# How can I prove Stokes theorem using Green's formula?

$$\int_{\partial \Omega} (u ~dx + v ~dy) = \iint_{\Omega} \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) ~dx ~dy$$ 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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I don't mean the implication. I want to know the proof of Stokes theorem by using Green's formula. –  Ann Jul 28 '12 at 2:16
Technically Green's theorem and Stokes are equivalent provided you view the fact that exterior derivatives commute with pullbacks as "trivial". The proof amounts to evaluating both sides of Stokes in a coordinate patch, where it reduces to Green's theorem. –  Ryan Budney Jul 28 '12 at 6:49

$\def\d{\mathrm{d}} \def\w{\omega}$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.
@Ann: Hi Ann. Stokes' theorem is about integration on $n$-dimensional manifolds and their boundaries, much more general spaces than considered in Green's theorem. –  user26872 Jul 30 '12 at 3:40