The integral of [tex]z[/tex] over [tex]S[/tex] is equal to the sum of the integrals of [tex]z[/tex] over [tex]S_1,S_2,S_3[/tex].
Parameterize the surface by
[tex]\vec r(u,v)=(9\cos u,9\sin u,v)[/tex]
with [tex]0\le u\le2\pi[/tex] and [tex]0\le v\le9+9\cos u[/tex]. Take the normal vector to [tex]S_1[/tex] to be
[tex]\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=(9\cos u,9\sin u,0)[/tex]
Then the integral of [tex]z[/tex] over [tex]S_1[/tex] is
[tex]\displaystyle\iint_{S_1}z\,\mathrm dS=\int_0^{2\pi}\int_0^{9+9\cos u}v\|(9\cos u,9\sin u,0)\|\,\mathrm du\,\mathrm dv=\frac{2187\pi}2[/tex]
Parameterize [tex]S_2[/tex] by
[tex]\vec s(u,v)=(u\cos v,u\sin v,0)[/tex]
with [tex]0\le u\le9[/tex] and [tex]0\le v\le2\pi[/tex]. Since [tex]z=0[/tex], the integral over [tex]S_2[/tex] is also 0.
Parameterize [tex]S_3[/tex] by
[tex]\vec t(u,v)=(u\cos v,u\sin v,9+u\cos v)[/tex]
with [tex]0\le u\le9[/tex] and [tex]0\le v\le2\pi[/tex]. The normal to [tex]S_3[/tex] is
[tex]\dfrac{\partial\vec t}{\partial u}\times\dfrac{\partial\vec t}{\partial v}=(-u,0,u)[/tex]
so that the integral over [tex]S_3[/tex] is
[tex]\displaystyle\iint_{S_3}z\,\mathrm dS=\int_0^{2\pi}\int_0^9(9+u\cos v)\|(-u,0,u)\|\,\mathrm du\,\mathrm dv=729\sqrt2\,\pi[/tex]
Putting the results together, the integral of [tex]z[/tex] over [tex]S[/tex] is
[tex]\iint_Sz\,\mathrm dS=\boxed{\left(\frac{2187}2+729\sqrt2\right)\pi}[/tex]
