= Solution
Take the outward orientation on the upper <hemisphere>; its boundary <circle> is then counterclockwise viewed from above. Parametrize that <circle> by $r(\phi)=(r\cos\phi,r\sin\phi,0)$, $0\leq\phi\leq2\pi$. Then $G\cdot dr=r^2(3\cos^2\phi-4\sin^2\phi)d\phi$, so <Stokes theorem> gives
$$
\boxed{\int_S(\nabla\times G)\cdot dS=\int_0^{2\pi}r^2(3\cos^2\phi-4\sin^2\phi)d\phi=-\pi r^2.}
$$
Equivalently $\nabla\times G=(0,0,-1)$, and the upward projected area of the <hemisphere> is $\pi r^2$. The opposite orientation gives the negative of this answer.
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