I'm studying distributions in sense of manifolds. First of all, I didn't study differential forms.
I'm trying to understand how to calculate integral submanifolds of involutive distributions.
So, given the vector fields
$$X=x \frac{\partial}{\partial y}-y \frac{\partial}{\partial x} \quad \text { and } \quad Y=\frac{\partial}{\partial z},$$
let $D$ the distribution generated by $X$ and $Y$.
I proved that the lie bracket $[X,Y]=0.$ Then $D$ is involutive and by Frobenius theorem we have $D$ integrable.
I'm trying to calculate the integrable submanifolds of $D$.
Let $p=(x,y,z).$ The integral curves of these vectors field with initial condition $p$ are:
For $X$:
$$\gamma(t)=(-y\sin(t)+x\cos(t), y\cos(t)+x\sin(t),z)$$
and for $Y$:
$$\lambda(t)=(x,y,z+t).$$
Now I'm trying to proceed in similar way of John Lee's book. (Proposition $19.28$ or Example $19.14$)
Let $$\eta(t,s,w)=\psi_{t}^{X}\circ \psi_{s}^{Y}(0,0,w)=(0,0,w+s),$$ where $\psi_{t}^{X}$ and $\psi_{s}^{Y}$ are the flow of $X$ and $Y$, respectively.
In this case, the flat chat should be the inverse of $\eta.$ But, $\eta$ is not inversible.
I did some examples when we have vector fields:
$$X=\dfrac{\partial}{\partial x}+\alpha(x,y,z)\dfrac{\partial}{\partial z} \text{ and } Y= \dfrac{\partial}{\partial y}+\beta(x,y,z)\dfrac{\partial}{\partial z}$$
and $\alpha,\beta$ are smooth real-valued functions defined in some open set of $\mathbb{R}^{3}$ satisfying
$$\frac{\partial \alpha}{\partial y}+\beta \frac{\partial \alpha}{\partial z}=\frac{\partial \beta}{\partial x}+\alpha \frac{\partial \beta}{\partial z}.$$
The method that I described above work well.
What is my error here? Have some problem is point $(0,0,0)$?
Can someone help me, please?
Just for further reference I offer a solution using a bit of differential forms.
A one-form that annihilates both vector fields is $$ \boldsymbol{\omega}=x\,dx+y\,dy\,. $$ From $d\boldsymbol{\omega}=0$ it follows that Frobenius' condition $d\boldsymbol{\omega}\wedge\boldsymbol{\omega}=0$ holds and that $\boldsymbol{\omega}$ is exact: $$ \boldsymbol{\omega}=df=d\frac{x^2+y^2}{2}\,. $$ The integral manifolds are then the level sets of the function $f\,.$ In $\mathbb R^3$ each such level set is the boundary of a cylinder whose axis of symmetry is the $z$-axis. The field $\partial_z$ is obviously tangent to that boundary. The normal vector of the boundary is the gradient of $f\,:$ $$ \boldsymbol{n}=\begin{pmatrix}x\\y\end{pmatrix} $$ which is obviously orthogonal to the other vector field $X\,.$ Hence, $X$ is also tangent.