# Combining Stokes and Divergence Theorem

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?

• What sort of things are you integrating over? Dec 21 '19 at 17:15
• 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? Dec 21 '19 at 17:25
• I mean, what are these things, $\gamma$, $S$ and $\tau$ that you are integrating over? Dec 21 '19 at 17:26
• 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? Dec 21 '19 at 17:59

## 1 Answer

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.

• Yeah that makes sense, thanks! If I had a closed surface can I cut it in half and apply Stokes separately to both parts? Dec 21 '19 at 18:04
• 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. Dec 21 '19 at 18:16
• 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. Dec 21 '19 at 18:19
• 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 $A$. Dec 21 '19 at 18:21
• I believe that @analysis1 's example implies your second case. Dec 21 '19 at 18:43