= Solution
Close the <hyperboloid> segment with the two disks at $z=\pm1$, and let $V$ be its interior. The <flux> through either disk is zero because $A_z=0$. Thus the desired <surface integral> is the total outward <flux> through the closed boundary. Since
$$
\nabla\cdot\mathbf A=3x^2+3y^2,
$$
the <divergence theorem> and <cylindrical coordinates> give
$$
\begin{aligned}
\int_{\widetilde S}\mathbf A\cdot\hat{\mathbf n}\,dS
&=\int_{-1}^1\int_0^{2\pi}\int_0^{\sqrt{1+z^2}}3r^2\,r\,dr\,d\theta\,dz\\
&=\frac{3\pi}{2}\int_{-1}^1(1+z^2)^2\,dz\\
&=\frac{3\pi}{2}\left(2+\frac43+\frac25\right)
=\boxed{\frac{28\pi}{5}}.
\end{aligned}
$$
The extra factor $r$ is the <volume integral> measure in <cylindrical coordinates>; omitting it would give the wrong <flux>.
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