Surface area by integration
In this lesson we compute the areas of surfaces of revolution, surfaces common in physical and biological applications. In any serious application, most of
the hard work goes into finding an approximate representation of these surfaces.
For now, we assume that work has been done.
Fig. 1 illustrates the construction of a surface of revolution. For now, assuming f (x) is positive for all x in the domain of f , the surface is obtained by
revolving the graph of y = f (x) about the x-axis.
y
y = f(x)
x
z
Figure 1: The surface obtained by revolving the graph y = f (x) about the
x-axis
A surface of revolution is made up of a family of circles, for each x a circle of
radius f (x). Guided by our calculation of the length of a curve, we approximate
portions of the curve by straight line segments. As illustrated in Fig. 2, revolving
each of these segments about the x-axis gives a frustrum, a portion of a cone.
So we need to find the area of these frustra, add them up, and take the limit as
widths of the frustra go to 0.
From geometry recall that the surface area of the cone with base radius r
and height h is
p
πr r2 + h2 = πrL
1
y
y = f(x)
a b
x
z
Figure 2: Calculating surface area with frustra
where L is called the slant height of the cone. See the left side of Fig. 3. Then
the area of the frustrum on the right side of Fig. 3 is the difference of the areas
of the large and small cones:
frustrum area = πr′ (L + L′ ) − πrL
= π(r′ − r)L + πr′ L′
= πrL′ + πr′ L′ = π(r + r′ )L′
The last line follows from the similarity of the triangles ∆ABC and ∆ADE.
Corresponding sides have the same ratios:
r′
r
=
L + L′
L
r′ L = r(L + L′ )
(r′ − r)L = rL′
Now rewrite the frustrum area as
2π
r + r′ ′
L
2
In the limit as r → r′ , the average
of r and r′ approaches r′ = f (x), the length
p
′
L approaches the arclength 1 + (f ′ (x))2 dx, and the sum of the areas of the
frustra approaches the integral
area =
Z
a
b
p
2πf (x) 1 + (f ′ (x))2 dx
2
(1)
A
L
h
B r
L
C
L’
r
D
r’
E
Figure 3: The area of a cone, the area of a frustrum
Examples
1. To illustrate this process, we compute the√area of a sphere of radius r, the
surface obtained by revolving the curve y = r2 − x2 , −r ≤ x ≤ r, about the
x-axis.
√
√
For this surface, f (x) = r2 − x2 . Then f ′ (x) = x/ r2 − x2 and so the area is
r
Z r
p
x2
2
2
2π r − x 1 + 2
area =
dx
r − x2
−r
Z r
p
r
dx
=
2π r2 − x2 √
2
r − x2
−r
Z r
rdx = 4πr2
= 2π
−r
We have recovered the result familiar from geometry.
2. Next, we compute the area of the surface obtained by revolving the curve
y = x3 , 0 ≤ x ≤ 1, about the x-axis.
Applying Eq. (1) we find
area =
Z
1
2πx3
0
p
1 + (3x2 )2 dx
Z 10
√
π
=
udu substituting u = 1 + 9x4
18 1
π
(103/2 − 1)
=
27
3