Calculus/Surface area

Suppose we are given a function $f$ and we want to calculate the surface area of the function $f$ rotated around a given line. The calculation of surface area of revolution is related to the arc length calculation.

If the function $f$ is a straight line, other methods such as surface area formulae for cylinders and conical frusta can be used. However, if $f$ is not linear, an integration technique must be used.

Recall the formula for the lateral surface area of a conical frustum:

$A=2\pi rl$

where $r$ is the average radius and $l$ is the slant height of the frustum.

For $y=f(x)$ and $a\le x\le b$ , we divide $[a,b]$ into subintervals with equal width $\delta x$ and endpoints $x_0,x_1,\dots ,x_n$ . We map each point $y_i=f(x_i)$ to a conical frustum of width Δx and lateral surface area $A_i$ .

We can estimate the surface area of revolution with the sum

$A=\sum _{i=0}^n{A}_i$

As we divide $[a,b]$ into smaller and smaller pieces, the estimate gives a better value for the surface area.

Definition (Surface of Revolution)

The surface area of revolution of the curve $y=f(x)$ about a line for $a\le x\le b$ is defined to be

$A=\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=0}^n{A}_i$

The Surface Area Formula

Suppose $f$ is a continuous function on the interval $[a,b]$ and $r(x)$ represents the distance from $f(x)$ to the axis of rotation. Then the lateral surface area of revolution about a line is given by

$A=2\pi \underset{a}{\overset{b}{\int }}r(x)\sqrt{1+f^{\prime }(x{)}^2}dx$

And in Leibniz notation

$A=2\pi \underset{a}{\overset{b}{\int }}r(x)\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$

Proof:

$A$	$=\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^n{A}_i$
	$=\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^{n}2\pi r_i{l}_i$
	$=2\pi \cdot \underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^n{r}_i{l}_i$

As $n\to \mathrm{\infty }$ and $\mathrm{\Delta }x\to 0$, we know two things:

1. the average radius of each conical frustum $r_i$ approaches a single value
2. the slant height of each conical frustum $l_i$ equals an infitesmal segment of arc length

From the arc length formula discussed in the previous section, we know that

$l_i=\sqrt{1+f^{\prime }(x_i{)}^2}$

Therefore

$A$	$=2\pi \cdot \underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^n{r}_i{l}_i$
	$=2\pi \cdot \underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^n{r}_i\sqrt{1+f^{\prime }(x_i{)}^2}\mathrm{\Delta }x$

Because of the definition of an integral $\underset{a}{\overset{b}{\int }}f(x)dx=\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^{n}f(c_i)\mathrm{\Delta }{x}_i$ , we can simplify the sigma operation to an integral.

$A=2\pi \underset{a}{\overset{b}{\int }}r(x)\sqrt{1+f^{\prime }(x{)}^2}dx$

Or if $f$ is in terms of $y$ on the interval $[c,d]$

$A=2\pi \underset{c}{\overset{d}{\int }}r(y)\sqrt{1+f^{\prime }(y{)}^2}dy$