Example 3. Calculate the area of the part of the surface of the paraboloid of revolution $2z = x^2 + y^2$, enclosed within the cylinder $x^2 + y^2 = R^2$.
Solution. We apply the formula $S=\iint d \sigma$, where $\sigma-$ is the surface of the paraboloid $z=\frac{x^{2}+y^{2}}{2}$ above the circle $x^{2}+y^{2} \leqslant R^{2}$, and we will perform the calculations using a different scheme, unlike in Example 2. We transition to the double integral $S=\iint_{x^{2}+y^{2} \leqslant R^{2}} \frac{d x d y}{\cos \gamma}$, where $\cos \gamma=\frac{1}{\sqrt{1+{z_{x}^{\prime}}^{2}+{z_{y}^{\prime}}^{2}}}$. We have $z_{x}^{\prime}=x, z_{y}^{\prime}=y, \frac{1}{\cos \gamma}=\sqrt{1+x^{2}+y^{2}} \cdot$ Therefore,

$$
\begin{aligned}
& S=\iint_{x^{2}+y^{2} \leqslant R^{2}} \sqrt{1+x^{2}+y^{2}} d x d y=\int_{0}^{2 \pi} d \varphi \int_{0}^{R} \sqrt{1+r^{2}} r d r= \\
&=\frac{2 \pi}{3}\left(\sqrt{\left(1+R^{2}\right)^{3}}-1\right)
\end{aligned}
$$

Answer. $S=\frac{2 \pi}{3}\left(\sqrt{\left(1+R^{2}\right)^{3}}-1\right)$.