Solution
= Solution
Close the surface with the unit disk $D$ in the plane $z=0$. Its outward normal is $-e_z$, but $F\cdot(-e_z)=-z^4=0$ there. Also
$$
\nabla\cdot F=2x-2y+4z^3.
$$
The $x$ and $y$ terms integrate to zero by symmetry. The <divergence theorem> therefore gives
$$
\begin{aligned}
\int_SF\cdot dS
&=\int_0^{2\pi}\int_0^1\int_0^{1-r^2}
4z^3r\,dz\,dr\,d\theta\\
&=2\pi\int_0^1r(1-r^2)^4\,dr
=\boxed{\frac\pi5}.
\end{aligned}
$$