Use Stokes's Theorem to evaluate $\iint_S ( ∇ × F ) ⋅ d S$ on the given surface 

By Stokes's Theorem 
\[ \iint_{S} ​(\nabla \times F)⋅dS=\int_{C}F \cdot dr,\]
where $C$ is a circle of radius $\sqrt{3}$ on $xz$-plane with positive orientation with respect to the surface $S$.  Indeed we can parametrize $C$ by $C=(\sqrt{3}\cos \theta, 0, \sqrt{3} \sin \theta)$, $-2\pi \leq \theta \leq 0$. Hence

\[\int_{C}F \cdot dr=\int_C (-x)^3dx+z^3dz=\int_{-2\pi}^{0}-(\sqrt{3})^3 \cos ^3 \frac{dx}{d \theta}+(\sqrt{3})^3 \sin^3 \theta \frac{dz}{d \theta}   d \theta\]
\[=(\sqrt{3})^4\int_{-2\pi}^{0}\cos ^3 \sin \theta +\sin ^3 \theta \cos \theta     d \theta\]
\[=9 \int_{-2\pi}^{0} \sin \theta \cos \theta (\sin^2 \theta +\cos ^2 \theta) d \theta\]
\[=9 \int_{-2\pi}^{0} \sin \theta \cos \theta  d \theta=\frac{9}{2} \int_{-2\pi}^{0} \sin 2\theta d \theta\]
\[\frac{9}{4} (-\cos 2 \theta)|_{-2 \pi}^{0}=\frac{9}{4} (-1+1)=0.\]