Length of Arc of Cycloid

Theorem
Let $$C$$ be a cycloid generated by the equations:
 * $$x = a \left({\theta - \sin \theta}\right)$$
 * $$y = a \left({1 - \cos \theta}\right)$$

Then the length of one arc of the cycloid (i.e. where $$0 \le \theta \le 2 \pi$$) is $$8a$$.

Proof
Let $$L$$ be the length of one arc of the cycloid. Then:


 * $$L = \int_0^{2\pi} \sqrt {\left({\frac{\mathrm{d}{x}}{\mathrm{d}{\theta}}}\right)^2 + \left({\frac{\mathrm{d}{y}}{\mathrm{d}{\theta}}}\right)^2} \mathrm{d} \theta$$

where, from the definition of the cycloid:
 * $$x = a \left({\theta - \sin \theta}\right)$$
 * $$y = a \left({1 - \cos \theta}\right)$$

we have:

$$ $$

Thus:

$$ $$ $$ $$

Thus:

$$ $$ $$

So $$L = 8a$$ where $$a$$ is the radius of the generating circle.