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 $a\leq x\leq b$ , we divide $[a,b]$ into subintervals with equal width $\delta x$ and endpoints $x_{0},x_{1},\ldots ,x_{n}$ . 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 $a\leq x\leq b$ is defined to be

$A=\lim _{n\to \infty }\sum _{i=0}^{n}A_{i}$

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

$A=2\pi \int _{a}^{b}r(x){\sqrt {1+f'(x)^{2}}}\,dx$

And in Leibniz notation

$A=2\pi \int _{a}^{b}r(x){\sqrt {1+\left({\tfrac {dy}{dx}}\right)^{2}}}\,dx$

Proof:

$A$	$=\lim _{n\to \infty }\sum _{i=1}^{n}A_{i}$
	$=\lim _{n\to \infty }\sum _{i=1}^{n}2\pi r_{i}l_{i}$
	$=2\pi \cdot \lim _{n\to \infty }\sum _{i=1}^{n}r_{i}l_{i}$

As $n\to \infty$ and $\Delta x\to 0$, 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

$l_{i}={\sqrt {1+f'(x_{i})^{2}}}$

Therefore

$A$	$=2\pi \cdot \lim _{n\to \infty }\sum _{i=1}^{n}r_{i}l_{i}$
	$=2\pi \cdot \lim _{n\to \infty }\sum _{i=1}^{n}r_{i}{\sqrt {1+f'(x_{i})^{2}}}\Delta x$

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

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

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

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