Rearrange from $f(r)$ to $r(f)$? I'm trying to understand a set of equations from a paper. The Photo shows the problem. The aim is to solve for $r$.
Can anyone explain how it is possible to go from $f(r)$ to $r(f)$ or have I completely misunderstood what they have done? I only have GCSE math so an intuitive explanation and a point at the right place to look up more information would be great.
These are the equations:
$f(r) = \frac{V(r)}{V_{total}}$
$f(r)= \frac{\frac{1}{3}(r^3-R^3)d\Omega}{\frac{1}{3}((R+T)^3-R^3}$
$f(r)= \frac{r^3-R^3}{(R+T)^3-R^3}$
Generally, we are interested in $r$ as a function of $f$, and rearranging the equation gives us the following::
$r(f)^3= ((R+T)^3-R^3)f -R^3$
Taking cube root gives the formula:
$r(f)= \sqrt[3]{((R+T)^3-R^3)f -R^3}$
Thank you!

Original paper - this is from Appendix A
 A: There are a couple of misprints in some of the equations, so I will present an account of the calculation that I believe is required (I do not have access to the paper, so only the picture in the question is used).
There are two concentric spheres with radii $R$ and $R+T$ (hence $T$ is thickness of region) and a solid angle $d\Omega$ steradians which enables one to write an expression for the corresponding volume between the two spheres. This volume is denoted by $V_{total}$ in your first expression, and is the difference between the volume  at $R+T$ and the volume at $R$ for the solid angle $d\Omega$:
$V_{total} = \frac {1}{3} ((R+T)^{3} - R^{3}) d\Omega$
What is required also is the fraction of this volume occupied at radius $r$ into the region ($r$ is measured from the origin). This means that $r>R$ and is denoted in the first equation as $V(r)$
$V(r) = \frac {1}{3} (r^{3} - R^{3}) d\Omega$ 
$f(r)$ is the fraction of the volume of the region contained up to $r$ and is given (with some cancellations) by:
$f(r) = \frac { r^{3} - R^{3}}{(R+T)^{3} - R^{3} }$
Note that $f(R) = 0$ and $f(R+T) = 1$ demonstrating this this is a fraction (essentially proportion).
The task is to find the fraction $f$ given a radius $R < r < R+T$, which involves inverting this equation:
$f \times [(R+T)^{3} - R^{3}] = r^{3} - R^{3}$
giving
$r^{3} = R^{3} + f ((R+T)^{3} - R^{3})$
or 
$r = \sqrt[3]{(R^{3} + f ((R+T)^{3} - R^{3}))}$
We see from this that for $f=0$ $r=R$ and for $f=1$ $r=R+T$.
