Evaluate the surface integral ∫∫ F.ds for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x,y,z)=xi+yj+5k S is the coundary of the region enclosed by the cylinder x^2+z^2=1 and the planes y=0 and x+y=2

Use the divergence theorem.[tex]\vec F(x,y,z)=x\,\vec\imath+y\,\vec\jmath+5\,\vec k\implies\mathrm{div}\vec F(x,y,z)=2[/tex]By the divergence theorem,[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F\,\mathrm dV[/tex]where [tex]R[/tex] is the region with boundary [tex]S[/tex].Compute the latter integral in cylindrical coordinates, taking[tex]\begin{cases}x=r\cos\theta\\y=y\\z=r\sin\theta\end{cases}\implies\mathrm dV=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dy[/tex][tex]\displaystyle\iint_R2\,\mathrm dV=2\int_0^{2\pi}\int_0^1\int_0^{2-r\cos\theta}r\,\mathrm dy\,\mathrm dr\,\mathrm d\theta=\boxed{4\pi}[/tex]