Find the volume of a revolved cycloid

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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.

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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 $$ V=V_2-V_1=-\pi \int_{0}^{2\pi}a^3(t - \sin t)^2 \sin t\, dt. $$

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$\pi\int_{a}^{b}f(x)^2\,dx$ 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 $a=1$:

$\hspace{2cm}$enter image description here

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 $$ V=2\pi(\pi a)(3\pi a^2) = \color{red}{6\pi^3 a^3}.$$