$$ \int_{\partial \Omega} (u ~dx + v ~dy) = \iint_{\Omega} \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) ~dx ~dy $$ Then I want to prove that$$ \int_{\partial \Omega} w = \iint_{\Omega} ~dw, \;(w = u ~dx + v ~dy) $$ Would you give me an elementary proof for this?
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$\begingroup$$\def\d{\mathrm{d}} \def\w{\omega}$Green's theorem is a special case of Stokes' theorem, not the other way around.
Let $\w$ be the differential one-form $u \d x + v \d y$. The exterior derivative of $\w$ is $$\d \w = \left(\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\right)\d x\wedge \d y.$$ Stokes' theorem takes the form $$\int_{\partial\Omega} (u \d x + v \d y) = \int_\Omega \left(\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\right)\d x\wedge \d y.$$ Since the manifold is $\mathbb{R}^2$ this can be rewritten as $$\int_{\partial \Omega} (u dx + v dy) = \int_{\Omega} \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) dx dy.$$ This is Green's theorem.
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