Definition:
Suppose $D \subset TM$ is a smooth distribution of rank $k$. An embedded submanifold $S^{k} \subset M$ is called an integral submanifold of $D$ if $$T_p S=D_p \quad \forall p \in S.$$
My problem is:
Consider the smooth vector fields \begin{equation} X:=z\partial_y+y\partial_z, \quad Y:=-x\partial_z-z\partial_x \quad \mbox{e}\quad Z:=y\partial_x-x\partial_y\end{equation} in $M:=\mathbb{R}^{3}-\{0\}$.Describe all integral submanifolds of the distribution $$D=\langle X,Y, Z \rangle \quad \mbox{(spanned by the vector fields)}.$$
I checked that $D$ is an involutive distribution of rank $2$. However I don't know how to describe all integral submanifolds of $D$. How can I do this?
Thank you for your answer.
In general, this can be a very hard question. However, in this particular case, as it is of rank 2, you could try and find a differential $1$-form whose kernel coincides with $D_p$ at every point. Quite easily, if you set $\omega=a(x,y,z)dx+b(x,y,z)dy+c(x,y,z)dz$ and ask that $\omega(X)=\omega(Y)=\omega(Z)=0$, you get the conditions $$zb+yc=0\qquad -xc-za=0\qquad ya-xb=0$$ which implies that $a(x,y,z)=xh(x,y,z)$, $b(x,y,z)=yh(x,y,z)$, $c(x,y,z)=-zh(x,y,z)$ for some function $h(x,y,z)$. Therefore, we obtain an ideal (as we expect!) which is principal and generated by $\omega_0=xdx+ydy-zdz$.
So, we have a differential $1$-form such that $D_p=\ker(\omega_0)\vert_p$ for every $p\in\mathbb{R}^3$. As we know that $D$ is involutive, we expect to find a potential for $\omega_0$ and indeed $2\omega_0=d(x^2+y^2-z^2)$.
So, the level sets of $f(x,y,z)=x^2+y^2-z^2$ are the integral submanifolds of $D$.