Use the divergence theorem.
[tex]\vec F(x,y,z)=x\,\vec\i\math+y\,\vec\jmath+10\,\vec k\implies\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(x)}{\partial x}+\dfrac{\partial(y)}{\partial y}+\dfrac{\partial(10)}{\partial z}=2[/tex]
The div theorem says the integral of [tex]\vec F[/tex] across [tex]S[/tex] is equal to the integral of [tex]\nabla\cdot\vec F[/tex] over the region with boundary [tex]S[/tex] (call it [tex]R[/tex]):
[tex]\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F(x,y,z))\,\mathrm dx\,\mathrm dy\,\mathrm dz[/tex]
[tex]=\displaystyle2\iiint_R\mathrm dx\,\mathrm dy\,\mathrm dz[/tex]
Convert to cylindrical coordinates:
[tex]\begin{cases}x=r\cos\theta\\y=y\\z=r\sin\theta\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dy[/tex]
Then the integral is
[tex]=\displaystyle2\int_0^{2\pi}\int_0^1\int_0^{8-r\cos\theta}r\,\mathrm dy\,\mathrm dr\,\mathrm d\theta=\boxed{16\pi}[/tex]
