 Solution

The differential-forms version of the Frobenius theorem says that a constant-rank distribution $D={\bigcap }_i\ker {\alpha }^i$ is integrable exactly whenfor suitable 1-forms ${\beta }^i{}_j$. For a plane distribution $D=\ker \alpha$ on ${\mathbb{R}}^3$, this reduces to the integrability criterion for a plane distribution $\alpha \wedge d\alpha =0$.

For example, $\ker dz$ is integrable: its integral surfaces are the horizontal planes $z=\text{constant}$. In contrast, for $\alpha =dz-xdy$,Thus $\ker (dz-xdy)$ is not integrable; it is the standard contact structure on ${\mathbb{R}}^3$.