Example 7. Calculate the moment of inertia about the $O y$ axis of the hemisphere $x^{2}+y^{2}+z^{2}=R^{2}, y \geqslant 0$.
Solution. We will use the formula $I_{y}=\iint\left(x^{2}+z^{2}\right) d \sigma$, where $\sigma$ can be explicitly defined as: $y=\sqrt{R^{2}-x^{2}-z^{2}}$. In this case, $d \sigma=\sqrt{1+y_{x}^{\prime 2}+{y_{z}^{\prime}}^{\prime 2}} d y d z=\frac{R}{\sqrt{R^{2}-x^{2}-y^{2}}} d y d z$, and $I_{y}=$ $=\iint_{x^{2}+z^{2} \leqslant R^{2}} \frac{x^{2}+z^{2}}{\sqrt{R^{2}-x^{2}-y^{2}}} d y d z$. We switch to polar coordinates:

\[
\begin{aligned}
& I_{y}=\iint_{r \leqslant R} \frac{r^{3} d r d \varphi}{\sqrt{R^{2}-r^{2}}}=\int_{0}^{2 \pi} d \varphi \int_{0}^{R} \frac{r^{3} d r}{\sqrt{R^{2}-r^{2}}}= \\
& =\left\{\begin{aligned}
r^{2} & =u, & & d u=2 r d r \\
v & =-\sqrt{R^{2}-r^{2}}, & & d v=\frac{v}{\sqrt{R^{2}-r^{2}}} d r
\end{aligned}\right\}= \\
& =4 \pi \int_{0}^{R} r \sqrt{R^{2}-r^{2}} d r=\frac{4 \pi}{3} R^{3}
\end{aligned}
\]

Answer. $\frac{4}{3} \pi R^{3}$.