If the length of the cycloid is $4$ times the diameter of a rotating circle, then the area under the arch traced out by that cycloid is how many times the area of the rotating circle?
I tried using the parametric equations for a cycloid: $x = a(t-\sin{t})$, and $y = a(1-\cos{t})$.
The answer is three. Can someone please explain this? I am preferably looking for an answer that does not involve calculus.
To use the parametric equation:
Render the area as $\int y dx$ between the appropriate limits, which are clearly from $t=0$ to $t=2\pi$. Plug in the parametric equations to get
$a^2\int_0^{2\pi}(1-\cos t)d(t-\sin t)$
To get an integral with just $dt$ differentiate the $t-\sin t$ function and combine the derivative with the preexisting factor $1-\cos t$. Thus
$a^2\int_0^{2\pi}(1-\cos t)^2dt$
This may be simplified by putting in $(1-\cos t)^2=1-2\cos t+\cos^2t$ and then $\cos^2 t=(1+\cos 2t)/2$. Then the cosine integrate to zero over full periods, allowing those terms to be canceled out and leaving just a constant ($3/2$) remaining in the inte-grand. Thus the final answer $3\pi a^2$ where the parametric equations had defined $a$ as the radius.