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 $$x_0,x_1,\ldots,x_n$$. 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
 * $$A=2\pi\int_a^b r(x) \sqrt{1 + \left(\tfrac{dy}{dx}\right)^2}\,dx$$

Proof:




 * $$A$$
 * $$=\lim_{n\to\infty}\sum_{i=1}^n A_i$$
 * $$=\lim_{n\to\infty}\sum_{i=1}^n 2\pi r_il_i$$
 * $$=2\pi\cdot\lim_{n\to\infty}\sum_{i=1}^n r_il_i$$
 * }
 * $$=2\pi\cdot\lim_{n\to\infty}\sum_{i=1}^n r_il_i$$
 * }
 * $$=2\pi\cdot\lim_{n\to\infty}\sum_{i=1}^n r_il_i$$
 * }

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


 * $$A$$
 * $$=2\pi\cdot\lim_{n\to\infty}\sum_{i=1}^n r_il_i$$
 * $$=2\pi\cdot\lim_{n\to\infty}\sum_{i=1}^n r_i\sqrt{1+f'(x_i)^2}\Delta x$$
 * }
 * $$=2\pi\cdot\lim_{n\to\infty}\sum_{i=1}^n r_i\sqrt{1+f'(x_i)^2}\Delta x$$
 * }

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 $$[c,d]$$


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