Use the divergence theorem. Let [tex]R[/tex] be the cylindrical region, then
[tex]\displaystyle\iint_{\partial R}\mathbf F\cdot\mathbf n\,\mathrm dS=\iiint_R\nabla\cdot\mathbf F\,\mathrm dV[/tex]
(where [tex]\mathbf n[/tex] denotes the unit normal vector to [tex]\partial R[/tex], but we don't need to worry about it now)
We have
[tex]\mathrm{div }\mathbf F=(\nabla\cdot\mathbf F)(x,y,z)=\dfrac{\partial\mathbf F}{\partial x}+\dfrac{\partial\mathbf F}{\partial y}+\dfrac{\partial\mathbf F}{\partial z}[/tex]
[tex]\nabla\cdot\mathbf F=3x^2+3y^2+3z^2[/tex]
For the solid [tex]R[/tex] with boundary [tex]\partial R[/tex], we can set up the following volume integral in cylindrical coordinates for ease:
[tex]\displaystyle3\iiint_R(x^2+y^2+z^2)\,\mathrm dV=3\int_{z=0}^{z=4}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=2}(r^2+z^2)r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz[/tex]
[tex]=\displaystyle6\pi\int_{z=0}^{z=4}\int_{r=0}^{r=2}(r^3+rz^2)\,\mathrm dr\,\mathrm dz[/tex]
[tex]=\displaystyle12\pi\int_{z=0}^{z=4}(2+z^2)\,\mathrm dz[/tex]
[tex]=352\pi[/tex]
