Parameterize the triangle [tex]\mathcal S[/tex] by the vector-valued function [tex]\mathbf s(u,v)[/tex] with
[tex]\mathbf s(u,v)=(1-u)((1-v)(1,0,0)+v(0,1,0))+u(0,1,1)[/tex]
[tex]\implies\mathbf s(u,v)=(\underbrace{(1-u)(1-v)}_{x(u,v)},\underbrace{v(1-u)+u}_{y(u,v)},\underbrace{u}_{z(u,v)})[/tex]
where [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Then the integral becomes
[tex]\displaystyle\iint_{\mathcal S}xz\,\mathrm dS=\int_{v=0}^{v=1}\int_{u=0}^{u=1}x(u,v)z(u,v)\left\|\frac{\partial\mathbf s}{\partial u}\times\frac{\partial\mathbf s}{\partial v}\right\|\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\sqrt2\int_{v=0}^{v=1}\int_{u=0}^{u=1}(1-u)^2u(1-v)\,\mathrm du\,\mathrm dv=\frac1{12\sqrt2}[/tex]
