Let $A$ be any vector field, then by Stokes Theorem we have:

$$ \oint_{\gamma} \mathbf{A} \cdot d \mathbf{r}=\int_{S} \operatorname{curl} \mathbf{A} \cdot d \mathbf{S} $$

We can now apply Divergence Theorem to $\operatorname{curl}\mathbf{A}$, which using the fact that divergence of curl is $0$, gives:

$$ \int_{S} \operatorname{curl} \mathbf{A} \cdot d \mathbf{S} = \int_{\tau} \operatorname{div} ( \operatorname{curl} \mathbf{A} ) d \tau = 0 $$

So we could conclude for any vector field $A$:

$$ \oint_{\gamma} \mathbf{A} \cdot d \mathbf{r} = 0$$

What is wrong here?

  • 1
    $\begingroup$ What sort of things are you integrating over? $\endgroup$ Dec 21, 2019 at 17:15
  • $\begingroup$ What do you mean? If it adds any context this is a very open ended question - I didn't have any specific region or any restriction in mind, are there any specific cases when this is true/untrue? $\endgroup$
    – analysis1
    Dec 21, 2019 at 17:25
  • $\begingroup$ I mean, what are these things, $\gamma$, $S$ and $\tau$ that you are integrating over? $\endgroup$ Dec 21, 2019 at 17:26
  • $\begingroup$ Ah ok, in Divergence Theorem the assumption is that S is a closed surface and tau is the region inside and in Stokes we have an open surface S and gamma is the boundary curve. So the two don't play together. Is that right? $\endgroup$
    – analysis1
    Dec 21, 2019 at 17:59

1 Answer 1


Your second equation is the reason for such a discrepancy: $$ \int_{S} \operatorname{curl} \mathbf{A} \cdot d \mathbf{S} = \int_{\tau} \operatorname{div} ( \operatorname{curl} \mathbf{A} ) d \tau = 0 $$

Here you combined Stokes' theorem with the three-dimensional divergence theorem, remember that the divergence theorem is only applicable on "closed surfaces" meanwhile the surface obtained from stokes theorem is usually not closed in three-dimensions, except for very special cases in which your statement holds true.

  • 1
    $\begingroup$ Yeah that makes sense, thanks! If I had a closed surface can I cut it in half and apply Stokes separately to both parts? $\endgroup$
    – analysis1
    Dec 21, 2019 at 18:04
  • $\begingroup$ You need to recall that stokes theorem ensures surface independence, which means that as long as two surfaces have the same boundary they have the same line integral over the boundary regardless to how different they are as surfaces. $\endgroup$
    – Malzahar
    Dec 21, 2019 at 18:16
  • $\begingroup$ so to answer your question, there is no need to apply stokes on both parts but if you did both would be of the same magnitude but different sign so they would add up to zero, which verifies your statement. $\endgroup$
    – Malzahar
    Dec 21, 2019 at 18:19
  • 1
    $\begingroup$ The surface $S$ could be closed, if $\gamma$ is a constant loop, for instance. Or a loop that goes from $A$ to $B$ and then back to $A$ along the exact same points. But then the integral over $\gamma$ is necessarily $0$, at least for any differentiable $A$. $\endgroup$
    – Arthur
    Dec 21, 2019 at 18:21
  • $\begingroup$ I believe that @analysis1 's example implies your second case. $\endgroup$
    – Malzahar
    Dec 21, 2019 at 18:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .