Rolling Disc Point

Very beautiful problem, though I'm curious as to whether it can be done without calc.

Begin with the parametric equation for a cycloid

$x=r(t-\sin(t))$

$y=r(1-\cos(t)$

The length of the curve is then given by

$L=\displaystyle \int_0^{2\pi} \sqrt{(\Bbb{d}x)^2+(\Bbb{d}y)^2} ~\Bbb{d}t$

$L=r\displaystyle \int_0^{2\pi} \sqrt{(1-\cos(t))^2+(\sin(t))^2}~ \Bbb{d}t$

$L=r\displaystyle \int_0^{2\pi} \sqrt{2-2\cos(t)}~ \Bbb{d}t$

$L=2r\displaystyle \int_0^{2\pi} \sqrt{\dfrac{1-\cos(t)}{2}}~ \Bbb{d}t$

$L=2r\displaystyle \int_0^{2\pi} \sin \left( \frac{t}{2} \right)~ \Bbb{d}t$

$L=4r( -\cos\left(\pi\right)+\cos\left(0\right))$

$L=8r$

Here, the radius is 5 and thus the answer is 40.