The area of a frustum

In order to perform the approximation step, we first need to discuss the surface area of a frustrum.

A frustum of a cone is a section of a cone bounded by two planes, where both planes are perpendicular to the height of the cone.

To compute the area of a surface of revolution, we approximate that this area is equal to the sum of areas of basic shapes that we can lay out flat. The argument for this goes way back to the great physicist and mathematician, Archimedes of Alexandria. To follow his argument, we have to begin by computing the area of a ‘lamp shade’ or frustum.

And of course, few things are more interesting than the area of a frustum:

The surface area of the frustum; meaning the lateral sides, not the top nor the bottom;

is given by $$\text {area of frustum} = 2\pi \cdot \frac {r+R}{2}\cdot s.$$

We can think of a frustum as approximated by an arrangement of $n$ congruent trapezoids:

To proceed, we make the following conventions:

• $n$ denote the number of trapezoids,
• $t_{n}$ denote the length of the top of each trapezoid,
• $h_{n}$ denote the height of each trapezoid,
• $b_{n}$ denote the length of the bottom of each trapezoid,

then from geometry, we have that each of the trapezoids, one of which is shown below:

has area: $$\left ( \frac {t_{n}+b_n}{2}\right )\cdot h_n.$$ The area of the frustum approximated by the sum of these $n$ trapezoids, so the area of the frustum is approximated by $$A_n = \answer [given]{n}\cdot \left (\frac {t_{n}+b_n}{2}\right )\cdot h_n =\left (\frac {n\cdot t_{n}+n\cdot b_n}{2}\right )\cdot h_n.$$ As $n$ goes to infinity, the congruent trapezoids “smooth” out:

Hence, $$\lim _{n\to \infty } \left (\left (\frac {n\cdot t_{n}+n\cdot b_n}{2}\right )\cdot h_n\right ) = \text {area of the frustum.}$$ However, if

• $c$ is the circumference of the top circle,
• $s$ is the slant height of the frustum as shown in the above figure, and
• $C$ is the circumference of the bottom circle,

then

$$\begin{align*} \lim_{n\to \infty} n \cdot t_{n} & = \answer[given]{c},\\ \lim_{n\to \infty} h_{n} & = \answer[given]{s},\\ \lim_{n\to \infty} n\cdot b_{n} & = \answer[given]{C}, \end{align*}$$ and by way of limit laws we find $$\lim _{n\to \infty } \left (\left (\frac {n\cdot t_{n}+n\cdot b_n}{2}\right )\cdot h_n\right ) = \answer [given]{\frac {c + C}{2}}\cdot s.$$ Now, letting

• $r$ be the radius of the circle defining the top of the frustum,
• $s$ be the slant height of the frustum, and
• $R$ be the radius of the circle defining the base of the frustum,

we see that:

$$\begin{align*} \text{area of frustum} &= \answer[given]{\frac{c + C}{2}\cdot s} \\ &= \frac{\answer[given]{2\pi r + 2\pi R}}{2}\cdot s\\ &=2\pi \cdot \frac{r+R}{2}\cdot s. \end{align*}$$

The area of a surface of revolution

Let’s consider a function $f$ with a continuous derivative, and form a surface of revolution formed by this curve by rotating the portion of the curve from $x=a$ to $x=b$ about the $x$-axis:

We can find a formula that gives the surface area of this surface of revolution using the procedure of “Slice, Approximate, Integrate”!

Step 1: Slice Since we have the curve to be revolved expressed as a function of $x$, we choose to slice with respect to $x$:

Step 2: Approximate We have seen how to find the surface area of a frustrum, so we should thus approximate each slice as a frustrum.

