Distribution of spheres radii from uniform distribution of volume I want to compute the density function for the radii of spheres given that the volume distribution must be uniform. This implies that it should be more likely to have small radii than large ones (otherwise the volume will concentrate on the larger spheres). So, I start from the distribution of volume, where the probability density is 
$$f(V) = \frac{1}{V_b - V_a} , $$
for $V_a \le V \le V_b$, where $V_b$ ($V_a$) are the maximum (minimum) volumes, respectively. Taking into account that 
$$V = \frac{4\pi}{3}r^3,$$
then I can compute $g(r)$ (the density for the radii) as
$$g(r) = f(V) \left|\frac{dV}{dr}\right| = \frac{1}{V_b - V_a} 4\pi r^2 = \frac{3r^2}{r_b^3 - r_b^3}.$$
Unfortunately this last expression favors large radii, not small ones. So it means that I am misunderstanding something or using the transformation in the wrong way but I cannot find the error. Sorry if this is a rookie mistake, I am not an expert on probability. Thanks for any advice. 
 A: Your density function $g(r) = \frac{4\pi}{V_b - V_a}r^2,$ for 
$(3V_a/4\pi)^{1/3} < r < (3V_b/4\pi)^{1/3},$ is correct. [I do not
understand the notation in the final expression for the density.]
I use a million simulated observations of the volume as a device to
make histograms of the uniform distribution of $V$ and of the transformed
distribution of $r.$ This is for the special case $V_a = 3$ and $V_b = 8.$
Each histogram bar
at the left represents about 100,000 values of $V$, and each histgram
bar at the right represents the images of the values in a bar of
the same color at the left. 
Ordinarily, it is not a good idea to
make a histogram with bins of variable widths, but here the variable widths
illustrate the effect of the factor $|dV/dr|$ in the density function
for $r.$ Your density function for $r$ is the black curve superimposed
in the histogram at the right. I believe, as @stochasticboy321 Comments,
your argument about the relative sizes of $V$'s and $r$'s fails to
take the effect of the factor $|dV/dr|$ into account. 

Note: The simulation is mainly a device to avoid writing code to make a lot of
rectangles. In case you are familiar with R statistical software and want to try this
illustration for different intervals $(V_a, V_b),$ I am including the 
R code for the histograms below [where a and b denote interval endpoints].
set.seed(424)
a = 3;  b = 8;  v = runif(10^6, a,b);  cutp=seq(a, b, len=11)
par(mfrow=c(1,2)) # enables 2 panels per plot
 hist(v, prob=T, br=cutp, col=rainbow(11))
 r = (3*v/(4*pi))^(1/3); cutpr = (3*cutp/(4*pi))^{1/3}
 hist(r, prob=T, br=cutpr, col=rainbow(11))
  curve(4*pi*x^2/(b-a), add=T, lwd=2)
par(mfrow=c(1,1))

