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.

You are correct with the graph parameterization $$\left(x, y, 1- {x\over2} -{y\over3}\right)$$ as observed from the given equation $$z = 1 - \frac{x}{2} - \frac{y}{3}.$$ 


*

*We then take the partial derivatives with respect to $x$ and $y$ and take the cross product of $T_x$ and $T_y$ to get the normal of $$n = \left({1\over2}, {1\over3}, 1\right).$$

*You've calculated $\operatorname{curl}({\mathbf F})$ using the determinant of the matrix as: $\operatorname{curl}({\mathbf F}) = (0, y-1, -z)$.


You now simply dot $\operatorname{curl}({\mathbf F})$ with the normal vector $n$ and integrate the product over the boundary domain.
If my basic math is correct, I get $\left({y\over3} - {7\over3}\right)$ for the dot product. We then take the double integral of the above and evaluate over the positive domain which is as previously mentioned a triangle subset in $\mathbb{R}^2$.
edit: working on the formatting of the post. I'm new to this haha.
