[tex]\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+10\,\mathbf k[/tex]
[tex]\implies\nabla\cdot\mathbf f(x,y,z)=1+1+0=2[/tex]
By the divergence theorem, the surface integral along [tex]S[/tex] is equivalent to the triple integral over the region [tex]R[/tex] bounded by [tex]S[/tex]:
[tex]\displaystyle\iint_S\mathbf f(x,y,z)\,\mathrm dS=\iiint_R\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=2\iiint_R\mathrm dV[/tex]
Convert to cylindrical coordinates, setting
[tex]\begin{cases}x=r\cos\theta\\y=Y\\z=r\sin\theta\end{cases}\implies\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dY[/tex]
The triple integral is then equivalent to
[tex]=\displaystyle2\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{Y=0}^{Y=2-r\cos\theta}r\,\mathrm dY\,\mathrm dr\,\mathrm\theta[/tex]
[tex]=\displaystyle2\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r(2-r\cos\theta)\,\mathrm dr\,\mathrm\theta[/tex]
[tex]=\displaystyle\frac23\int_{\theta=0}^{\theta=2\pi}(3-\cos\theta)\,\mathrm dr\,\mathrm\theta[/tex]
[tex]=4\pi[/tex]
