How to find the maximal integral submanifold in a concrete case? Take $M=\mathbb{R}^3$ be a smooth manifold. Consider a distribution
$\Delta_{(x,y,z)} = Span\{y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y},
z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z} \}$.
1) Show that the distribution is integrable.
2) Describe the maximal integral submanifolds.
Here is some of my drafts:
Take $V = y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}$ and $W = z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z}$, then the Lie bracket
$[V,W]=VW-WV=-y\frac{\partial}{\partial z}+z\frac{\partial}{\partial y}$.
If $[V_p, W_p]\in \Delta_p$, then the distribution is involutive and further integrable. However, for $p=(0,y,z)$, $V_p=y\frac{\partial}{\partial x}$, $W_p = z\frac{\partial}{\partial x}$, and $[V_p, W_p]=-y\frac{\partial}{\partial z}+z\frac{\partial}{\partial y}$. Easy to see the Lie bracket is not in the span. If this is true, then the distribution is not integrable. Could anyone remind me what is wrong?
As to the second part, I did not have a clear clue of how to complete. A concrete computation would be appreciated.
 A: Let $f\in C^\infty(\Bbb R^3).$ We have 
\begin{align*} [V,W]f &= V(Wf)-W(Vf) \\
&= (y\frac{\partial}{\partial x}-x\frac{\partial}{\partial z})(z\frac{\partial f}{\partial x}-x\frac{\partial f}{\partial z}) - (z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z})(y\frac{\partial f}{\partial x}-x\frac{\partial f}{\partial z}) \\
&= y\left(0\frac{\partial f}{\partial x} + z\frac{\partial^2 f}{\partial x^2} - 1\frac{\partial f}{\partial z} - x\frac{\partial^2 f}{\partial x\partial z}\right) - x\left(1\frac{\partial f}{\partial x} + z\frac{\partial^2 f}{\partial z\partial x} - 0\frac{\partial f}{\partial z} - x\frac{\partial^2 f}{\partial z^2}\right) \\ &~~~- z\left(0+y\frac{\partial^2 f}{\partial x^2} -1\frac{\partial f}{\partial z} -x\frac{\partial^2 f}{\partial x\partial z}\right) + x\left(0+y\frac{\partial^2 f}{\partial z\partial x} - 0 - x\frac{\partial^2 f}{\partial z^2}\right) \\
&= (yz-zy)\frac{\partial^2 f}{\partial x^2} + (-yx-xz+zx+xy)\frac{\partial^2 f}{\partial z\partial x} + (x^2-x^2)\frac{\partial^2 f}{\partial z^2} \\ 
&~~~+ \left(-y\frac{\partial f}{\partial z} -x\frac{\partial f}{\partial x} +z\frac{\partial f}{\partial z} \right) \\
&= (z-y)\frac{\partial f}{\partial z} -x\frac{\partial f}{\partial x}
\end{align*}
which shows that your calculation for the Lie bracket is incorrect. The Lie bracket should be $[V,W]=(z-y)\frac{\partial}{\partial z}-x\frac{\partial}{\partial x}.$
A: The tangent distribution $\Delta$ on $\mathbb R^3$ is spanned by $E_2=-x\partial_z+z\partial_x$ and $E_3=-y\partial_x+x\partial_y,$ which are the infinitesimal generators of the rotations of $\mathbb R^3$ respectively around the $y$-axis and the $z$-axis.  
Computing the minors of $\begin{pmatrix}z & 0 & -x\\ -y & x & 0\end{pmatrix},$ we observe that $\Delta$ is a singular tangent distribution on $\mathbb R^3,$ because $$\textrm{rank}\Delta_{(x,yz)}=\begin{cases}
0 & \textrm{ if } x=y=z=0,\\
1 & \textrm{ if } x=0 \textrm{ and } y^2+z^2\neq 0,\\
2 & \textrm{ if } x\neq 0.
\end{cases}$$
As you have computed $[E_2,E_3]=-y\partial_z+z\partial_y\equiv-E_1,$ where $E_1$ is the infinitesimal generator of the rotations around the $x$-axis.
Observe that $xE_1=yE_2+zE_3,$ so $\Delta$ is involutive on $\{(x,y,z)\mid x\neq 0\}.$
The Lie algebra generated by $E_2$ and $E_3$ is $\mathfrak{so}(3)=\mathrm{span}\{E_1,E_2,E_3\},$ the Lie algebra of the infinitesimal generators of the natural action $\mathrm{SO}(3)$ on $\mathbb R^3:$
$$(A,p)\in\mathrm{SO}(3)\times\mathbb R^3\to A.p\in\mathbb R^3.$$
Therefore the maximal integral manifolds of $\mathrm{span}\{E_1,E_2,E_3\}$ are the $\mathrm{SO}(3)$-orbits, i.e. the spheres centered at the origin.
