β is revolved about the y-axis,
then the area of the resulting surface is given by
s 2
Z β
2
dx
dy
SA = 2π
x(t)
+
dt.
dt
dt
α
Example: Find the area of the surface obtained by revolving the curve y = x3 , 0 ≤ x ≤ 2
about the x-axis.
Since y 0 = 3x2 , the surface area is
Z
2
SA = 2π
√
x3 1 + 9x4 dx.
0
Let u = 1 + 9x4 , du = 36x3 dx. Then
Z
π 145 √
udu
SA =
18 1
145
π 2 3/2 =
u
18 3
1
√
π
=
(145 145 − 1).
27
1
Example: Find the area of the surface obtained by revolving the curve x = (y 2 + 2)3/2 ,
3
1 ≤ y ≤ 2 about the x-axis.
Since x0 = y
p
y 2 + 2, the surface area is
Z
SA = 2π
=
=
=
=
=
=
2
y
p
1 + y 2 (y 2 + 2)dy
Z1 2 p
y y 4 + 2y 2 + 1dy
2π
1
Z 2 p
y (y 2 + 1)2 dy
2π
Z1 2
2π
y(y 2 + 1)dy
Z1 2
2π
(y 3 + y)dy
1
2
1 4 1 2 2π
y + y 4
2
1
21π
.
2
Example: Find the area of the surface obtained by revolving the parametric curve defined
by x(t) = 3t − t3 , y(t) = 3t2 , 0 ≤ t ≤ 1 about the x-axis.
Since
dx
dy
= 3 − 3t2 and
= 6t, the surface area is
dt
dt
Z 1 p
3t2 (3 − 3t2 )2 + (6t)2 dt
SA = 2π
Z0 1 √
t2 9 − 18t2 + 9t4 + 36t2 dt
= 6π
Z0 1 √
= 6π
t2 9 + 18t2 + 9t4 dt
Z0 1 p
= 6π
t2 9(1 + t2 )2 dt
Z0 1
t2 (1 + t2 )dt
= 18π
Z0 1
(t2 + t4 )dt
= 18π
0
1
1 3 1 5 t + t = 18π
3
5
0
48π
=
.
5
Example: Find the area of the surface obtained by revolving the curve y = 1 − x2 , 0 ≤ x ≤ 1
about the y-axis.
Since y 0 = −2x, the surface area is
Z
SA = 2π
1
√
x 1 + 4x2 dx.
0
Let u = 1 + 4x2 , du = 8xdx. Then
Z
π 5√
SA =
udu
4 1
5
π 2 3/2 =
u
4 3
1
π √
=
(5 5 − 1).
6
Example: Find the area of the surface obtained by revolving the curve x =
0 ≤ y ≤ 1 about the y-axis.
p
2y − y 2 ,
1−y
Since x0 = p
, the surface area is
2y − y 2
Z
1
SA = 2π
v
u
u
p
2
2y − y t1 +
0
Z
s
1
p
2y − y 2
= 2π
1+
0
Z
1
= 2π
p
2y − y 2
Z0 1
= 2π
r
1−y
p
2y − y 2
!2
dy
1 − 2y + y 2
dy
2y − y 2
1
dy
2y − y 2
1dy
0
= 2π.
Example: Find the area of the surface obtained by revolving the parametric curve defined
by x(t) = et − t, y(t) = 4et/2 , 0 ≤ t ≤ 1 about the y-axis.
Since
dy
dx
= et − 1 and
= 2et/2 , the surface area is
dt
dt
Z 1
q
SA = 2π
(et − t) (et − 1)2 + (2et/2 )2 dt
Z0 1
√
= 2π
(et − t) e2t − 2et + 1 + 4et dt
Z0 1
√
= 2π
(et − t) e2t + 2et + 1dt
Z0 1
p
= 2π
(et − t) (et + 1)2 dt
Z0 1
= 2π
(et − t)(et + 1)dt
Z0 1
= 2π
e2t + et − tet − tdt
0
Z 1
1 2t
1 2
t
t
te dt − t .
= 2π e + e −
2
2
0
Let u = t and dv = et dt. Then du = dt and v = et . So
1
1 2t
1 2 t
t
t
SA = 2π e + e − te + e − t 2
2
0
2
= πe + 2πe − 6π.
Example: (Gabriel’s Horn)
Consider revolving the region under the graph of y =
1
, 1 ≤ x ≤ ∞ about the x-axis.
x
(a) Find the volume of the resulting solid.
Using the Disk Method,
Z
V
∞
= π
1
1
dx
x2
Z N
1
dx
x2
1
N !
1 = π lim − N →∞
x 1
1
= π lim 1 −
N →∞
N
= π.
= π lim
N →∞
(b) Find the surface area of the solid.
Since y 0 = −
1
, the surface area is
x2
r
1
1
2π
1 + 4 dx
x
x
Z1 ∞
1
2π
dx
x
1
Z N
1
2π lim
dx
N →∞ 1
x
N
2π lim ln |x|
Z
SA =
≥
=
=
∞
N →∞
1
= 2π lim ln N
N →∞
= ∞.
The resulting solid has finite volume, but infinite surface area. Thus, the horn can be filled
with a finite amount of paint but the surface can never be completely covered!