# Calculus/Surface area

 ← Arc length Calculus Work → Surface area

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

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

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

${\displaystyle A=2\pi rl}$

where ${\displaystyle r}$ is the average radius and ${\displaystyle l}$ is the slant height of the frustum.

For ${\displaystyle y=f(x)}$ and ${\displaystyle a\leq x\leq b}$ , we divide ${\displaystyle [a,b]}$ into subintervals with equal width ${\displaystyle \delta x}$ and endpoints ${\displaystyle x_{0},x_{1},\ldots ,x_{n}}$ . We map each point ${\displaystyle y_{i}=f(x_{i})}$ to a conical frustum of width Δx and lateral surface area ${\displaystyle A_{i}}$ .

We can estimate the surface area of revolution with the sum

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

As we divide ${\displaystyle [a,b]}$ 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 ${\displaystyle y=f(x)}$ about a line for ${\displaystyle a\leq x\leq b}$ is defined to be

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

## The Surface Area Formula

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

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

And in Leibniz notation

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

Proof:

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

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

1. the average radius of each conical frustum ${\displaystyle r_{i}}$ approaches a single value
1. the slant height of each conical frustum ${\displaystyle l_{i}}$ equals an infitesmal segment of arc length

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

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

Therefore

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

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

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

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

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