Arc Length and Surface Area

Determining the length of an irregular arc segment is also called rectification of a curve. Historically, many methods have been used for specific curves. The advent of infinitesimal calculus led to a general formula, which we will learn in this atom. We will also use integration to calculate the surface area of a three-dimensional object.

Arc Length

Consider a real function $f(x)$ such that $f(x)$ and $f'(x)=\frac{dy}{dx}$ (its derivative with respect to $x$) are continuous on $[a, b]$. The length $s$ of the part of the graph of $f$ between $x = a$ and $x = b$ can be found as follows.

Consider an infinitesimal part of the curve $ds$ (or consider this as a limit in which the change in $s$ approaches $ds$). According to Pythagoras's theorem $ds^2=dx^2+dy^2$, from which:

 $\displaystyle{\frac{ds^2}{dx^2}=1+\frac{dy^2}{dx^2} \\ ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx \\ s = \int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }\, dx}$

Approximating Deltas

For a small piece of curve, $\Delta s$ can be approximated with the Pythagorean theorem.

If a curve is defined parametrically by $x = X(t)$ and y = Y(t), then its arc length between $t = a$ and $t = b$ is:

 $\displaystyle{s = \int_{a}^{b} \sqrt { [X'(t)]^2 + [Y'(t)]^2 }\, dt}$

This is more clearly a consequence of the distance formula, where instead of a $\Delta x$ and $\Delta y$, we take the limit. A useful mnemonic is:

$\displaystyle{s = \int_{a}^{b} \sqrt { dx^2 + dy^2 } = \int_{a}^{b} \sqrt { \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 }\,dt}$

Surface Area

For rotations around the $x$- and $y$-axes, surface areas $A_x$ and $A_y$ are given, respectively, as the following:

$\displaystyle{A_x = \int 2\pi y \, ds , \,\, ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx \\ \\ A_y = \int 2\pi x \, ds, \,\, ds=\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy }$

Example 

For a circle $f(x) = \sqrt{1 -x^2}, 0 \leq x \leq 1$, calculate the arc length.

The curve can be represented parametrically as $x=\sin(t), y=\cos(t)$ for $0 \leq t \leq \frac{\pi}{2}$. Therefore:

$\displaystyle{s = \int_0^{\frac{\pi}{2}}\sqrt{\cos^2(t)+\sin^2(t)} = \frac{\pi}{2}}$

Now, calculate the surface area of the solid obtained by rotating $f(x)$ around the $x$-axis:

$\displaystyle{A_x = \int_{0}^{1} 2\pi \sqrt{1-x^2}\cdot \sqrt{1+\left(\frac{-x}{\sqrt{1-x^2}}\right)^2} \, dx = 2\pi}$