Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$ \int_{\partial \Omega} (u ~dx + v ~dy) = \iint_{\Omega} \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) ~dx ~dy $$ 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?

share|cite|improve this question
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
up vote 5 down vote accepted

$\def\d{\mathrm{d}} \def\w{\omega}$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.

share|cite|improve this answer
+1 for : "Green's theorem is a special case of Stokes' theorem, not the other way around"! – Arjang Jul 28 '12 at 1:18
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.