Is Stokes' Theorem always surface independent? In my textbook it says that if F has a vector potential A such that curl(A)=F, then the following is true:
$$\iint F \cdot dS =\int A \cdot dr$$ Excuse the lack of formal notation, I couldn't figure out how to write the circulation symbol.
Anyway, I don't really understand how this is any different than the regular Stokes' theorem. My book says that if a vector field has a vector potential then the result of Stokes' theorem is surface independent give the same boundary and orientation. However, if you substitute in the curl of A for F, you get:
$$\iint (\nabla\times A) \cdot dS = \int A \cdot dr$$
Compared to the regular Stokes' Theorem this seems exactly the same, just that we replaced the letter F with A.
Regular Theorem:
$$\iint (\nabla \times F) \cdot dS = \int F \cdot dr$$
Vector Potential Version:
$$\iint (\nabla \times A) \cdot dS = \int A \cdot dr$$
Am I missing something here or does this mean that Stokes' Theorem is always surface independent? If so, why would the book bring this up as a different thing? Any help explaining this would be very helpful, thanks.