Can Stoke's Theorem Be Applied to Different Surfaces with the Same Boundary?

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Stoke's Theorem states that the line integral of a vector field F around a closed curve C is equal to the surface integral of the curl of F over any surface sigma bounded by C. This theorem applies regardless of the specific surface chosen, provided it has the same boundary. The discussion highlights that the equality holds even if the divergence of F is not zero, as the theorem specifically relates to the curl of F, not F itself.

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Hello: I usually think of Stoke's Theorem as:

\oint F\bullet dr = \int \int curl F \bullet dS
where dr is over a curve C and dS is over a surface sigma. But today in class the instructor said that Stoke's Theorem can also be used to change the surface over which one is intergrating, so that if sigma has a well defined boundary, say, C, then the surface integral of function F over sigma = surface integral F over any surface with C as the boundary. A more concrete example, so then say you are integrating some F over a paraboloid z = sqrt(1-x^2-y^2) above the xy plane. So then would it be true that my surface integral over the paraboloid would be the same as if I integrated over the disk formed by x^2+y^2 = 1? Thanks.
 
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Well, Stokes says the two surface integrals are equal to the same line integral. Why shouldn't they be equal?
 
they would be, but I thought then you must have that F is the curl of some other function. So if div F = 0 then it would be true. But would it be true if div F is not zero? Thanks.
 
The version of Stokes I'm thinking of says that the surface integral of curl(F).ds is equal to the line integral of F.dr. It doesn't say require div(F)=0. And it doesn't say anything about the surface integral of F itself, just curl(F).
 

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