The solid angle for an arbitrary oriented surface $S$ subtended at a point $P$ is equal to the solid angle of the projection of the surface $S$ to the unit sphere with center $P$ (https://en.wikipedia.org/wiki/Solid_angle)
So let $S\subseteq \mathbb{R}^3$ be a smooth oriented surface, and to start, suppose it is parametrized by $\phi: V \rightarrow S$. How do I get $\Omega$, the aforementioned solid angle covered by $S$ with respect to $P$?
My try
The projected surface $S'$ is $f(S)=S'$, where $f(x)\displaystyle=P+\frac{x-P}{||x-P||}$. Hence a parametrization of $S'$ is $\psi:=f \circ \phi$, and $\Omega=\displaystyle\int_V ||\psi_{,u} \wedge \psi_{,v}||dudu$. I want to get to Wikipedia's formula $\Omega=\displaystyle \int_S \frac{\hat{r}\cdot n}{r^2}dS$, where "${\displaystyle {\hat {r}}={\tfrac {\vec {r}}{r}}}{\displaystyle {\hat {r}}={\tfrac {\vec {r}}{r}}}$ is the unit vector corresponding to ${\displaystyle {\vec {r}}}$, the position vector of an infinitesimal area of surface $dS$ with respect to point $P$, and where ${\displaystyle {\hat {n}}}$ represents the unit normal vector to $dS$".
So, by computations, $\psi_{,u}=\phi_{,u}-(\hat {r}\cdot \phi_{,u})\hat {r}$, and therefore $\psi_{,u} \wedge \psi_{,v}=\phi_{,u} \wedge \phi_{,v}+\hat {r}\wedge ((\hat {r} \cdot\phi_{,v})\phi_{,u}-(\hat {r} \cdot\phi_{,u})\phi_{,v})=||\phi_{,u} \wedge \phi_{,v}||(n+\hat {r}\wedge(\hat {r}\wedge n))$, so that $||\psi_{,u} \wedge \psi_{,v}||=||n+\hat {r}\wedge(\hat {r}\wedge n)||||\phi_{,u} \wedge \phi_{,v}||$.
So we need to prove $||n+\hat {r}\wedge(\hat {r}\wedge n)||=\displaystyle \frac{\hat{r}\cdot n}{r^2}$.
Do you have any idea on how to proceed? How do you generalize this to an $S$ not necessarily globally parametrizable?