= Solution
The enclosed volume lies above $z=0$ and below $z=4-x^2-y^2$. In <cylindrical coordinates>, its limits are $0\le r\le2$, $0\le\theta\le2\pi$, $0\le z\le4-r^2$, with volume element $r\,dz\,dr\,d\theta$. The <divergence theorem> gives
$$
\oiint\mathbf F\cdot d\mathbf S=2\int_0^{2\pi}\int_0^2\int_0^{4-r^2}r\,dz\,dr\,d\theta=4\pi\int_0^2(4r-r^3)\,dr.
$$
Thus the volume is $8\pi$, and the outward <flux> is $\boxed{16\pi}$.
A direct check separates the boundary into its cap and base. On the base, $\mathbf F\cdot(-\hat{\mathbf z})=0$. On the graph $z=4-x^2-y^2$, the upward vector area element is $(2x,2y,1)\,dx\,dy$, so the cap flux is $\iint_{x^2+y^2\le4}2(x^2+y^2)\,dx\,dy=16\pi$, in agreement with the volume calculation.
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