May 22, 2014

The method for finding an arc length by integrating comes from extending the Pythagorean theorem to infinitesimal quantities. What this gives you is a differential equation, which, when solved and integrated, results in the total arc length.

The key to understanding why arc length by integration works is realizing that you aren't just integrating the curve itself, you're integrating a different curve whose area represents the arc length of the other curve. Integration, in its most fundamental form, is the summation of an infinite number of infinitesimal quantities. This summation can be interpreted to mean area under the curve, at least for the curve which you're integrating; but it can also be interpreted to mean something else, depending on its context.

It's hard for me to explain what I mean, but I'll give you an example.

Let's say we have a simple curve, just a line, defined by $y = 2 x$.

I won't go into how to derive it here, but the arc length by integration formula is:

$l = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Since the derivative $\frac{\mathrm{dy}}{\mathrm{dx}} = 2$, the integral is rather easy to solve.

Note, however, that the curve which we'll end up integrating isn't the one we started with, it's actually just a horizontal line:

$l = {\int}_{a}^{b} \sqrt{1 + {\left(2\right)}^{2}} \mathrm{dx} = {\int}_{a}^{b} \sqrt{5} \mathrm{dx}$

Let's say we go from $a = 0$ to $b = 2$. The integral evaluates to $2 \sqrt{5}$.

Note that since our original function is a line, we can use the Pythagorean theorem to verify our answer:

$l = \sqrt{{4}^{2} + {2}^{2}} = \sqrt{20} = 2 \sqrt{5}$

So, ultimately, what we're doing here is integrating a function whose area represents the arc length of a different function. Because the arc length formula is essentially the Pythagorean theorem extended to infinitesimals, we're allowed to use integrals via the Fundamental Theorem of Calculus as an algebraic operation, in order to "cancel" the differentials. This works in much the same way that when you're solving a linear equation, you add or multiply to "cancel" terms and solve for the other variable.

Hope that helps.