= Solution
Parametrize the upward-oriented paraboloid by
$$
X(r,\theta)=(r\cos\theta,r\sin\theta,1-r^2),
\quad0\leq r\leq1,\quad0\leq\theta\leq2\pi.
$$
Then
$$
X_r\times X_\theta=(2r^2\cos\theta,2r^2\sin\theta,r).
$$
On the surface,
$$
F=(r^2\cos^2\theta,-r^2\sin^2\theta,(1-r^2)^4).
$$
The cubic trigonometric terms integrate to zero over $0\leq\theta\leq2\pi$, so the <surface integral> is
$$
\begin{aligned}
\int_SF\cdot dS
&=2\pi\int_0^1r(1-r^2)^4\,dr\\
&=\boxed{\frac\pi5}.
\end{aligned}
$$
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