I'm tasked with computing the circulation of the vector field $\vec F = <y^2, z, xy>$ along the triangle with vertices $(1,0,0), (0,1,0), (0,0,1)$ with the orientation of the curve following this order.
My first step is to compute the 1-Form of $\vec F$: $\alpha_{\vec F} = y^2dx+zdy+xydz$. Knowing that Stokes's Theorem states: $\int_{\partial D}\alpha_{ \vec F} = \int_Dd\alpha_{\vec F}$ for a bounded surface $D \subset \mathbb R^3 $ with boundary $\partial D$.
Next, I compute $d\alpha_{\vec F} = -2ydx\land dy - dz\land dx$. Using this result and applying Stokes' Theorem, I have:
$\int_{\partial D}y^2dx+zdy+xydz = \int_D-2ydx\land dy-dz\land dx$.
Now, I need to parametrise the surface $D$, and I do so knowing that the equation of the triangle in the plane is $x+y+z=1$ for $0 \le x,y,z \le 1$. I call this parametrisation $G(x,y) = <x,y,1-x-y>$ and compute its Jacobian:
$$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ -1 & -1 \\ \end{bmatrix} $$
So, using the fact that $dx\land dy = $$ \begin{vmatrix} 1 & 0 \\ 0 & 1 \\ \end{vmatrix} = 1 $ and that $dz\land dx = $$ \begin{vmatrix} -1 & -1 \\ 1 & 0 \\ \end{vmatrix} = -1 $, we have:
$\int_D-2ydx\land dy-dz\land dx$ = $\int_D-2y-1 $.
Now, the region of integration is what's tripping me up. I think the bounds should be $0\le y \le 1 $ and $0\le x\le 1-y $, and if I use those bounds, I get $\frac {-5} 6 $, which doesn't really make any sense.
Where did I go wrong here? Was it in the parametrisation of the curve or the region of integration? Any help and errors pointed out would be much appreciated.
Stokes theorm says
$\oint F\cdot dr = \iint \nabla \times F\cdot n dS\\ \nabla \times F = (x-1,-y,-2y)\\ n = \frac {(1,1,1)}{\sqrt 3}\\ dS= \sqrt 3\ dy\ dx\\ \int_\limits0^1\int_\limits0^{1-x} x-3y-1\ dy\ dx$
Based on the symmetry of the region it would appear that:
$\int_\limits0^1\int_\limits0^{1-x} x\ dy\ dx = \int_\limits0^1\int_\limits0^{1-x} y\ dy\ dx$
And so,
$\int_\limits0^1\int_\limits0^{1-x} x-3y-1\ dy\ dx = \int_\limits0^1\int_\limits0^{1-x} -2y-1\ dy\ dx$
We should have the same result.
Supposing we didn't use stokes.
$x = 1-t, y=t, z = 0$
$\int_\limits0^1 (t^2,0,t-t^2)\cdot (-1,1,0) \ dt\\ \int_0^1 -t^2 \ dt$
$y = 1-t, z=t, x = 0$
$\int_\limits0^1 (t^2-2t + 1,t,0)\cdot (0,-1,1) \ dt\\ \int_\limits0^1 -t \ dt$
$z = 1-t, x=t, y = 0$
$\int_\limits0^1 (0,1-t,0)\cdot (1,0,-1) \ dt\\ \int_\limits0^1 0 \ dt$
$\int_\limits0^1 -t^2 - t\ dt = -\frac 56$