Finding an integral submanifold of an involutive distribution I've been given the following vector field on the manifold $M = \mathbb{R}^3 \setminus \{\text{$z$-axis}\}$ $$X = (x-2y)\,\frac{\partial}{\partial x} + (2x+y)\, \frac{\partial}{\partial y} + z\, \frac{\partial}{\partial z},\quad Y = -y\,\frac{\partial}{\partial x} + x\, \frac{\partial}{\partial y}$$
It was straightforward to show that they're linearly independent and commute, therefore they define an involutive rank $2$ distribution, say $D$. The question is find and describe its integral submanifolds. I tried the following methods:

*

*I was able to figure out that the form $\omega = - xz\,dx - yz\,dy + (x^2 + y^2)\,dz$ annihilates $X$ and $Y$ and $d\omega = 3x\,dx\,dz + 3y\,dy\,dz$. Since our distribution is involutive, naturally $d\omega\wedge \omega = 0$. But I've been unable to figure out an $\alpha$ such that $d\omega = \omega\wedge\alpha$. This could then help me figure out the integral submanifolds


*Alternatively, we can the flows of $X$ and $Y$, which I did after a painstaking computations but I'm not sure how to proceed after that.
Any help would be appreciated.
 A: I'll assume you mean $\require{color}Y=-y\dfrac{\partial}{\partial{\color{red}x}}+x\dfrac{\partial}{\partial y}$ instead of $Y=-y\dfrac{\partial}{\partial{\color{green}z}}+x\dfrac{\partial}{\partial y}$ (which doesn't give $[X,Y]=0$ in fact $[X,Y]\notin\operatorname{span}\{X,Y\}$).  Then using cylindrical polars, we see
$$
X=\rho\frac{\partial}{\partial\rho}+2\frac{\partial}{\partial\theta}+z\frac{\partial}{\partial z},\quad Y=\frac{\partial}{\partial\theta}
$$
So $X,Y$ commutes and the integral submanifolds are given by $z/\rho=\text{constant}$, i.e., (one-sided) cone with the $z$-axis being its axis of symmetry.
A: Note that there is an easy integrating factor to guess for your $\omega$: Multiply through by $f=\dfrac1{(x^2+y^2)z}$ and the resulting form is exact. Indeed, $f\omega = d(-\log r + \log z)$, where $r=\sqrt{x^2+y^2}$. Integral manifolds are then $z=cr$ for various constants $c$.
Following Sophus Lie, there is a good way to look for such integrating factors. If you see an obvious one-parameter group leaving the distribution invariant (or sending $\omega$ to a functional multiple of $\omega$), let $V$ be its infinitesimal generator and let $f=1/\omega(V)$. Then you can check that $f\omega$ will be closed. In your case, the obvious one-parameter group is given by radial stretches, and $V=x\dfrac{\partial}{\partial x}+y\dfrac{\partial}{\partial y}+z\dfrac{\partial}{\partial z}$. Unfortunately, the numbers work out to $\omega(V)=0$ — in other words, $\mathscr L_V\omega = 0$ rather than just a general functional multiple of $\omega$. But, still, we will see the appearance of $(x^2+y^2)z$ when we do this computation.
EDIT: A natural way to avoid the issue with $z=0$ about which you're worried is to introduce a homogenizing variable $u$ with $z=ur$, $r=\sqrt{x^2+y^2}$. Then
$$\omega = -zr\,dr + r^2\,dz = -ur^2\,dr + r^2(u\,dr + r\,du) = r^3\,du,$$
and so, naturally, $\dfrac 1{r^3}\omega = du$ is exact. (So, here, in the new coordinate system, $1/r^3$ does become the integrating factor!) Oh, and finally, the integral manifolds are patently obvious: They're given by $u=c$ for all constants $c$.
A: Just for the record, here's a way to do it with minimal guessing: The key observations needed are seeing that $\omega$ looks simpler written in cylindrical coordinates and recognizing the formula for the exterior derivative of a quotient.
\begin{align*}
  \omega &= -xz\,dx - yz\,dy + (x^2+ y^2)\,dz\\
         &= -z(x\,dx + y\,dy) + (x^2+y^2)\,dz\\
         &= -\frac{z}{2}(2x\,dx + 2y\,dy) + (x^2+y^2)\,dz\\
         &= -\frac{z}{2}d(x^2+y^2) + (x^2+ y^2)\,dz\\
         &= -\frac{z}{2}d(r^2) + r^2\,dz\\
         &= -z(r\,dr) + r^2\,dz\\
         &= r(-z\,dr + r\,dz)
\end{align*}
where $x^2 + y^2 = r^2$. Now observe that the second factor looks like the numerator of the exterior derivative of a quotient, namely
$$
  d\left(\frac{z}{r}\right) = \frac{r\,dz - z\,dr}{r^2}.
$$
Therefore,
\begin{align*}
  \omega &= r^3\,d\left(\frac{z}{r}\right)
\end{align*}
It follows that the integral manifolds are the level sets given by
$$
  \frac{z}{r} = c
$$
outside the $z$ axis. These are of course just the cones given by
$$
  z = cr
$$
