$S$ is the upper surface of hemisphere, $S=\{(x,y,z)\mid x^2+y^2+z^2=1, z>0\}$. There is a vector field $\mathbf{F}=(u,v,w)=(\frac{x^3}{3}+z^2,x^2+y^2-\frac{z^3}{3},x^2 y+\frac{y^3}{3})$. Find the value of surface integral $$\int_{S}(wy-vz)\ \mathrm{d}S$$ My first instinct is substitute $w$ and $v$ in and calculate it with polar coordinates. That is \begin{equation} \int_{S}\left(x^2y^2+\frac{y^4}{3}-(x^2+y^2)z+\frac{z^4}{3}\right)\ \mathrm{d}S \end{equation} It's a bit annoying but is OK to get the answer.
I noticed the surface $S$ has the normal vector $\mathbf{n}=(x,y,z)(z>0)$ at every point on $S$. So I tried to write the integral in the form: $$\int_{S}(\frac{1}{2}xy^2-xz,\frac{1}{2}x^2 y+\frac{y^3}{3}-yz,\frac{z^3}{3})\cdot\mathbf{n}\ \mathrm{d}S=\int_{S}\mathbf{G}\cdot\mathbf{n}\ \mathrm{d}S$$ Add another surface $S_{0}=\{x^2+y^2\leq1,z=0\}$ i.e. the bottom of $S$, so $S\cup S_{0}$ is closed surface and can apply Gauss divergence theorem on it. $$=\int_{V}\text{div}\mathrm{G}\ \mathrm{d}V$$
The question I have is, the $u$ seems no use with the question, can we have a more simple way to get the result without huge calculations?