Split up the surface [tex]\mathcal S[/tex] into three sub-surfaces [tex]\mathcal S_1,\mathcal S_2,\mathcal S_3[/tex], where:
(1) [tex]\mathcal S_1[/tex] is the part of [tex]\mathcal S[/tex] in the plane [tex]x=0[/tex] parameterized by
[tex]\mathbf s_1(u,v)=\langle0,u\cos v,u\sin v\rangle[/tex]
with [tex]0\le u\le5[/tex] and [tex]0\le v\le2\pi[/tex];
(2) [tex]\mathcal S_2[/tex] is the part of [tex]\mathcal S[/tex] in the plane [tex]x+y=10[/tex] parameterized by
[tex]\mathbf s_2(u,v)=\langle10-u\cos v,u\cos v,u\sin v\rangle[/tex]
with [tex]0\le u\le5[/tex] and [tex]0\le v\le2\pi[/tex]; and
(3) [tex]\mathcal S_3[/tex] is the part of [tex]\mathcal S[/tex] on the cylinder [tex]y^2+z^2=25[/tex] parameterized by
[tex]\mathbf s_3(u,v)=\langle u,5\cos v,5\sin v\rangle[/tex]
with [tex]0\le u\le10-5\cos v[/tex] and [tex]0\le v\le2\pi[/tex].
In each respective case, we get surface elements
[tex]\mathrm dS_1=\|{{\mathbf s}_1}_u\times{{\mathbf s}_1}_v\|\,\mathrm du\,\mathrm dv=u\,\mathrm du\,\mathrm dv[/tex]
[tex]\mathrm dS_2=\|{{\mathbf s}_2}_u\times{{\mathbf s}_2}_v\|\,\mathrm du\,\mathrm dv=\sqrt2u\,\mathrm du\,\mathrm dv[/tex]
[tex]\mathrm dS_3=\|{{\mathbf s}_3}_u\times{{\mathbf s}_3}_v\|\,\mathrm du\,\mathrm dv=5\,\mathrm du\,\mathrm dv[/tex]
So the surface integral is
[tex]\displaystyle\iint_{\mathcal S}xz\,\mathrm dS=\sum_{i=1}^3\iint_{\mathcal S_i}x(u,v)z(u,v)\,\mathrm dS_i[/tex]
The first integral is
[tex]\displaystyle\iint_{\mathcal S_1}xz\,\mathrm dS_1=0[/tex]
The second is
[tex]\displaystyle\iint_{\mathcal S_2}xz\,\mathrm dS_2=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=5}(10-u\cos v)u\sin v(\sqrt2u)\,\mathrm du\,\mathrm dv=0[/tex]
The third is
[tex]\displaystyle\iint_{\mathcal S_3}xz\,\mathrm dS_3=5\int_{v=0}^{v=2\pi}\int_{u=0}^{u=10-5\cos v}\mathrm du\,\mathrm dv=0[/tex]
and so
[tex]\displaystyle\iint_{\mathcal S}xz\,\mathrm dS=0[/tex]
