How to use Stokes Theorem to evaluate $\int_{S} \text{curl} F\cdot d\mathbf{S}$ Let F = $( yz, 0, x)$ and $S$ is the portion of the plane ${x\over2} + {y\over3} + z = 1$ where $x, y, z \ge 0$, oriented with an upward pointing normal then prove:
$$\int_{S} \text{curl} F\cdot d\mathbf{S} = -1 = \int_{\partial S} F\cdot d\mathbf{s}$$
I figured out that $\text{curl} F = (0, y-1, -z)$ then I think the parametrization of the plane is $(x, y, 1- {x\over2} -{y\over3})\implies n = (\frac 12, \frac13,1)$. So
$$\int_{S} \text{curl} F\cdot d\mathbf{S} = \int_{S} \left(0, y-1, -1 + {x\over2} +{y\over3}\right)\cdot\left(\frac 12, \frac 13,1\right)dA$$
I don't know how to find the boundary of the plane and I'm not sure if I did right?
Thanks for your helping.

$S$ will be a triangular region with vertices at $(0,0,1),(2,0,0),(0,3,0).$ We could try to parameterize this surface considering two vectors with tail at one of these vertices, for instance $(0,0,1)$, pointing to the other two vertices. These vectors are $\langle0,3,-1\rangle$ and $\langle 2,0,-1\rangle$. Note that a point on $S$ will be described by a linear combination of these vectors: $\mathbf r(u,v)=\langle0,3,-1\rangle  u+ \langle 2,0,-1\rangle v + \langle 0,0,1 \rangle = \langle 2v,3u,1-u-v\rangle$. Verify that $S$ is generated by this parameterization where $0\leq u \leq 1$ and $0\leq v\leq 1-u.$
The integral is done by:
$$\iint_S \text{curl}\mathbf F\cdot d\mathbf S=\iint_D\text{curl}\mathbf F(\mathbf r(u,v))\cdot (\mathbf r_u\times \mathbf r_v)\,dA$$
where $(u,v)\in D$ with the mentioned limits.
