Let
$$\mathbf{F}(x,y,z) = \left({-x\,{\bf i}-3\,x\,y^2\,{\bf j}+z\,{\bf k}}\right).$$
so then the $\nabla\cdot\mathbf{F}= -6yx$
then
$$V = \left\{(x,y,z)\mid 0\le z\le 2xy,~ x^2+y^2\le {1}\right\},$$
I want to evaluate the surface integral:
$$\displaystyle\iint_{S}\left(\mathbf{F}\cdot\mathbf{n}\right)dS$$
I know that $S=\delta V$
So parametrization is
$$x = r\cos(\theta)$$ $$y=r\sin(\theta)$$ $$z = z$$
and then $0 \le \theta \le2\pi$ and $0\le z \le 2xy$ and $r=1$
But how do I calculate the integral?
We can use divergence theorem, and we have
$$\displaystyle\iint_{S}\left(\mathbf{F}\cdot\mathbf{n}\right)dS)=\iiint_V \nabla\mathbf{F}\,dV$$
with $\nabla\mathbf{F}= -6yx$.
By cylindrical coordinates the set up should be
$$\iiint_V \nabla\mathbf{F}\,dV=\int_0^{2\pi}d\theta\int_0^1dr\int_0^{2r^2\cos \theta \sin \theta}-6r^3\cos\theta \sin \theta \,dz$$