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Stoke's Theorem

  1. Dec 12, 2009 #1
    Hello: I usually think of Stoke's Theorem as:

    [tex]\oint F\bullet dr = \int \int curl F \bullet dS[/tex]
    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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  3. Dec 12, 2009 #2

    Dick

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    Well, Stokes says the two surface integrals are equal to the same line integral. Why shouldn't they be equal?
     
  4. Dec 13, 2009 #3
    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.
     
  5. Dec 13, 2009 #4

    Dick

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    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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