Given the curve described by
$x(t)=a(t-sin(t))$ and $y(t)=a(1-cos(t))$ for $t$ in the interval $[0,2\pi]$
How do I find the volume of a Solid $S$ result of making the area between the curve and $y=0$ rotate around the x axis
So far I have just gotten that the curve is similar to the positive side of an elipse but don't know exactly how to describe the curve in cartesian coordinates or if there is a way to integrate the given curve as it is
This can be done with a simple change of variables.
$$dx=a(1-\cos t)\,dt$$
$$V=\int_{x(0)}^{x(2\pi)}\pi y^2\,dx=\int_0^{2\pi}\pi(a(1-\cos t))^2a(1-\cos t)\,dt=\pi a^3\int_0^{2\pi}(1-\cos t)^3\,dt$$
Do you think you can take it from here?