I've looked everywhere online and through my textbooks but I can't find anything to clear up my confusion on this problem!
I've been giving a vector field $F = (yz,xy,xz)$ and the surface is simply the $1 \times 1$ square $0<x<1, 0<y<1$, at $z = 2$.
I have determined the $\text{curl}(F)$ as $(0,y-z,y-z)$, and the Normal Vector as $(0,0,1)$, but I can't figure out what I am supposed to use as the bounds for my integral.
In my mind it only makes sense to integrate $x$ and $y$ each from $0$ to $1$, but then the expression being integrated is $y-z$, and that integral doesn't make any sense.
Am I supposed to simply substitute $2$ for $z$, since the square is located at $z = 2$?
In Stokes you have surface integral = line integral: $$\int_S\text{curl }F\,dS = \int_{\partial S}F.$$ For the surface integral you must parametrize the surface. For example: $$\Phi(x,y) = (x,y,2).$$ Now, for this parametrization you must calculate the normal vector $N(x,y)$ and $$ \int_S\text{curl }F\,dS = \int_0^1\int_0^1\text{curl }F(\Phi(x,y))\cdot N(x,y)\,dxdy. $$