I understand how to compute the area under the whole cycloid. But how do you compute a partial area. Let say, we want to know the area of the intersection between a cycloid generated by a circle of radius $r$ that stretches from $-r\pi$ to $+r\pi$ and a square of side 1 that stretches from $-1/2$ to $+1/2$. With: $1/(2\pi) < r < 1/2$
I tried using the parametric equation the same way I did to compute the whole cycloid area. But I am stuck when I switch from an integral over x to an integral over t as t run from and to an angle than depends on $r$.
How do you proceed to compute the area ?
Let the square have sidelength $2a$ in order to make things dimensionally correct. Parametrize the cycloid by $$\gamma:\quad x(t)=r(t+\sin t), \quad y(t)=r(1+\cos t)\qquad(-\pi\leq t\leq\pi)\ .$$ The vertical sides of the square intersect $\gamma$ in two points $P$, $Q$. The area $S$ in question is then bounded by four arcs, among them a part $\gamma'$ of $\gamma$ which belongs to a certain subinterval $-t_0\leq t\leq t_0$. The value $t_0\in\>]0,\pi[\>$ has to be determined numerically from the equation $$x(t)=r(t+\sin t)=a\ .$$ In order to compute the area of $S$ we use Green's theorem $${\rm area}(S)=-\int_{\partial S} y\>dx=\int_{\gamma'}y\>dx=\int_{-t_0}^{t_0}y(t)\,\dot x(t)\>dt\ .$$ Here we have made use of the fact that $y\,dx=0$ along three of the four arcs constituting $\partial S$, and that $\gamma'$ has the wrong sense of direction in $\partial S$.