# 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

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