# Confusion about differential forms and spherical coordinates

Let $$x,y,z$$ denote the standard coordinates of $$\Bbb R^3$$, and let $$r,\phi,\theta$$ denote the spherical coordinates on $$\Bbb R^3-\{0\}$$. It is well-known that $$dx \wedge dy\wedge dz=r^2 \sin \phi ~dr\wedge d\phi \wedge d\theta$$. But, since $$\phi=0$$ on the positive $$z$$-axis and $$\phi=\pi$$ on the negative $$z$$-axis, this means that $$dx\wedge dy\wedge dz$$ vanishes on the $$z$$-axis. However, $$dx\wedge dy\wedge dz$$ is a top form on $$\Bbb R^3$$ and it must be a basis of $$\bigwedge^* (T_p\Bbb R^3)$$ for each point $$p\in \Bbb R^3$$. What went wrong?

Spherical coordinates don't cover all of $$\mathbb{R}^3\setminus\{0\}$$. Typcally, one restricts to the ranges $$r\in(0,\infty),\ \ \ \ \ \theta\in(0,2\pi),\ \ \ \ \ \phi\in(0,\pi)$$ which maps to a domain (which notably does not include the $$z$$-axis) $$U=\left\{(x,y,z)\in\mathbb{R}^3:y\neq0\text{ or }x<0\right\}$$ It isn't possible to extend the coordinates further without running into discontinuities. The volume form $$\omega=r^2\sin(\phi)dr\wedge d\phi\wedge d\theta$$ is only defined on $$U$$ as written, where it is equal to $$dx\wedge dy\wedge dz$$. While it's possible to smoothly extend $$\omega$$ to all of $$\mathbb{R}^3$$, one cannot assume that the coordiante functions $$r,\theta,\phi$$ (or their derivatives) can be extended: if you want to describe the behavior outside of the domain $$U$$, it would be best to change to a new set of coordinates which cover the points of interest.
The behavior around the $$z$$-axis gives an example of the pitfalls of assuming $$r,\theta,\phi$$ extend smoothly beyond their domains: even though $$\sin(\phi)$$ approaches zero, $$d\theta$$ has a singularity, and it's not obvious in polar coordinates that these two divergences "cancel out" to produce a nonzero limit.