Another geometric question that arises naturally is: "What is the surface area of a volume?'' For example, what is the surface area of a sphere? More advanced techniques are required to approach this question in general, but we can compute the areas of some volumes generated by revolution.

As usual, the question is: how might we approximate the surface area?
For a surface obtained by rotating a curve around an axis, we can take
a polygonal approximation to the curve, as in the last section, and
rotate it around the same axis. This gives a surface composed of many
"truncated cones;'' a truncated cone is called a
**frustum** of a cone.
Figure 9.10.1 illustrates this approximation.

So we need to be able to compute the area of a frustum of a cone. Since the frustum can be formed by removing a small cone from the top of a larger one, we can compute the desired area if we know the surface area of a cone. Suppose a right circular cone has base radius $r$ and slant height $h$. If we cut the cone from the vertex to the base circle and flatten it out, we obtain a sector of a circle with radius $h$ and arc length $2\pi r$, as in figure 9.10.2. The angle at the center, in radians, is then $2\pi r/h$, and the area of the cone is equal to the area of the sector of the circle. Let $A$ be the area of the sector; since the area of the entire circle is $\ds \pi h^2$, we have $$ \eqalign{{A\over\pi h^2}&={2\pi r/h\over 2\pi}\cr A &= \pi r h.\cr} $$

Now suppose we have a frustum of a cone with slant height $h$ and radii $\ds r_0$ and $\ds r_1$, as in figure 9.10.3. The area of the entire cone is $\ds \pi r_1(h_0+h)$, and the area of the small cone is $\ds \pi r_0 h_0$; thus, the area of the frustum is $\ds \pi r_1(h_0+h)-\pi r_0 h_0=\pi((r_1-r_0)h_0+r_1h)$. By similar triangles, $${h_0\over r_0}={h_0+h\over r_1}.$$ With a bit of algebra this becomes $\ds (r_1-r_0)h_0= r_0h$; substitution into the area gives $$ \pi((r_1-r_0)h_0+r_1h)=\pi(r_0h+r_1h)=\pi h(r_0+r_1)=2\pi {r_0+r_1\over2} h = 2\pi r h. $$ The final form is particularly easy to remember, with $r$ equal to the average of $\ds r_0$ and $\ds r_1$, as it is also the formula for the area of a cylinder. (Think of a cylinder of radius $r$ and height $h$ as the frustum of a cone of infinite height.)

Now we are ready to approximate the area of a surface of revolution. On one subinterval, the situation is as shown in figure 9.10.4. When the line joining two points on the curve is rotated around the $x$-axis, it forms a frustum of a cone. The area is $$ 2\pi r h= 2\pi {f(x_i)+f(x_{i+1})\over2} \sqrt{1+(f'(t_i))^2}\,\Delta x. $$ Here $\ds \sqrt{1+(f'(t_i))^2}\,\Delta x$ is the length of the line segment, as we found in the previous section. Assuming $f$ is a continuous function, there must be some $\ds x_i^*$ in $\ds [x_i,x_{i+1}]$ such that $\ds (f(x_i)+f(x_{i+1}))/2 = f(x_i^*)$, so the approximation for the surface area is $$\sum_{i=0}^{n-1} 2\pi f(x_i^*)\sqrt{1+(f'(t_i))^2}\,\Delta x.$$ This is not quite the sort of sum we have seen before, as it contains two different values in the interval $\ds [x_i,x_{i+1}]$, namely $\ds x_i^*$ and $\ds t_i$. Nevertheless, using more advanced techniques than we have available here, it turns out that $$\lim_{n\to\infty} \sum_{i=0}^{n-1} 2\pi f(x_i^*)\sqrt{1+(f'(t_i))^2}\,\Delta x= \int_a^b 2\pi f(x)\sqrt{1+(f'(x))^2}\,dx$$ is the surface area we seek. (Roughly speaking, this is because while $\ds x_i^*$ and $\ds t_i$ are distinct values in $\ds[x_i,x_{i+1}]$, they get closer and closer to each other as the length of the interval shrinks.)

Example 9.10.1 We compute the surface area of a sphere of radius $r$. The sphere can be obtained by rotating the graph of $\ds f(x)=\sqrt{r^2 - x^2}$ about the $x$-axis. The derivative $f'$ is $\ds -x/\sqrt{r^2-x^2}$, so the surface area is given by $$\eqalign{ A&=2\pi \int_{-r }^r \sqrt{r^2 - x^2}\sqrt{1+{x^2\over r^2-x^2}}\,dx\cr &=2\pi \int_{-r }^r \sqrt{r^2 - x^2}\sqrt{r^2\over r^2-x^2}\,dx\cr &=2\pi \int_{-r }^r r\,dx=2\pi r\int_{-r }^r 1\,dx=4\pi r^2\cr}$$ $\square$

If the curve is rotated around the $y$ axis, the formula is nearly identical, because the length of the line segment we use to approximate a portion of the curve doesn't change. Instead of the radius $\ds f(x_i^*)$, we use the new radius $\ds \bar x_i= (x_i+x_{i+1})/2$, and the surface area integral becomes $$\int_a^b 2\pi x\sqrt{1+(f'(x))^2}\,dx.$$

Example 9.10.2 Compute the area of the surface formed when $\ds f(x) =x^2$ between $0$ and $2$ is rotated around the $y$-axis.

We compute $f'(x)= 2x$, and then $$2\pi\int_0^2 x\sqrt{1+4x^2}\,dx={\pi\over6}(17^{3/2}-1),$$ by a simple substitution. $\square$

## Exercises 9.10

You can use Sage to help check your work.

**Ex 9.10.1**
Compute the area of the surface formed when $\ds f(x)=2\sqrt{1-x}$
between $-1$ and $0$ is rotated around the $x$-axis.
(answer)

**Ex 9.10.2**
Compute the surface area of example 9.10.2 by rotating $\ds f(x)=\sqrt x$ around the $x$-axis.

**Ex 9.10.3**
Compute the area of the surface formed when
$\ds f(x)=x^3$ between $1$ and $3$ is rotated around the $x$-axis.
(answer)

**Ex 9.10.4**
Compute the area of the surface formed when
$\ds f(x)=2 +\cosh (x)$ between $0$ and $1$ is rotated around the
$x$-axis.
(answer)

**Ex 9.10.5**
Consider the surface obtained by rotating the graph of $\ds
f(x)=1/x$, $x\geq 1$, around the $x$-axis. This surface is called
**Gabriel's horn** or **Toricelli's
trumpet**.
In exercise 13 in
section 9.7 we saw that Gabriel's horn has
finite volume.
Show that Gabriel's horn has
infinite surface area.

**Ex 9.10.6**
Consider the circle $\ds (x-2)^2+y^2 = 1$. Sketch the
surface obtained by rotating this circle about the $y$-axis. (The
surface is called a **torus**.) What is the surface area?
(answer)

**Ex 9.10.7**
Consider the ellipse with equation $\ds x^2/4+y^2 = 1$.
If the ellipse is rotated around the $x$-axis it forms
an **ellipsoid**.
Compute the surface area.
(answer)

**Ex 9.10.8**
Generalize the preceding result: rotate the ellipse
given by $\ds x^2/a^2+y^2/b^2=1$ about the
$x$-axis and find the surface area of the resulting ellipsoid. You
should consider two cases, when $a>b$ and when $a< b$. Compare to the
area of a sphere.
(answer)