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bodensee9

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[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.