 Solution

Stokes theorem gives ${\int }_S(\nabla \times \mathbf{F})\cdot \hat{\mathbf{n}}dS={\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{r}$ for a sufficiently smooth vector field and consistently oriented surface and boundary. Choose the hemisphere's outward normal vector; its boundary circle is then traversed counterclockwise viewed from $+z$. Parametrize it by $\mathbf{r}=(2\cos t,2\sin t,0)$, $0\le t\le 2\pi$. On this curve, $\mathbf{F}=(8\sin t,2\cos t,0)$ and $d\mathbf{r}=(-2\sin t,2\cos t,0)dt$. ThereforeAs a check, $\nabla \times \mathbf{F}=-3\hat{\mathbf{z}}$, whose flux is $-3$ times the projected disk area $4\pi$, not $-3$ times the hemisphere's surface area. Reversing orientation gives $+12\pi$.