Consider the set $S = \{(x,y,z) \in \mathbb{R}^3 \mid x^2+y^2+z^2=1, z\geq 0\}$.
We need to compute $$\int_S \omega $$ on $S$ where $\omega$ is given as follows:
$\omega = dydz/x$ when $x \neq 0$, $\omega = dzdx/y$ when $y \neq 0$, $\omega = dxdy/z$ when $z \neq 0$.
My guess is to cover $S$ with 3 coordinate charts such that they match the conditions of domain of definition of $\omega$.
Can someone show the explicit computation?
Since you're only integrating over the upper hemisphere, you can work just with the formula $\omega = \dfrac{dx\wedge dy}z$. (Officially, you are missing a set of measure $0$ when you integrate over the open hemisphere, rather than the closed hemisphere. However, the fact that $\omega$ is given by the other formulas when $z=0$ shows you that in fact $\omega$ is smooth everywhere and there is no problem. By the way, you should check why $$\frac{dx\wedge dy}z = \frac{dy\wedge dz}x = \frac{dz\wedge dx}y \quad\text{when } x,y,z\ne 0.)$$ At any rate, this gives an improper integral, but it converges just fine. Switching to polar coordinates, we get $$\int_0^{2\pi}\int_0^1 \frac r{\sqrt{1-r^2}}dr\wedge d\theta.$$ You can finish from here.