= Solution
For a bounded volume $\Omega$ with piecewise smooth boundary and a continuously differentiable vector field on a neighborhood of its closure, the <divergence theorem> states
$$
\oint_{\partial\Omega}\mathbf F\cdot\mathbf n\,dS=\int_\Omega\nabla\cdot\mathbf F\,dV,
$$
where $\mathbf n$ is the outward unit normal.
The boundary here consists of a paraboloid cap over the unit disk, a cylindrical wall of height one, and the bottom disk. Their joined surface encloses
$$
\Omega=\{(x,y,z):x^2+y^2\leq1,\ 0\leq z\leq3-2(x^2+y^2)\}.
$$
The cap meets the wall at $r=1,z=1$, and the wall meets the base at $r=1,z=0$. The sketches show each boundary piece and their closed union.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-3-closed-surface.png]
{title=Paraboloid cap, cylindrical wall, base disk, and their closed union}
Differentiate the field components to find the <divergence>:
$$
\nabla\cdot\mathbf F=(y+6x^5)+(-y+8y^7)+1=1+6x^5+8y^7.
$$
The volume is invariant under reflecting $x$ or $y$, so <odd-integrand cancellation by reflection> removes both odd-power contributions. The outward flux equals the volume:
$$
\begin{aligned}
\oint_S\mathbf F\cdot d\mathbf S
&=\int_0^{2\pi}\int_0^1\int_0^{3-2r^2}r\,dz\,dr\,d\theta\\
&=2\pi\int_0^1(3-2r^2)r\,dr
=\boxed{2\pi}.
\end{aligned}
$$
The corners between the three pieces are permitted by the piecewise smooth boundary hypothesis.
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