Surface integral on an inclined ellipse/Stokes theorem

Question

I have encountered a problem related to Stoke´s Theorem. We are given the intersection between a circular base cylinder that is parallel to the z axis (R3), and a plane that cuts it obliquely, and therefore we end up with an ellipse, that from the top looks like a circle (the contour of the cylinder). If we apply the Theorem on the intersection, we can instead of computing the line integral of a vector field we are given, compute the flux of the curl of the vector field through the surface it encloses, that in cylindrical coordinates ranges between (0,2pi) and (0,r), a fixed r, which is the same as the area of a circle. Does this imply that every cross-section of the cylinder has the same area, or where am I losing it?

Answer 1

You forgot the $z$ values. The intersection cannot be represented only using $x,y$, hence besides $\theta,r$, you have another $z$ component in the cylindrical system, which can be represented by the function of the plane.

What you have so far is only the projection of the surface onto $xy$-plane.

Answer 2

The fact that the endpoints are always fixed in cylindrical coordinates does not mean that the areas of the cross-sections are the same (as you probably suspected). Don't forget that the endpoints are not everything, the function that you integrate also matters. When you integrate the constant function $1$ between these endpoints, yes, you get an area. But who guarantees that in every problem you will integrate $1$? For one, the scalar product between the curl of your vector field and the unit normal field to your cross-section, when expressed in cylindrical coordinates, might lead to a very complicated function. Next, also keep in mind that you also add to the mix the modulus of the Jacobian of the change of variables (from cylindrical to Cartesian coordinates). Therefore, save some very simple cases, you end up integrating a function that is not $1$, and therefore the result most of the time is not the area of the cross-section.

To remember: the integral on a piece of surface is in general not the area of that piece of surface. This is a common mistake among beginners.