= Solution
The <divergence theorem> states $\oint_{\partial V}\mathbf F\cdot\hat{\mathbf n}\,dS=\int_V\nabla\cdot\mathbf F\,dV$ with outward orientation. For $\mathbf F=(x,y,0)$, its <divergence> is $1+1=2$. The enclosed region has cylindrical bounds $0\leq r\leq2$, $0\leq z\leq4-r^2$, $0\leq\phi\leq2\pi$, so its volume is
$$
V=2\pi\int_0^2(4-r^2)r\,dr=8\pi.
$$
Therefore
$$
\boxed{\oint_{\partial V}\mathbf F\cdot d\mathbf S=2V=16\pi.}
$$
The planar base contributes zero because its normal is vertical and $F_z=0$; the curved part accounts for the entire outward flux.
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