Compute $\int_S \mathbf{F}\cdot d \mathbf{S}$ where $\mathbf{F}(x, y, z) = (x, y, z)$ and S is given by $$g(u,v) = \begin{pmatrix} u-v \\ u + v \\ uv \end{pmatrix}$$ when $0 \le u \le 1, 0 \le v \le 2$.
We can define the surface integral of F over S by $$\int _d \mathbf{F}(g(u,v)) \cdot \left(\frac{\partial g}{\partial u}(u,v)\space \times \space \frac{\partial g}{\partial v}(u,v) \right)dudv$$
We can calculate the following as $$\left(\frac{\partial g}{\partial u}(u,v)\space \times \space \frac{\partial g}{\partial v}(u,v) \right)= (u-v, -u-v, 2)$$
However I'm unsure on how to proceed from here as $\mathbf{F}(x,y,z)$ throws me off, so I'd really appreciate some guidance from the community!
The parametrized surface is $g(u ,v) = (u-v, u + v, uv), 0 \le u \le 1, 0 \le v \le 2$
The vector field $\vec F = (x, y, z)$
So, $ \vec F(g(u, v)) = (u-v, u+v, uv)$
As you obtained, $g_u \times g_v = (u-v, -u-v, 2)$
$ \big[$ Note that the orientation of the surface is not given so it could either be $g_u \times g_v$ or $g_v \times g_u \big]$
Then, $\vec F(g(u, v)) \cdot (g_u \times g_v) = - 2uv$
So the surface integral is,
$ \displaystyle \iint_D \vec F(g(u, v)) \cdot (g_u \times g_v) ~ du ~ dv = \int_0^2 \int_0^1 - 2uv ~ du ~ dv$
$ = - 2$