37. Area of Surface of Revolution

Dated: 30-10-2024

Surface Area Problem

Let us have a continuous function¹ \(f(x)\) defined over the interval² \([a, b]\)

If we revolve it about x-axis then we get a solid as

If we divide this solid into slices, we will get a collection of frustums which look like this

The formula for finding surface area of a frustum is

\[S = \pi (r_1 + r_2) \cdot l\]

Where \(l\) is the line segment (the arc length) on the \(f(x)\).
Therefore, for any arbitrary frustum, if

\[r_1 = f(x_{i - 1})\]

\[r_2 = f(x_{i})\]

Using the pythagorus theorem and mean value theorem³

\[l=\sqrt{(\Delta x)^2+\left[f(x_k)-f(x_{k-1})\right]^2} = \sqrt{1+\left[f'(x_i^*)\right]^2}\Delta x_i\]

then

\[S_i=\pi\left(f(x_{i-1})+f(x_i)\right)\sqrt{1+\left(f'(x_i^*)\right)^2}\Delta x_i\]

Then using the intermediate value theorem,⁴

\[\frac{1}{2}\left(f(x_{i-1})+f(x_i)\right)=f(x_i^{**})\]

Here \(x_i^{**}\) is a value existing within \([x_{i - 1}, x_i]\)
Hence, we can rewrite our equation to

\[S_i=2\pi f(x_i^{**})\sqrt{1+\left(f'(x_i^*)\right)^2}\Delta x_i\]

Adding all the frustums up, we get

\[\sum_{i=1}^{n}S_i=\sum_{i=1}^{n}2\pi f(x_i^{**})\sqrt{1+\left(f'(x_i^*)\right)^2}\Delta x_i\]

Then to make our approximations better,

\[S = \lim_{\max(\Delta x_i\to 0)}\sum_{i=1}^{n}2\pi f(x_i^{**})\sqrt{1+\left(f'(x_i^*)\right)^2}\Delta x_i\]

Which of course can be written in integral⁵ form.

\[S = \int_{a}^{b}2\pi f(x)\sqrt{1+\left(f'(x)\right)^2}dx\]

Example

Find the surface area of the portion of the sphere generated from \(y = \sqrt{1 - x^2}\) bounded within \(0 \le x \le \frac 1 2\).

Solution

Using our equation

\[S = \int_{a}^{b}2\pi f(x)\sqrt{1+\left(f'(x)\right)^2}dx\]

\[= \int_0^{\frac 1 2} 2 \pi \sqrt{1 - x^2} \cdot \sqrt{1 + \left(\frac{d}{dx} \sqrt{1 - x^2}\right)^2} dx\]

After simplifying the integrand,⁵ we get

\[\int_0^{\frac 1 2} 2 \pi dx = 2 x \pi \bigg]_0^{\frac 1 2} = \pi\]

37. Area of Surface of Revolution

Surface Area Problem

Example

Solution

References