Is $\zeta=\frac{x dy \wedge dz+y dz \wedge dx+z dx \wedge dy}{r^3}$ exact in the complement of every line through the origin?

Question

$r=\sqrt{x^2+y^2+z^2}$ of course.

If the line is the $z$ axis, it is given in the book (Rudin) that $\zeta=d \left( -\dfrac{z}{r} \dfrac{xdy-ydx}{x^2+y^2} \right)$

I've managed to figure out 2 similar cases by myself:

$x$ axis: $\zeta=d \left( -\dfrac{x}{r} \dfrac{ydz-zdy}{y^2+z^2} \right)$

$y$ axis: $\zeta=d \left( -\frac{y}{r} \dfrac{zdx-xdz}{x^2+z^2} \right)$

I have a strong feeling that it's true for any line, but I'm having great difficulties of proving so. Any suggestions?

Answer 1

Let $A,B,C$ be an orthonormal (and positively oriented/ordered) basis of $\mathbb R^3$ with $A$ in the direction of your line. Given position $p = (x,y,z),$ define new coordinate functions $a = p \cdot A, \; b = p \cdot B, \; c = p \cdot C.$ Then you get $da, \; db, \; dc.$ Then try $$ d \left( - \frac{a}{r} \; \frac{b dc - c db}{b^2 + c^2} \right) $$