Originally Posted by

**mr fantastic** A couple of things:

1. Since you're integrating on the surface (a plane), you can substitute z = 4x + y (which is the equation of the plane). This solves your technical infelicity of three variables in the integrand ......

2. It looks like you're using a unit vector normal to the surface (plane). That's incorrect. $\displaystyle F \cdot dS \neq F \cdot \hat{n} \, dx \, dy$ in general.

In fact, $\displaystyle F \cdot dS = F \cdot n \, dx \, dy$:

$\displaystyle F \cdot dS = F \cdot \left( \frac{\partial z}{\partial x} \, i + \frac{\partial z}{\partial y} \, j - k\right) \, dx \, dy$.

3. Are you OK with the region in the xy-plane you integrate over (and hence the integral terminals) .....? The region is the right angle triangle with vertices at (0, 0), (2, 0) and (2, 2).