Parameterize the region [tex]\mathcal S[/tex] by the vector-valued function,
[tex]\mathbf s(u,v)=((1-u)(1-v),-2u(1-v),10v)[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Then the surface element is
[tex]\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=6\sqrt{14}(1-v)\,\mathrm du\,\mathrm dv[/tex]
So the surface integral is
[tex]\displaystyle\iint_{\mathcal S}x\,\mathrm dS=6\sqrt{14}\int_{v=0}^{v=1}\int_{u=0}^{u=1}\underbrace{(1-u)(1-v)}_x(1-v)\,\mathrm du\,\mathrm dv=\sqrt{14}[/tex]
