The divergence of the vector field [tex]\vec F[/tex] is
[tex]\nabla\cdot\vec F=y^2+0+x^2=x^2+y^2[/tex]
By the divergence theorem,
[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\iiint_V(x^2+y^2)\,\mathrm dV[/tex]
where [tex]V[/tex] denotes the region with boundary [tex]S[/tex]. Convert to cylindrical coordinates:
[tex]x=u\cos v[/tex]
[tex]y=u\sin v[/tex]
[tex]z=z[/tex]
The integral is then
[tex]\displaystyle\int_0^{2\pi}\int_0^2\int_{u^2}^4u^3\,\mathrm du\,\mathrm dv=\frac{32\pi}3[/tex]