By Frobenius Theorem, in $\mathbb{R}^3$ there exists a smooth surface whose tangent space is spanned by the vector fields $V(x,y,z)=(x^2+y^2,0,-y)$ and $W(x,y,z)=(0,x^2+y^2,x)$. How can I find this surface? Is there in general a way to find it when the vector fields are algebraic?
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Note that both vector fields vanish on the $z$-axis, so you really want to talk about $\Bbb R^3 - {x=y=0}$. – Ted Shifrin Jul 17 '16 at 00:18
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Note that $\omega=y\,dx - x\,dy+(x^2+y^2)\,dz$ annihilates both $V$ and $W$. Also note that if we set $f(x,y,z)=1/(x^2+y^2)$, then $d(f\omega) = 0$, so $f\omega$ is locally exact. Indeed, the integral manifolds are the graphs $z=\theta+c$ of the multivalued function $\theta(x,y)$ (defined away from $x=y=0$).
Ted Shifrin
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