Approach:
The parametrisation of the surface is given. So I just have to doubly integrate the standard formula for surface area over the required region.
Problem:
(Please refer to below): My double integral involves a square root of a square.
$$\iint \sqrt{2(u-v)^2} \ dA $$
My tutor pointed out that we "must consider both cases", i.e.
$$= \iint \sqrt{2} (\color{red}{u-v}) \ dA$$
AND
$$= \iint \sqrt{2} (\color{red}{v-u}) \ dA$$
But I am stuck as to what to do next - what is the end result? What is the actual surface area?
Thank-you kindly for reading, and any help is greatly appreciated :)
Now you just need to compute the intergral!
Use the linearity property of integrals: $\iint\sqrt2(u-v)dA=\sqrt2(\iint (u-v)dA$. I'm just going to assume that $dA=dudv$ here. So $\sqrt2\iint (u-v)dA=\sqrt2\iint(u-v)dudv=\sqrt2\int (\frac{u^2}{2}-uv+C)dv$ and I leave the rest to you!
Hope this helps!