# Calculus/Surface area

 ← Arc length Calculus Work → 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 formulas for cylinders and conical frustra 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:

$A = 2 \pi r l \,$

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 Δx 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 $a\le x\le b$ 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

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

Proof:

 $A \$ = $\lim_{n \to \infty} \sum_{i=1}^n A_i$ = $\lim_{n \to \infty} \sum_{i=1}^n 2 \pi r_i l_i$ = $2 \pi \lim_{n \to \infty} \sum_{i=1}^n r_i l_i$

As $n \rightarrow \infty$ and $\Delta x \rightarrow 0$, 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 \$ = $2 \pi \lim_{n \to \infty} \sum_{i=1}^n r_i l_i$ = $2 \pi \lim_{n \to \infty} \sum_{i=1}^n r_i \sqrt{ 1+ (f'(x_i))^2} \Delta x$

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]

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