Unable to simplify dS (Stokes' theorem)

In summary, surface B is a cylinder stretching in the y direction and surface C is a plane going at a 45 degree angle across the x-y plane. Using Stokes' theorem, we can integrate over the surface instead of the line, with the normal vector being (1, 1, 0)/√2. To solve the integral, a function g(u,v) must be found to cover the area S. This can be done either by finding the cross-product of the derivatives of g(u,v) or by using a geometric approach and finding an affine transformation.
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?

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.

1. What is the meaning of "unable to simplify dS (Stokes' theorem)"?

Stokes' theorem is a mathematical tool used to evaluate integrals over a surface by converting them into line integrals along the boundary of the surface. "Unable to simplify dS" means that the surface integral cannot be reduced to a simpler form using this theorem.

2. Why is it important to be able to simplify dS using Stokes' theorem?

Simplifying dS using Stokes' theorem allows for the evaluation of complex surface integrals in a more efficient and accurate manner. It also provides a deeper understanding of the underlying mathematical concepts and relationships between different types of integrals.

3. What are some common scenarios in which dS cannot be simplified using Stokes' theorem?

Some common scenarios include when the surface is not closed, when the surface is not smooth, and when the vector field is not conservative. These conditions violate the assumptions required for the application of Stokes' theorem.

4. Are there any alternative methods to simplify dS if Stokes' theorem cannot be used?

Yes, there are alternative methods such as Green's theorem and the Divergence theorem that can be used to simplify surface integrals. However, these theorems have their own set of limitations and may not be applicable in all cases.

5. How can one determine if dS can be simplified using Stokes' theorem?

In order to use Stokes' theorem, the surface must be closed, smooth, and the vector field must be conservative. To determine if these conditions are met, one can check for the existence of a boundary curve, the differentiability of the surface, and the curl of the vector field, respectively.

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