1.20. Find the area of the part of the surface of the paraboloid $z=x^{2}-y^{2}$, bounded by the paraboloids $z=3 x^{2}+$ $+y^{2}-2$ and $z=3 x^{2}+y^{2}-4$.
Solution. To calculate the area, we also use the formula (1.83)

$$
\mu(S)=\iint_{\Omega_{X O Y}} \sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y
$$

where $\Omega_{X O Y}$ is the projection of the part of the hyperbolic paraboloid onto the $X O Y$ plane, determined by the conditions:

$$
\begin{aligned}
& x^{2}-y^{2}=3 x^{2}+y^{2}-4 \\
& x^{2}-y^{2}=3 x^{2}+y^{2}-2
\end{aligned}
$$

Thus, we obtain $\Omega_{\text {XoY }}=\left\{(x, y): 1 \leqslant x^{2}+y^{2} \leqslant 2\right\}$.

$$
\mu(S)=\iint_{\Omega_{X O Y}} \sqrt{1+4 x^{2}+4 y^{2}} d x d y
$$

To compute the double integral, we switch to polar coordinates $x=\rho \cos \varphi ; y=\rho \sin \varphi$:

where

$$
\mu(S)=\iint_{D} \sqrt{1+4 \rho^{2}} \rho d \rho d \varphi,
$$

$$
\begin{aligned}
& D=\{(\rho, \varphi): \rho \in[1, \sqrt{2}] ; \varphi \in[0,2 \pi]\} ; \\
& \mu(S)=2 \pi \int_{1}^{\sqrt{2}} \sqrt{1+4 r^{2}} r d r=\left.\frac{\pi}{6}\left(1+4 r^{2}\right)^{\frac{3}{2}}\right|_{1} ^{\sqrt{2}}=\frac{\pi}{6}(27-5 \sqrt{5}) .
\end{aligned}
$$