Find the volume of a revolved cycloid I have to find the volume of the area surrounded by the first arc of the cycloid given by $x(t) = a(t - \sin t), y(t) = a(1 - \cos t)$ when revolved around the $y$-axis. I know that:
$$ V = \pi \int\limits_{a}^{b}[f(x)]^2dx$$
The problem here is that I do not know which bounds I need to take to integrate.
I did the following:
$$ V = \pi \int\limits_{0}^{2\pi a}[x(t)]^2dy(t) = \pi \int\limits_{0}^{2\pi a}a^3(t - \sin t)^2 cos t dt$$
but this yields the wrong result. I chose the bounds this way because  when $t = 2\pi a$, then $y = 0$ again. Can someone provide any help?
EDIT: I would like to know a solution using the formula I listed above.
 A: $\pi\int_{a}^{b}f(x)^2\,dx$ is the volume of the region generated by revolving the graph of $f(x)$ around the $x$-axis. Here you have a parametric curve and a different axis. The first arc of the given cycloid is given by $t\in[0,2\pi]$; here it is its graph for $a=1$:
$\hspace{2cm}$
The area enclosed by the $x$-axis and such arc is $3\pi a^2=\int_{0}^{2\pi}a^2(1-\cos t)^2\,dt$ by well-known formulas. The centroid of such region lies at $(x_G,y_G)$ with $x_G=\pi a$ by symmetry. By Pappus' centroid theorem, by revolving such region around the $y$-axis we get an object whose volume is
$$ V=2\pi(\pi a)(3\pi a^2) = \color{red}{6\pi^3 a^3}.$$ 
A: Your formula works only if a line parallel to the $x$-axis intersects the curve at a single point. But that is not the case here, because for every $y$ between $0$ and $a$ there are two intersections.
The integral
$$ 
V_1 = \pi \int_{0}^{2a}x_{int}^2 dy
= \pi \int_{0}^{\pi}a^3(t - \sin t)^2 \sin t\, dt
$$
gives the volume of the space comprised between the $y$ axis and the rotated cycloid (notice that $dy=a\sin t\,dt$). 
The integral
$$ 
V_2 = \pi \int_{0}^{2a}x_{ext}^2 dy
= -\pi \int_{\pi}^{2\pi}a^3(t - \sin t)^2 \sin t\, dt
$$
gives the volume of the rotated outer half-cycloid. The volume of the rotated cycloid is then
$$
V=V_2-V_1=-\pi \int_{0}^{2\pi}a^3(t - \sin t)^2 \sin t\, dt.
$$
