Asymmetry parameter of a phase function

[Mimmo Briganti]

Hi everybody.
On chapter 12, p.581, the first formula on the page represents the
asymmetry parameter, g, for an arbitrary phase function.
In my opinion, the second integral misses a sin(theta) term, due to
the Jacobian of the transformation.
So

2\pi \int_0^{\pi} p(\cos \theta) \cos\theta d\theta

should be

2\pi \int_0^{\pi} p(\cos \theta) \cos\theta \sin\theta d\theta.

Am I wrong?

Regards,
Mimmo Briganti

[Matt Pharr]

(Very late reply, sorry. :-( ).

I think that you are wrong in fact. That sin theta jacobian term
comes in when integrating over the full sphere, in particular from the
phi part of integrating over (theta phi). Toward the top/bottom of
the sphere, a full trip around the sphere in phi covers a much smaller
(differential) area than toward the middle.

Here, the function is constant over phi, so the 2pi term can be pulled
out. Intuitively, if we still integrated over (theta, phi) here and
had a function like p() that is independent of phi, we'd evaluate a
lot more samples of it towards the top/bottom of the sphere and the
sin theta term would be needed to compensate for that. Hope this
helps...

Thanks,
-matt