Solution. Given the symmetries of the integration region $\sigma$: $x^{2}+\frac{y^{2}}{4} \leqslant 1$, the equation of the surface, and the density function, it is sufficient to compute the integral over one quarter of the region and multiply the result by 4: $m=4 \int_{D} \rho(x, y, z) d \sigma$, where $D$ is the quarter of the region $\sigma$ lying in the first quadrant of the $O x y$ plane. Further, we have:

$$
\begin{gathered}
z=x y, z_{x}^{\prime}=y, z_{y}^{\prime}=x, \sqrt{1+{z_{x}^{\prime}}^{2}+z_{y}^{\prime 2}}=\sqrt{1+x^{2}+y^{2}} \\
\rho d \sigma=\frac{|z| \sqrt{1+x^{2}+y^{2}}}{\sqrt{1+x^{2}+y^{2}}} d x d y=x y d x d y
\end{gathered}
$$

$$
\begin{array}{ll}
m=4 \cdot \int_{\substack{x^{2}+y^{2} / 4 \leqslant 1 \\
x \geqslant 0, y \geqslant 0}} x y d x d y=4 \int_{0}^{1} x d x \int_{0}^{2 \sqrt{1-x^{2}}} y d y= & \left.2 \int_{0}^{1} x d x \cdot y^{2}\right|_{0} ^{2 \sqrt{1-x^{2}}}= \\
& =8 \int_{0}^{1}\left(1-x^{2}\right) x d x=2 .
\end{array}
$$

Answer. $m=2$.
Example 3. Calculate the mass of the surface $z=x y$, located inside the cylinder $x^{2}+\frac{y^{2}}{4}=1$, if the density is $\rho=\frac{|z|}{\sqrt{1+x^{2}+y^{2}}}$.