Parameterize S by
[tex]\vec r(s, t) = (1-s)(1-t)\,\vec\imath + 6s(1-t)\,\vec\jmath + 6t\,\vec k[/tex]
with 0 ≤ s ≤ 1 and 0 ≤ t ≤ 1.
Take the normal vector to S to be
[tex]\vec n = \dfrac{\partial\vec r}{\partial s}\times\dfrac{\partial\vec r}{\partial t} = (36-36t)\,\vec\imath+(6-6t)\,\vec\jmath+(6-6t)\,\vec k[/tex]
with magnitude
[tex]\|\vec n\| = 6\sqrt{38}(1-t)[/tex]
Then the surface integral is
[tex]\displaystyle \iint_S xy\,\mathrm dS = 6\sqrt{38} \int_0^1\int_0^1 (1-t)\,\mathrm dt\,\mathrm ds = \boxed{3\sqrt{\frac{19}2}}[/tex]
