C is the curve arising from the intersection of the elliptic cylinder $\frac{(x-2)^2}{36} + \frac{(y-1)^2}{9} = 1$ and the plane $3x + y + z = 1$, which circulates counterclockwise as seen from above. $\vec F = (2z^2 + y^2 + sin(x^2))\hat{i} + (2xy + z)\hat{j} + (xz + 2yz)\hat{k}$
I want to get the flux through the surface enclosed by the boundary $C$. So as to do so, let's use Stoke's Theorem:
$$\iint_{S} (\nabla \times \vec F) \cdot n \frac{\mathrm dx\,\mathrm dy}{|n \cdot k|} = \oint_{C} \vec F \cdot \mathrm d \vec l$$
I want to solve the problem evaluating the left hand side of ST. I have tried the following:
the projection of the enclosed surface on the $xy$ plane is the given ellipse. So we indeed need to get a unit normal vector of that ellipse:
$$n = \nabla \left(\frac{(x-2)^2}{4} + (y-1)^2\right) = \frac{x-2}{2} \hat{i} + 2(y-1) \hat{j}$$
$$\hat{n} = \frac{\frac{x-2}{2} \hat{i} + 2(y-1) \hat{j}}{\sqrt{\frac{(x-2)^2}{4}+ 4(y-1)^2}}$$
I guess there has to be a way of simplifying $\hat{n}$ but don't see it :(
So now we just have to set up the integral:
$$\iint (\nabla \times \vec F) \cdot \mathrm d \vec a = \iint \left[(2z - 1)\hat{i} + 3z\hat{j}\right]\left[\frac{\frac{x-2}{2}\hat{i} + 2(y-1)\hat{j}}{\sqrt{\frac{(x-2)^2}{4}+ 4(y-1)^2}}\right] \mathrm da$$
Now we gotta make a change of variables so as to go from an ellipse to a circle centred at the origin.
Rearranging the curve we get:
$$\frac{(x-2)^2}{36} + \frac{(y-1)^2}{9} = 1$$
Using the following change of variables and using it:
\begin{align} x - 2 &= 6u \\ y - 1 &= 3v \\ u^2 + v^2 &= 1 \end{align}
Don't forget the Jacobian! :)
$$\frac{\partial(x, y)}{\partial (u, v)} = 18$$
$$-18\iint \left[(36u+6v+13)\hat{i} + (36u+9v+12)\hat{j}\right]\left[\frac{\frac{6u}{2}\hat{i} + 2(3v)\hat{j}}{\sqrt{\frac{(6u)^2}{4}+ 4(3v)^2}}\right]\mathrm du\, \mathrm dv$$
There's just one step left: apply the last change of variables (polar coordinates):
$$u = r\cos\theta\\ v = r\sin\theta$$
I've been told this is wrong, but don't get why. Please explain how to evaluate $\iint (\nabla \times \vec F) \cdot \mathrm d \vec a$.