Sampling uniformly distributed random points inside a spherical volume

Generate a set of points uniformly distributed within a cube, then discard the ones whose distance from the center exceeds the radius of the desired sphere.

While I prefer the discarding method for spheres, for completeness I offer the exact solution.

In spherical coordinates, taking advantage of the sampling rule:

phi = random(0,2pi)
costheta = random(-1,1)
u = random(0,1)

theta = arccos( costheta )
r = R * cuberoot( u )

now you have a (r, theta, phi) group which can be transformed to (x, y, z) in the usual way

x = r * sin( theta) * cos( phi )
y = r * sin( theta) * sin( phi )
z = r * cos( theta )

I agree with Alleo. I translated your Matlab code to Python and it can generate thousands of points very fast (a fraction of second in my computer for 2D and 3D). I've even ran it for up to 5D hyperspheres. I found your code so useful that I'm applying it in a study. Tim McJilton, who should I add as reference?

import numpy as np
from scipy.special import gammainc
from matplotlib import pyplot as plt
def sample(center,radius,n_per_sphere):
    r = radius
    ndim = center.size
    x = np.random.normal(size=(n_per_sphere, ndim))
    ssq = np.sum(x**2,axis=1)
    fr = r*gammainc(ndim/2,ssq/2)**(1/ndim)/np.sqrt(ssq)
    frtiled = np.tile(fr.reshape(n_per_sphere,1),(1,ndim))
    p = center + np.multiply(x,frtiled)
    return p

fig1 = plt.figure(1)
ax1 = fig1.gca()
center = np.array([0,0])
radius = 1
p = sample(center,radius,10000)
ax1.scatter(p[:,0],p[:,1],s=0.5)
ax1.add_artist(plt.Circle(center,radius,fill=False,color='0.5'))
ax1.set_xlim(-1.5,1.5)
ax1.set_ylim(-1.5,1.5)
ax1.set_aspect('equal')

There is a brilliant way to generate uniformly points on sphere in n-dimensional space, and you have pointed this in your question (I mean MATLAB code).

Why does it work? The answer is: let us look at the probability density of n-dimensional normal distribution. It is equal (up to constant)

exp(-x_1*x_1/2) *exp(-x_2*x_2/2)... = exp(-r*r/2), so it doesn't depend on the direction, only on the distance! This means, after you normalize vector, the resulting distribution's density will be constant across the sphere.

This method should be definitely preferred due to it's simplicity, generality and efficiency (and beauty). The code, which generates 1000 events on the sphere in three dimensions:

size = 1000
n = 3 # or any positive integer
x = numpy.random.normal(size=(size, n)) 
x /= numpy.linalg.norm(x, axis=1)[:, numpy.newaxis]

BTW, the good link to look at: http://www-alg.ist.hokudai.ac.jp/~jan/randsphere.pdf

As for having uniform distribution within a sphere, instead of normalizing a vector, you should multiply vercor by some f(r): f(r)*r is distributed with density proportional to r^n on [0,1], which was done in the code you posted