Original motivation: How can I apply Stokes' Theorem to the annulus $1 < r < 2$ in $\mathbb{R}^2$?


  • Since the annulus is a manifold without boundary, it would seem that Stokes' Theorem would always return an answer of $\int_M d\omega = \int_{\partial M} \omega = 0$ for compactly supported forms $\omega$. Is this correct?
  • What about the annulus $1 < r \leq 2$? This seems like a manifold-with-boundary to me, yet an application of Stokes' Theorem will return a different answer. And what about $1 \leq r \leq 2$?

For instance, consider $\omega = -y\,dx + x\,dy$ on the annulus $1 < r \leq 2$, so that $d\omega = 2\,dx\,dy$. Then $$\int_M d\omega = 2\,\text{Area}(M) = 6\pi,$$ whereas $$\int_{\partial M} \omega = \int_0^{2\pi} 4\,dt = 8\pi,$$ where $\partial M$ is the circle $r = 2$.

What explains this discrepancy?

A friend of mine has suggested that this can be explained by the fact that $\omega = -y\,dx + x\,dy$ is not compactly supported on $1 < r \leq 2$, and hence Stokes' Theorem can't really be applied. Is this correct?

For reference, I am using the following version of the theorem:

Stokes' Theorem: Let $M$ be a smooth, oriented $n$-manifold with boundary, and let $\omega$ be a compactly supported smooth $(n-1)$-form on $M$. Then $$\int_M d\omega = \int_{\partial M} \omega.$$

  • 6
    $\begingroup$ I completely agree with your friend. Stokes is only applicable to the form on the closed annulus $1 \leq r \leq 2$. Note also that the discrepancy of $2\pi$ is obtained by subtracting the integral of the form over the circle with radius $1$ -- you need to be careful about orientation! $\endgroup$
    – t.b.
    May 15, 2011 at 22:13
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    $\begingroup$ Well, the support of $\omega$ is the whole of your open annulus, which is not compact. So the Stokes theorem doesn't apply as stated. $\endgroup$
    – Zhen Lin
    May 15, 2011 at 22:13
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    $\begingroup$ OK, I understand where the discrepancy of $2\pi$ comes from, yes, but I suppose it just seems odd to me that the requirement of being compactly supported can carry such weight. Especially for forms like $\omega = -y\,dx + x\,dy$, which don't blow up at the boundary or anything... $\endgroup$ May 15, 2011 at 22:17
  • 7
    $\begingroup$ The point is not about blowing up at the boundary, but about the non-vanishing of the form at the boundary, which complicates the integration by parts. $\endgroup$
    – Henri
    May 15, 2011 at 22:37
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    $\begingroup$ Dear Jesse, "It just seems odd to me that the requirement ... can carry such weight". The example in your post shows exactly why this condition is important: otherwise there can be "missing" boundary contributions. The form $-y dx + x dy$, which is not compactly supported, "feels" the missing boundary along $r = 1$ in the way that a compactly supported form wouldn't. Regards, $\endgroup$
    – Matt E
    May 16, 2011 at 3:24

1 Answer 1


Annulus with $1 < r < 2$ does not have a boundary and the form you pick is not compactly supported there. The form $\omega$ only vanishes at the origin so its support is in fact open in $\mathbb{R}^2$.


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