Over the cylinder's lateral face: parameterize this part of [tex]S[/tex] (call it [tex]S_1[/tex]) by
[tex]\mathbf r(u,v)=(5\cos u,5\sin u,v)[/tex]
with [tex]0\le u\le2\pi[/tex] and [tex]0\le v\le4[/tex], then the surface element is
[tex]\mathrm dS=\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv=5\,\mathrm du\,\mathrm dv[/tex]
[tex]\displaystyle\iint_{S_1}x^2+y^2+z^2\,\mathrm dS=5\int_0^{2\pi}\int_0^4(25+v^2)\,\mathrm dv\,\mathrm du=\frac{3640\pi}3[/tex]
Over the bottom disk: parameterize this part ([tex]S_2[/tex]) by
[tex]\mathbf r(u,v)=(u\cos v,u\sin v,0)[/tex]
with [tex]0\le u\le5[/tex] and [tex]0\le v\le2\pi[/tex], then
[tex]\mathrm dS=u\,\mathrm du\,\mathrm dv[/tex]
(this will be the same for the top disk) and the integral is
[tex]\displaystyle\iint_{S_2}x^2+y^2+z^2\,\mathrm dS=\int_0^{2\pi}\int_0^5u^3\,\mathrm du\,\mathrm dv=\frac{625\pi}2[/tex]
Over the top disk ([tex]S_3[/tex]): the only change is replacing [tex]z=0[/tex] with [tex]z=4[/tex], so that the integral is
[tex]\displaystyle\iint_{S_3}x^2+y^2+z^2\,\mathrm dS=\int_0^{2\pi}\int_0^5(u^2+16)u\,\mathrm du\,\mathrm dv=\frac{1425\pi}2[/tex]
Then the total surface integral is the sum of the partial surface integrals:
[tex]\displaystyle\iint_Sx^2+y^2+z^2\,\mathrm dS=\frac{6715\pi}3[/tex]
