I suppose my question is really how to parametrize points of the triangle. But for context, suppose the three vertices of a triangle are $(-5, 1, 0), (0, -5, 1), (1, 0, -5)$, the coordinate functions of a vector field $F$ are hard/impossible to integrate and $\nabla \times F = \begin{bmatrix}-1 & 1 & 2\end{bmatrix}^T$.
We want to find the line integral $\oint_A F \cdot dr$ over the triangle, but by Stoke's theorem we know that $\oint_{\partial P} F \cdot dr = \int\int_P (\nabla \times F)\cdot n dS$. Therefore if we know how to evaluate the surface integral, we are done.
My original idea was to parametrize the triangle with vectors $\overrightarrow{a} = (5, -6, 1), \overrightarrow{b} = (6, -1, -5)$ by $\begin{cases}x &= 5 + 5u + 6v\\ y &= 1 - 6u - v\\ z &= u - 5v\end{cases}, u, v \in [0, 1]$. With $\overrightarrow{a}$ and $\overrightarrow{b}$ the normal vector is $n = (31, 31, 31)$, hence the unit normal vector is $\hat{n} = (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$. Therefore I thought the integral would be: $\int\int (\nabla \times F)\cdot \hat{n} dS = \int\int (\nabla \times F)\cdot \frac{n}{||n||} ||n||dA = \int\int (\nabla \times F)\cdot n dudv = \int_0^1\int_0^1(-1, 1, 2)\cdot (31, 31, 31)^Tdudv$.
But wouldn't this integral integrate a rectangle instead of the triangle? How can I determine the limits so that I integrate the triangle?
I am assuming positive (anticlockwise) orientation of the triangle.
Vertices of the triangle $A (-5, 1, 0), B(0, -5, 1), C(1, 0, -5)$. So $\vec AB = (5, -6, 1), \vec {BC} = (1, 5, -6)$
Please note that $\frac{1}{2} \|\vec {AB} \times \vec {BC}|$ gives you area of the triangular surface. As the components of the curl are constant, it is easy to see that the answer should evaluate to $I = \frac{1}{2} \ (-1, 1, 2) \cdot (31, 31, 31) = 31$.
But here is a standard approach -
$\vec {AB} \times \vec {BC} = (31, 31, 31)$ and so equation of the plane is,
$31(x+5) + 31(y-1)+31z = 0 \ $ i.e $ x + y + z + 4 = 0$
So the normal vector to the plane that the triangular surface resides in is $(1, 1, 1)$.
So, $\displaystyle \iint_S (\nabla \times \vec F) \cdot \hat {n} \ dS = \iint_A (-1, 1, 2) \cdot (1, 1, 1) \ dA = 2 \iint_A dA \ $
where $A$ is the region of the projection of $S$ in XY plane formed by $(-5, 1), (0, -5)$ and $(1, 0)$.
If you want to find the area using integral, find equation of lines and you will have to split the integral into two parts -
$ A = \displaystyle \int_{-5}^0 \int_{-(6x+25)/5}^{(1-x)/6} dy \ dx + \int_0^1 \int_{(5x-5)}^{(1-x)/6} dy \ dx = \frac{31}{2}$
So $I = 2 A = 31$.