Calculate the volume of a torus I want calculate the volume of the solid obtained by rotating around the $z$-axis $$C=\{(x,y,z) \in \mathbb{R}^3\;:\; (x-R)^2+z^2\le r^2,y=0\}$$ with $0<r<R$.
Can I use polar coordinates?
 A: By Pappus's centroid theorem, the volume of the torus is given by
$2\pi R\cdot \pi r^2$.
The same result can be obtained by using integration and cylindrical coordinates: the torus is generated by the disc $C$ is
$$T=\left\{(x,y,z) \in \mathbb{R}^3\;:\; \left(\sqrt{x^2+y^2}-R\right)^2+z^2\le r^2\right\}$$
and therefore
$$\begin{align}\text{Vol}(T)&=\int_{\theta=0}^{2\pi}\int_{\rho=R-r}^{R+r}\int_{z=-\sqrt{r^2-(\rho-R)^2}}^{\sqrt{r^2-(\rho-R)^2}} dz \rho d\rho  d \theta\\
&=4\pi\int_{\rho=R-r}^{R+r}\sqrt{r^2-(\rho-R)^2}\, \rho d\rho\\
&=4\pi r^2\int_{s=-1}^{1}\sqrt{1-s^2}\, (rs+R) ds\\
&=4\pi r^2R\int_{s=-1}^{1}\sqrt{1-s^2}\,ds=2\pi R \cdot \pi r^2
\end{align}$$
where $s=(\rho-R)/r$ (the integral of $rs$ is zero by symmetry).
A: Using cylindrical coordinates (As $r$ is usually used in this coordinates I will use $S$ instead): by symmetry, $(x-R)^2 + z^2 = S^2$ gives
$$
(r-R)^2 + z^2 = S^2 \implies r = R\pm\sqrt{S^2-z^2},\quad z = \sqrt{S^2-(r-R)^2}.$$
$$
V = \int_0^ {2\pi}\int_{-S}^S\int_{R-\sqrt{S^2-z^2}}^{R+\sqrt{S^2-z^2}}r\,drdzd\theta = \cdots
$$
$$V = \int_0^ {2\pi}\int_{R-S}^{R+S}\int_{-\sqrt{S^2-(r-R)^2}}^{\sqrt{S^2-(r-R)^2}}r\,dzdrd\theta = \cdots
$$
(What is better?)
Alternatively, you can do a cov adapted to the problem:
$$
\eqalign{
r &= R + s\cos\psi\cr
x &= (R + s\cos\psi)\cos\theta\cr
y &= (R + s\cos\psi)\sin\theta\cr
z &= s\sin\psi
}
$$
$$(s,\theta,\psi)\in[0,S]\times[0,2\pi]\times[0,2\pi]$$
$$
\eqalign{J
&=
\left|\matrix{
\cos\psi\cos\theta &     -(R+s\cos\psi)\sin\theta & -s\sin\psi\cos\theta\cr
\cos\psi\sin\theta &\hfill(R+s\cos\psi)\cos\theta & -s\sin\psi\sin\theta\cr
\sin\psi           &           0              &  s\cos\psi
}\right|\cr
&= 
(R+s\cos\psi)s\left|\matrix{
\cos\psi\cos\theta &     -\sin\theta & -\sin\psi\cos\theta\cr
\cos\psi\sin\theta &\hfill\cos\theta & -\sin\psi\sin\theta\cr
\sin\psi           &      0      &  \cos\psi
}\right| = (R+s\cos\psi)s.
}
$$
$$
V =
\int_0^{2\pi}\!\!\int_0^S\!\int_0^{2\pi}(R+s\cos\psi)s\,d\psi ds d\theta =
\int_0^{2\pi}\!\!\!\int_0^S 2\pi Rs\,ds d\theta = \int_0^{2\pi}\pi RS^2\,d\theta = 2\pi^2 R S^2.
$$
