831. Find the mass of a circular ring if at each of its points the surface density is inversely proportional to the square of the distance from it to the center of the ring.
Solution. Let the radii of the circles bounding the ring be denoted by $r_{1}$ and $r_{2}\left(r_{1}<r_{2}\right)$, and place the pole of the polar coordinate system at the center of the ring; then the equations of the circles will be $\rho=r_{1}$ and $\rho=r_{2}$, and the surface density at point $M(\varphi, \rho)$ of the ring $\delta(M)=: \frac{k}{\rho^{2}}$.

The mass of the entire ring can be found using formula (1), transforming it into polar coordinates:

$$
\begin{aligned}
& m=\iint_{D} \delta \rho d \varphi d \rho= \iint_{r_{1} \leqslant r_{2}} \frac{k}{\rho} d \varphi d \rho=k \int_{0}^{2 \pi} d \varphi \int_{r_{1}}^{r_{2}} \frac{d \rho}{\rho}=\left.k \int_{0}^{2 \pi} \ln \rho\right|_{r_{1}} ^{r_{2}} d \varphi= \\
&=k \ln \frac{r_{2}}{r_{1}} \int_{0}^{2 \pi} d \varphi=2 k \pi \ln \frac{r_{2}}{r_{1}} .
\end{aligned}
$$