Surface area by integration

In this lesson we compute the areas of surfaces of revolution, surfaces com-mon in physical and biological applications. In any serious application, most ofthe 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, as-suming f(x) is positive for all x in the domain of f , the surface is obtained byrevolving the graph of y = f(x) about the x-axis.

x

y

z

y = f(x)

Figure 1: The surface obtained by revolving the graph y = f(x) about thex-axis

A surface of revolution is made up of a family of circles, for each x a circle ofradius f(x). Guided by our calculation of the length of a curve, we approximateportions of the curve by straight line segments. As illustrated in Fig. 2, revolvingeach 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 aswidths of the frustra go to 0.

From geometry recall that the surface area of the cone with base radius rand height h is

πr√

r2 + h2 = πrL

1

x

y

z

y = f(x)

a b

Figure 2: Calculating surface area with frustra

where L is called the slant height of the cone. See the left side of Fig. 3. Thenthe area of the frustrum on the right side of Fig. 3 is the difference of the areasof 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′

L + L′=

r

L

r′L = r(L + L′)

(r′ − r)L = rL′

Now rewrite the frustrum area as

2πr + r′

2L′

In the limit as r → r′, the average of r and r′ approaches r′ = f(x), the lengthL′ approaches the arclength

√

1 + (f ′(x))2dx, and the sum of the areas of thefrustra approaches the integral

area =

∫ b

a

2πf(x)√

1 + (f ′(x))2dx (1)

2

Lh

r

A

B C

D Er’

L’

r

L

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, thesurface 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

area =

∫ r

−r

2π√

r2 − x2

√

1 +x2

r2 − x2dx

=

∫ r

−r

2π√

r2 − x2r

√r2 − x2

dx

= 2π

∫ r

−r

rdx = 4πr2

We have recovered the result familiar from geometry.

2. Next, we compute the area of the surface obtained by revolving the curvey = x3, 0 ≤ x ≤ 1, about the x-axis.

Applying Eq. (1) we find

area =

∫

1

0

2πx3√

1 + (3x2)2dx

=π

18

∫ 10

1

√udu substituting u = 1 + 9x4

=π

27(103/2 − 1)

3