Let $\alpha$ be a $1$-form defined on a manifold $M$ and $\Delta = ker (\alpha)$. The classical theorem of Frobenius says that $\Delta$ is integrable if $\alpha \wedge d\alpha =0$ i.e if $d\alpha$ is zero restricted to $\Delta$. Using such things as Lie brackets, it is more or less easy to show the equivalance of this condition to integrability.
However I have always felt that there must be some relation between Stokes' theorem and integrability but I have never been able to construct this relationship. To see this relation though, I believe it is sufficient to prove the following lemma using Stokes' theorem only and no Lie brackets!
Lemma. Let $M$ be a three dimensional manifold, $\alpha$ a $1$-form, $\Delta$ it is $2$-dimensional kernel. Let $X,Y$ be two local vector fields in $\Delta$ defined over some nbd $U$. Let $\gamma_X(t,p),\gamma_Y(t,p)$ be it is integral curves for $p \in U$, $t\in \mathbb{R}$. Then $\alpha \wedge d\alpha=0$ implies the two dimensional surface
$$S(t,s)=\gamma_X(t,\gamma_Y(s,p))$$
is everywhere tangent to $\Delta$ (i.e its tangent space lies in $\Delta$).
Again as I said this is easy to prove using such notions as Lie brackets but I only want a proof using Stokes' theorem. The problem in applying Stokes' theorem is I don't a priori know if there are pieces of surfaces everywhere tangent to $\Delta$ (this is what I want to prove anyway) and without knowing that it is not very helpful to know $d\alpha$ restricted to $\Delta$ is $0$. Can any body give a proof of this using only Stokes' theorem or a reference for some book or an article proving this integrability condition with Stokes' theorem? Thanks.