Example 2. Calculate the area of the part of the paraboloid of revolution $3 y=x^{2}+z^{2}$, which is located in the first octant and bounded by the plane $y=6$.
Solution. Apply the formula $S=\int_{\sigma} \int d \sigma$. It is convenient to project $\sigma$ onto the plane $O x z$, in which case the surface $\sigma$ is explicitly given by the equation $y=\frac{1}{3}\left(x^{2}+z^{2}\right)$, and the projection of $\sigma$ onto the plane $O x z$ is a quarter circle $x^{2}+z^{2} \leqslant 18(x \geqslant 0, z \geqslant 0)$. Then
$y_{x}^{\prime}=\frac{2}{3} x, y_{z}^{\prime}=\frac{2}{3} z, d \sigma=\sqrt{1+{y^{\prime}}_{x}^{2}+{y^{\prime}}_{z}^{2}} d y d z=\sqrt{1+\frac{4}{9}\left(x^{2}+z^{2}\right)} d x d z$. The area $S$ of the surface $\sigma$ is (recall that $x \geqslant 0, z \geqslant 0$ )

$$
S=\iint_{x^{2}+z^{2} \leqslant 18} \sqrt{1+\frac{4}{9}\left(x^{2}+z^{2}\right)} d x d z
$$

We switch to generalized polar coordinates, setting $x=\frac{3}{2} r \cos \varphi, y=\frac{3}{2} r \sin \varphi$. Then $d x d y=\frac{9}{4} r d r d \varphi$, and the circle $x^{2}+z^{2} \leqslant 18$ transforms into the circle $\frac{9}{4} r^{2}\left(\cos ^{2} \varphi+\sin ^{2} \varphi\right) \leqslant 18$, i.e., $r \leqslant \sqrt{8}$. Our integral takes the form (in the integral $0 \leqslant \varphi \leqslant \frac{\pi}{2}$ )

$$
\begin{aligned}
S=\iint_{r \leqslant \sqrt{8}} \frac{9}{4} \sqrt{1+r^{2}} r d r d \varphi & =\frac{9}{4} \int_{0}^{\pi / 2} d \varphi \int_{0}^{\sqrt{8}} \sqrt{1+r^{2}} r d r= \\
& =\left.\frac{9 \cdot \pi}{4 \cdot 2} \cdot \frac{1}{3}\left(1+r^{2}\right)^{3 / 2}\right|_{0} ^{\sqrt{8}}=\frac{3 \pi\left[9^{3 / 2}-1\right]}{8}=\frac{39 \pi}{4}
\end{aligned}
$$

Answer. $\frac{39 \pi}{4}$.