The figure shows the surface created when the
I think to find the surface integral we need to do the double integral of ru cross rv, with respect to du and dv respectively. However, im not even sure how to get to that point.
How about doing the red part below and multiplying by 16 right? You could work through the cross product but that's not necessary. Just use the formula:
and that red one is and you can figure out R by equating so . So it's just that triangular region in the x-y plane above the red surface.