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Just a theory question. Does a surface necessary have to be closed for Stokes's theorem to apply? I know for it is true for Green's theorem and it is supposed to be a baby version of Stokes's theorem.

Also I know that it is necessary true for the Divergence theorem as well (from physics on Gauss's law)

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It seems to me that you are confusing two different uses of the word "closed".

A "closed surface", as in the formulation of Divergence Theorem, means a surface without boundary. In other words, a surface which has no "edge", like a sphere or a torus, and as opposed to a hemisphere (which has a circle as boundary). In this sense, the surface in the statement of Stokes' theorem need certainly not be closed. In fact, integrating the curl of a smooth vector field over a closed surface yields $0$, so it's not exactly the most interesting case. (Can you say why it's $0$? It follows from Stokes' theorem.)

On the other hand I think you might be thinking of a surface as "closed" as a topological subspace of $\mathbb{R}^3$, which is not at all the same thing.

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because of div(curlF) = 0 –  sidht Apr 20 '12 at 16:48
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