I have the following problem:
let M be the unit sphere and $\omega=x~dy\wedge dz+y~dz\wedge dx+z~dx\wedge dy$ be our $2$-Form. we want to compute $$\int_M \omega$$
In the lecture we did the following. Take the parametrisation $\phi(\theta,\rho)=(\cos\theta\sin\rho,\sin\theta\sin\rho,\cos\rho)$ where $\theta\in [0,2\pi]$ and $\rho \in [0,\pi]$. We know that this parametrisation is an orientable chart. Till this point everything is clear to me. But then we wrote that for this parametrisation the following point $$\{y=0,x>0\}$$ are missing, but since this set has measure zero one can "ignore" this fact.
And that's the thing I don't understand. First of all I don't get where I can see that exaclty this points are missing, and therefore I also can't see why this is a nullset.
Could someone explain this to me because the rest (computations) are clear afterwards.
Thanks for your help.
This is just a write-up of the comments on the question:
It depends on the definitions you are using, but in order to have a parametrization that is an inverse of a chart you need the domain to be an open set. So you want $\theta \in (0,2\pi)$ and $\rho \in (0,\pi)$. But with this open set you cannot reach points of the form
$$\phi(0,\rho)=(\cos(0)\sin(\rho),\sin(0)\sin(\rho),\cos(\rho)).$$
If you inspect those points you find that this is a curve on an equator from $(0,0,1)$ to $(0,0,-1)$. Since the sphere is two dimensional this gives a set of measure zero.