So the question is
Calculate the flux $F(x,y,z)=(x,y,xz)$ through the surface $$S= \{(x,y,z) \in \mathbb{R}^3: x^2+y^2=9, 0<x<3, 0<y<3, 0<z<9\}$$ away from the z-axis.
I calculated the divergence, which is $2+x$ and used spherical coordinates to get $2+r \cdot \cos \theta$ but I am not still getting the right answer, which should be $\frac{81\pi}{2}$.
Thank you for your answers!
$S$ is the surface of cylinder $x^2+y^2=3^2$ cut by the plane $z=0$ and $z=9$ and contained in the first octant. Note that $S$ is not closed and the normal vector at $(x,y,z)\in S$ is $\frac{(x,y,0)}{3}$. Hence the flux of $F$ through the surface $S$ is equal to $$\iint_S (x,y,xy)\cdot \frac{(x,y,0)}{3} dS=\iint_S 3 dS=3|S|=3\cdot \frac{2\cdot 3\cdot \pi\cdot 9}{4}=\frac{81\pi}{2}.$$