 Solution

The enclosed volume lies above $z=0$ and below $z=4-x^2-y^2$. In cylindrical coordinates, its limits are $0\le r\le 2$, $0\le \theta \le 2\pi$, $0\le z\le 4-r^2$, with volume element $rdzdrd\theta$. The divergence theorem givesThus 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 $(2x,2y,1)dxdy$, so the cap flux is ${\iint }_{x^2+y^2\le 4}2(x^2+y^2)dxdy=16\pi$, in agreement with the volume calculation.