
  [Contact forms; integral surfaces].
Let $\alpha$ be a non-vanishing one-form $\alpha$ on $M$. 
At each $m \in M$ the kernel of the linear functional  $\alpha (m): T_m M \to \R$ 
defines a linear hyperplane $D_m \subset T_m M$ by $D_m = \ker(\alpha(m))$.
An {\it integral hypersurface} for $\alpha$ (or for $D$)   is an embedded hypersurface $\Sigma \subset M$
 everywhere tangent to $D$: meaning that  $D_p =  T_p \Sigma$ for all $p \in \Sigma$.

(a)   Show that the embedded hypersurface $\Sigma$ is an integral hypersurface if and only
if $i^* \alpha = 0$ where  $i: \Sigma \to M$ denotes  the inclusion. 

(b)  Show that the one-form 
$\alpha = dz - ydx$ on $\R^3$ admits no integral surface passing through the origin.
Do so by using the result of problem 1 [copied from HW 1 above]  and obtaining a contradiction.

Follow the with standard coordinates $(x,y,z)$. 
{\bf Prove} that $\alpha$ admits no integral hypersurfaces.


\noindent{\bf 3.} Problem 6-7 (Chapter 6) on page 152 of the textbook.