Compute the surface integral over the given oriented surface: F=y3i+8j−xk, portion of the plane x+y+z=1 in the octant x,y,z≥0, downward-pointing normal

Parameterize the surface (call it [tex]S[/tex]) by[tex]\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(1-u-v)\,\vec k[/tex]with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Take the normal vector to [tex]S[/tex] to be[tex]\vec s_v\times\vec s_u=-\vec\imath-\vec\jmath-\vec k[/tex]Then the integral of [tex]\vec F[/tex] over [tex]S[/tex] with the given orientation is[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1(v^3\,\vec\imath+8\vec\jmath-u\,\vec k)\cdot(-\vec\imath-\vec\jmath-\vec k)\,\mathrm du\,\mathrm dv[/tex][tex]=\displaystyle\int_0^1\int_0^1(-v^3-8+u)\,\mathrm du\,\mathrm dv=\boxed{-\frac{31}4}[/tex]