Claim: the flux of $\mathbf G$ over the unit disk $\mathscr D = \{(x, y, z) \,|\, x^2 + y^2 \leq 1 \text{ and } z = 0\}$ is zero.
Proof. Using the usual polar coordinates $x = r \cos \theta$ and $y = r \sin \theta,$ we parametrize the unit disk $\mathscr D$ by $D(r, \theta) = \langle r \cos \theta, r \sin \theta, 0 \rangle$ over the domain $U = \{(r, \theta) \,|\, 0 \leq r \leq 1 \text{ and } 0 \leq \theta \leq 2 \pi \}.$ We have therefore that $D_r(r, \theta) = \langle \cos \theta, \sin \theta, 0 \rangle$ and $D_\theta(r, \theta) = \langle -r \sin \theta, r \cos \theta, 0 \rangle,$ from which it follows that $(D_r \times D_\theta)(r, \theta) = \langle 0, 0, r \rangle.$ We conclude that $\mathbf G(D(r, \theta)) \cdot (D_r \times D_\theta)(r, \theta) = \langle r \cos \theta, r \sin \theta, 0 \rangle \cdot \langle 0, 0, r \rangle = 0$ so that $\iint_{\mathscr D} \mathbf G \cdot d \mathbf S = \iint_U \mathbf G(D(r, \theta)) \cdot (D_r \times D_\theta)(r, \theta) \, dr \, d \theta = 0.$ QED.
Considering that flux is additive, the flux you seek is equal to the sum of the flux on the unit disk $\mathscr D$ and the flux on the top of the northern hemisphere, i.e., the flux on $\mathscr S = \{(x, y, z) \,|\, x^2 + y^2 + z^2 = 1 \text{ and } z \geq 0 \}.$ Can you compute this flux and finish the problem? (Hint: use spherical coordinates to parametrize $\mathscr S.$)
Best Answer
Use the change of variables:
$$(x,y,x) \rightarrow (x,y-1,z)$$
Note also that $\nabla \cdot F = 0$ so by the Divergence Theorem the surface integral is $0$.