I think the diagram is wrong: phi should be the polar angle.
If we use the diagram phi = 0 then z should be 0, but cos(0) = 1.
@jasonx Yes, correct, I mention this in an instructor note on slide 14. Apologies.
also if know the solid angle differential is $\sin \theta d\theta d\phi$ then that gives you $\sin \theta$ as PDF (up to scaling).
I think the diagram is wrong: phi should be the polar angle.
If we use the diagram phi = 0 then z should be 0, but cos(0) = 1.
@jasonx Yes, correct, I mention this in an instructor note on slide 14. Apologies.
also if know the solid angle differential is $\sin \theta d\theta d\phi$ then that gives you $\sin \theta$ as PDF (up to scaling).