= Solution
The <divergence> is $\nabla\cdot F=-1+0+1=0$. Take the hemisphere's outward orientation, and close it with the unit disk $D$ at $z=0$, whose outward normal is $-e_z$. The <divergence theorem> then gives $\int_SF\cdot dS+\int_DF\cdot dS=0$. On the disk, $F\cdot(-e_z)=-x^2$, and <polar coordinates> give
$$
\int_Dx^2dA=\int_0^1\int_0^{2\pi}r^2\cos^2\theta\,r\,d\theta\,dr=\frac\pi4.
$$
Consequently the curved hemisphere has <flux>
$$
\boxed{\int_SF\cdot dS=\frac\pi4.}
$$
The reversed hemisphere orientation would reverse the sign.
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