2.13. Find the mass of the cylindrical surface $x^{2}+$ $+y^{2}=R^{2}$, bounded by the planes $z=0, z=H$, if at each point the surface density is inversely proportional to the square of the distance from it to the origin.
Solution. According to the problem, the surface density $p=\frac{k}{r^{2}}(k=$ const $), r^{2}=x^{2}+y^{2}+z^{2}$. The surface of the cylinder projects onto a rectangle $y \in[-R, R], z \in[0, H]$ on the YOZ plane. The desired mass can be expressed as a surface integral of the first kind:

$$
M=\iint_{\sigma} \frac{k d S}{x^{2}+y^{2}+z^{2}}
$$

The element of area of the cylindrical surface $x= \pm \sqrt{R^{2}-y^{2}}$ is

$$
d S=\sqrt{1+x_{y}^{2}+x_{z}^{2}} d y d z=\frac{R}{\sqrt{R^{2}-y^{2}}} d y d z
$$

We compute the surface integral using a double integral over the rectangle $|y| \leqslant R, 0 \leqslant z \leqslant H$ on the YOZ plane (considering that $x^{2}=R^{2}-y^{2}$):

$$
\begin{gathered}
\iint_{\sigma} \frac{k d S}{x^{2}+y^{2}+z^{2}}=2 \int_{-R}^{R} d y \int_{0}^{H} \frac{k R d z}{\left(R^{2}+z^{2}\right) \cdot \sqrt{R^{2}-y^{2}}}= \\
=2 k R \int_{-R}^{R} \frac{d y}{\sqrt{R^{2}-y^{2}}} \int_{0}^{H} \frac{d z}{R^{2}+z^{2}}= \\
=2 k\left(\left.\arcsin \frac{y}{R}\right|_{-R} ^{R}\right) \cdot\left(\left.\operatorname{arctg} \frac{z}{R}\right|_{0} ^{H}\right)=2 k \pi \operatorname{arctg} \frac{H}{R}
\end{gathered}
$$