Wolfgang Wieser wrote:
> B: VRand_In_Sphere(), full sphere.
> 
> For the full sphere, there is not much difference, just the
> radius is chosen to be randomly distributed using
> pow(rand(Stream),1/3)).
> I don't believe that this is right until someone (me?) proves
> it to be right - ;)
> 
As shown last time, we already have a function for the uniform 
distribution on a sphere surface. For the VRand_In_Sphere() function, 
it is therefore sufficient to prove that the radius calculation 
(i.e. vector length) is correct. 

For that purpose, it must be shown that the probability for the 
vector being of a certain length range is equal to the volume 
enclosed by that radius range (divided by 4*pi/3, of course). 

The sphere volume between radii r1 and r2 (r2>r1) is: 
  V = 4*pi/3 * (r2^3 - r1^3)

The vector length l is calculated as l=a^(1/3) with a=rand(Stream), 
hence l^3 is uniformly distributed in the range 0..1. 

The probability for the vector to end in the volume V (i.e. radius 
range r1..r2) is: 
  P(vector in V) = P(r1 <= l < r2) = P(r1^3 <= l^3 < r2^3) = 
and as l^3=a:
  = P(r1^3 <= a < r2^3) = r2^3 - r1^3

Hence, the quotient (probability per volume) is: 
  P/V = 3/(4*Pi)
...which is what we want. 

- q.e.d. 

So, it works; VRand_In_Sphere() is uniformly distributed in the sphere. 

Any doubts or problems?

Wolfgang

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