In fact, for any R^2, assume that X^2 + Y^2 + Z^2 = R^2. Then by symmetry the distribution of X/R^2 is the same of the x-coordinate on [a sphere of radius 1 with the uniform distribution]. Since the area of this sphere lying between a and b for -1 <= a <= b <= 1 is proportional to b-a, it follows that the original X/(X^2+Y^2+Z^2) has a uniform distribution on [-1,1]. --Dan << << And here's a fun fact related to Archimedes' insight about the sphere and the cylinder: If X, Y, Z are independent Gaussians of mean 0 and variance 1, X/(X^2+Y^2+Z^2) is uniform on [0,1].
This doesn't seem possible, since there's 1/2 a chance that X, and thus X/(X^2+Y^2+Z^2), is < 0.
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