[tex]\mathbf F(x,y,z)=\langle x,y,4z\rangle[/tex]
By the divergence theorem, the surface integral taken over [tex]S[/tex],
[tex]\displaystyle\iint_S\mathbf F\,\mathrm dS[/tex]
is equivalent to the triple integral of the divergence of [tex]\mathbf F[/tex] over [tex]W[/tex], the space bounded [tex]S[/tex],
[tex]\displaystyle\iiint_W(\nabla\cdot\mathbf F)\,\mathrm dV[/tex]
We have
[tex]\nabla\cdot\mathbf F=\dfrac{\partial x}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial(4z)}{\partial z}=1+1+4=6[/tex]
The latter integral is then given by
[tex]\displaystyle6\iiint_W\mathrm dV=6\int_{|x|\le3}\int_{|y|\le3}\int_{x^2+y^2\le z\le9}\mathrm dz\,\mathrm dy\,\mathrm dx=648[/tex]
