Application Stokes's Theorem I am a bit unsure the way Stoke's theorem is applied in this case.

Evaluate $\oint\limits_C {xydx + yzdy + zxdz} $ around the triangle with vertices $(1,0,0), (0,1,0), and (0,0,1)$, oriented clockwise as seen from the point $(1,1,1)$

My reasoning
Since Stokes's theorem says that we can sum up individual microscopic rotations, i.e. we can turn a line integral into a double integral:
$$\oint\limits_C {xydx + yzdy + zxdz}  = \iint\limits_D {\nabla  \times \vec F \cdot \hat NdS} =  - \iint\limits_D {(y,z,x) \cdot \hat NdS}$$
Then, I used parametrization $r = (u,v,1 - u)$.
The surface element turns out to be $$d{\mathbf{S}} = \frac{{\partial (x,y,z)}}{{\partial (u,v)}}dudv  = (1,0,1)dudv$$
So, the integral is
$$\iint\limits_{D'} {u + vdudv} = \int\limits_0^1 {\int\limits_0^{1 - v} {u + vdudv = \int\limits_0^1 {(1 - v)v + \frac{{{{(1 - v)}^2}}}{2}dv}  = \frac{1}{2}\int\limits_0^1 {1 - {v^2}dv}  = \frac{1}{2}\int\limits_0^1 {1 - {v^2}dv}  = \frac{1}{3}} } $$
According to the answer, it should be $1/2$. Is this correct application of Stokes's theorem?
 A: Thanks to @Paul's comment, I was able to put all the pieces together to finally obtain the right answer.
The application of Stokes's theorem is correct, so everything is fine up to (and including):
$$\oint\limits_C {xydx + yzdy + zxdz}  = \iint\limits_D {\nabla  \times \vec F \cdot \hat NdS} =  - \iint\limits_D {(y,z,x) \cdot \hat NdS}$$
After this, the hard job is to find the right parametrization for the surface. I.e. we want to express $x,y,z$ in terms of $u,v$.

First, let's illustrate this in 3 space.
By projecting the pyramid like object onto $xy$ plane, we get a triangle. We could keep $x$ and $y$ as they are, i.e. by letting $x=u,\quad y=v$. Now the tricky part, the $z$-axis. If we look at this, we see a tilted triangle (from $(1,1,1)$). This is a plane with a normal 
$$n=(1,1,1)\quad\implies \quad 1=x+y+z$$
 The final parametrization is therefore
$$r=(u,v,1-u-v)$$
By finding the $$\left\| {\frac{{\partial r}}{{du}} \times \frac{{\partial r}}{{dv}}} \right\|$$ or the way I expressed in my question
$$ \frac{{\partial (x,y,z)}}{{\partial (u,v)}}  = (1,1,1)$$
If we perform the dot product of $\nabla \times F$ with the normal vector, we obtain $$\int_0^1 {\int_0^{1 - v} {dudv} }  = \frac{1}{2}$$
Note, the reason why we have $1-v$ as the upper bound is because the domain (the body projected onto xy plane) has a boundary $1=u+v$.