Thus the surface area, $\Delta SA$ of this frustum is: $$\Delta SA = 2\pi \frac {r_1+r_2}{2} \Delta s$$ Note that there is a value $r$ between $r_1$ and $r_2$ such that $\frac {r_1+r_2}{2} = \frac {2r}{2} = r$, so we write:

$$\Delta SA = 2\pi \frac {r_1+r_2}{2} \Delta s$$

and can find the total approximate surface area by using $n$ frustra by adding together all of the surface areas:

$$SA = \sum _{k=1}^n \Delta SA = \sum _{k=1}^n 2\pi r \Delta s$$

Step 3: Integrate The formula above has good conceptual meaning, it does not readily pass to an integral quite yet! and have seen that we can express $\Delta s$ free in terms of either $\Delta x$ or $\Delta y$, which allows us to express the infinitesimal $ds$ by:

$$ds = \sqrt {1+\left (\frac {dy}{dx} \right )^2} dx \qquad \textrm { or } ds = \sqrt {1+\left (\frac {dx}{dy} \right )^2} dy$$

Note also that as the slice widths shrink, the value $r$ above approaches the distance that the corresponding slice is away from the axis of rotation.

To make sure that we emphasize this freedom in expressing $ds$ as well as the inherent geometric results we used to build the surface area, we write:

Using the remark, and letting $A=(a,c)$ and $B=(b,d)$ we can therefore write:

An important concept to note is that the slice is located at a point $(x,y)$ on the curve. The choice of variable of integration may require that we express either $x$ or $y$ in terms of the other by using the equation that describes the curve. We will see this in the following examples.

Consider the surface of revolution found by revolving $f(x) = \sqrt {x}$ from $x=1$ to $x=4$ around the $x$-axis. We set up an integral with respect to $x$ and an integral with respect to $y$ that gives its surface area.

We begin by considering looking at a picture,

As a brief aside, note that this slice gives rise to the following frustrum when revolved about the $x$-axis:

Let’s first set up the integral with respect to $x$. In order to do this, we choose:

$ds = \sqrt {1+\left (\frac {dy}{dx} \right )^2} dx$ $ds = \sqrt {1+\left (\frac {dx}{dy} \right )^2} dy$

Note that $r$ here is a vertical distance that must be expressed in terms of $x$. Since the slice is located at a point $(x,y)$ on the curve $y=\sqrt {x}$:

$y_{top}=y$ $y_{top}=\sqrt {x}$ $y_{top}=x$ $y_{top}=0$

$y_{bot}=y$ $y_{bot}=\sqrt {x}$ $y_{bot}=x$ $y_{bot}=0$

Thus, $r= \answer [given]{\sqrt {x}}$.

Now, write with me: $$\frac {dy}{dx} = \answer [given]{\frac {x^{-1/2}}{2}}$$ So

$$\begin{align*} SA &= \int_{x=\answer{1}}^{x=\answer{4}} 2\pi \sqrt{x} \sqrt{1 + \left(\answer[given]{\frac{x^{-1/2}}{2}}\right)^2} dx\\ &= \int_1^4 2\pi \sqrt{x} \sqrt{1 + \frac{1}{4x}} dx\\ &= \int_1^4 2\pi \sqrt{x+\frac{1}{4}} dx\\ &= \bigg[\answer[given]{\frac{4\pi}{3}\left(x+\frac{1}{4}\right)^{3/2}}\bigg]_1^4\\ &= \frac{4\pi}{3}\left(\frac{17}{4}\right)^{3/2} - \frac{4\pi}{3}\left(\frac{5}{4}\right)^{3/2} \\ &= \frac{\pi}{6} \left((17)^{3/2}-(5)^{3/2} \right) \textrm{after some algebra} \end{align*}$$

As our final example, we will compute the surface area of the sphere.

Final thoughts

The key formulas in this section are:

$$SA=\int _{x=a}^{x=b} 2 \pi r ds \qquad \textrm { or } \qquad SA=\int _{y=c}^{y=d} 2 \pi r ds$$

We are free to choose the variable of integration here since we can express $ds$ in terms of either $dx$ or $dy$ easily. Once this choice of variable has been determined, we need to express the radius $r$ of the infinitesimal frustrum and the limits for the integral in terms of the variable of integration.

This radius $r$ is the distance from the axis of rotation to the slice at $(x,y)$, which is either a horizontal or vertical distance. We just need to make sure that we express it in terms of the variable of integration appropriately.

Many of the integrals that arise in the context of these problems can be difficult. Careful differentiation and algebra, as well as a good grasp of integration techniques can be vital when finding surface areas. As usual, this can be challenging and practice is the key here.