I am trying to evaluate
$$\int _C \vec F \cdot d\vec r$$
using Stoke's Thm. where $\vec F = (x+y^2, y+z^2, z+x^2)$ and $S$ is the triangle with vertices $(1,0,0), (0,1,0), (0,0,1)$
So, I can see why Stoke's Thm. is easier than working on the line integral, and I believe that the expression is equal to
$$\int \int_S \operatorname{curl} \vec F \cdot d\vec S = \int \int_S (\operatorname{curl} \vec F)\cdot \vec n \space dS$$
and because $\operatorname{curl} \vec F = -2(x,y,z)$ and the normal vector to the triangle is $(1,1,1)$,
the inside portion simplifies to $-2$ so all I have to do is to evaluate
$$\int \int_S dS$$
This is where I am a bit confused.
I though that this expression is equivalent to the surface area which is an equilateral triangle with side $\sqrt{2}$
whose area must be $\frac{\sqrt{3}}{2}$, so I want to say that the answer is $-\sqrt{3}$,
but supposedly it should be $-1$.
One possible thing that I thought of was that the projection of $S$ onto the $xy$-plane is a right triangle with area $\frac{1}{2}$ which would make the answer $-1$.
Can anyone help me out with this problem?
Remember that in the formula
$$ \iint_S \vec V \cdot d \vec S = \iint_S \vec V \cdot \vec n \, dS $$
the normal vector $n$ must be a unit normal vector. Your normal vector $(1,1,1)$ has magnitude $\sqrt{3}$, so the unit normal is $\vec n = \frac{1}{\sqrt{3}}(1,1,1)$. This seems to be your only mistake.