Unable to simplify dS (Stokes' theorem)

[Addez123]

Homework Statement

Calculate the line integral along the curve created by the intersection of surface B and C, over the vector field A, using stokes theorem.

A = (yz + 2z, xy - x + z, xy + 5y)
B: x^2 + z^2 = 4
C: x + y = 2

Relevant Equations

Stokes Theorem

Basically surface B is a cylinder, stretching in the y direction.
Surface C is a plane, going 45 degrees across the x-y plane.

Drawing this visually it's self evident that the normal vector is
$$(1, 1, 0)/\sqrt 2$$
Using stokes we can integrate over the surface instead of the line.
$$\int A(r) dr = \iint rot A \cdot n dS $$
$$\iint (x + 4, 2, y - z - 1) \cdot (1, 1, 0) /\sqrt 2 dS= \frac 1 {\sqrt 2} \iint x + 6 dS$$

Seems extremely simple to solve, except now I'm forced to find a good function g(u, v) that covers the area S.
Assume I find such a funciton, I need to either calculate the normal vector by doing $g'u \times g'v$
or find the non-existant jacobian. Since |d(x, y, z)/d(u, v)| is not a square matrix that would result in a single scalar number.

I just get stuck here. It's so SIMPLE, yet replacing a single variable x causes such headaces I never get it right.
Is there a simple way of solving the integral without digressing into g(u,v) and it's derivatives?

 

[pasmith]

Given $x^2 + z^2 = 4$, the obvious choice would be $(x,y,z) = (u\cos v, y(u,v), u\sin v)$.
Calculating the cross-product of the derivatives is straightforward, but there is another way.

On geometric grounds, the surface is an ellipse; there must therefore exist an affine transformation $\mathbf{x} = A\mathbf{X} + \mathbf{b}$ where $A$ is orthogonal such that the surface is given by $\mathbf{X}(r,\theta) = (ar\cos \theta, br\sin \theta, 0)$ for some constants $a > 0$ and $b > 0$. The area element is then $dX\,dY = abr\,dr\,d\theta$. Why not try finding such a transformation (which is not unique)?

I leave it to you to judge whether this was simpler than calculating a cross-product of derivatives.