I'm working on a vector calculus problem (provided below) and the issue is I'm getting two different answers, and I'm not sure which is right.
The question is as follows: Calculate the surface integral $\int_{S} \vec{F} \cdot d\vec{S} $ over a triangular surface bound by the points $(2,0,0), (0,2,0), (0,0,2)$, where $\vec{F} = (x, y, z)$ and $d\vec{S} = \vec{n} dS$, with $\vec{n} $ being the outward unit normal. (Hint: the plane equation is $ax+by+cz=1$, use it to find the normal, using the points on the surface)
I started with the plane points, and found that: $$\frac{1}{2} (x+y+z)=1$$ Yielding a normal of $\frac{1}{2} (1,1,1)$. Using the surface integral: $$\int_{S} \vec{F} \cdot d\vec{S} = \int_{S} (\vec{F} \cdot \vec{N} )dxdy$$
Which I computed to be $$\frac{1}{2} \int_{S} (x+y+z) dxdy = \frac{1}{2} \int_{S} 2 dxdy = 2$$ Using the area of the projected triangle on the $xy$ plane.
What I don't understand is, if I rewrite the plane as $$x+y+z=2$$ And compute the integral, I get an answer of $4$? Given the same plane equations (just rearranged), I don't understand where I'm losing the factor of two, or as to which answer is actually correct?
All help appreciated!