= Solution
Let $a=\sinh1$ and $c=\cosh1$. The region is the first-octant portion of the <ellipsoid> $(x^2+y^2)/a^2+z^2/c^2\leq1$. In <cylindrical coordinates>, write $\rho=\sqrt{x^2+y^2}$ and $0\leq\phi\leq\pi/2$. Away from the axis, the <divergence> is
$$
\nabla\cdot\mathbf F=\frac{2x+2y}{x^2+y^2}+\frac1\rho=\frac{2\cos\phi+2\sin\phi+1}{\rho}.
$$
The field is singular on the axis, so applying the classical <divergence theorem> there without qualification would be invalid. Instead remove $\rho<\varepsilon$ and apply it to the remaining region. On the added cylindrical boundary the <unit normal> is $-\mathbf e_\rho$, and
$$
\mathbf F\cdot(-\mathbf e_\rho)=-\log(\varepsilon^2)(\cos\phi+\sin\phi).
$$
Its area element is $\varepsilon\,d\phi\,dz$, so its <flux> is $O(\varepsilon|\log\varepsilon|)\to0$. The axial component of the field has zero <dot product> with this normal. The <volume integral> of the <divergence> is locally integrable, because its $1/\rho$ factor cancels the cylindrical measure. Thus <removing a singular axis in a flux integral> gives the desired improper <surface integral> with no extra axial term.
The <divergence theorem> now reduces the calculation to
$$
\begin{aligned}
I&=\int_0^{\pi/2}(2\cos\phi+2\sin\phi+1)\,d\phi\int_0^c\int_0^{a\sqrt{1-z^2/c^2}}d\rho\,dz\\
&=\left(4+\frac\pi2\right)ac\int_0^1\sqrt{1-s^2}\,ds\\
&=\left(4+\frac\pi2\right)\frac{\pi ac}{4}
=\boxed{\frac{\pi(8+\pi)}8\sinh1\cosh1=\frac{\pi(8+\pi)}{16}\sinh2.}
\end{aligned}
$$
The integral over $s$ is the area of a quarter unit disk. All boundary <fluxes> are interpreted as convergent <improper integrals> at the axis.
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