= Solution
Let $D$ be the disk $x^2+y^2\le3$. The upper boundary is the graph $z=3-x^2-y^2$, with outward vector <vector surface element> $(2x,2y,1)\,dx\,dy$. Thus its <flux> is
$$
\iint_D(2x^2+2y^2)\,dx\,dy
=\int_0^{2\pi}\int_0^{\sqrt3}2r^3\,dr\,d\theta=9\pi.
$$
The lower disk has outward <normal vector> $-\mathbf e_z$ and zero <flux>, since $F_z=0$. The total outward <surface integral> is therefore $9\pi$.
Independently, $\nabla\cdot\mathbf F=2$ and the enclosed <volume> is
$$
V=\int_0^{2\pi}\int_0^{\sqrt3}(3-r^2)r\,dr\,d\theta=\frac{9\pi}{2}.
$$
Consequently $\iiint_V\nabla\cdot\mathbf F\,dV=2V=\boxed{9\pi}$, agreeing with the <divergence theorem>. Both the orientation of the cap and inclusion of the closing disk have been accounted for.
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