I need to obtain the inverse, $g^{-1}$, of the following function, $g$.
\begin{align} u = g\left(x\right) = 2r^2 \cos^{-1} \left(\frac{x}{2r}\right) - \frac{x}{2}\sqrt{4r^2 - x^2}, {\rm{~~}} 0 \le x \le 2r. \end{align}
As you can see, $u$ is the area of the 'intersection' between two circles with the equal radius of $r$, while $x$ indicates the 'distance' between the two circles.
WolframAlpha is not able to compute the inverse. I wonder if there is any of you who can help me out.
Background: $x$ is a uniform random variable; and I am trying to obtain the pdf of $U$, via transformation of $f_{U}(u) = f_{X}(x = g^{-1}(u)) \left|\frac{\text{d}}{\text{d}u}g^{-1}(u)\right|$, for which the inverse $g^{-1}$ is needed.
Additional questions: Is there a way to APPROXIMATE the square root term $\sqrt{4r^2 - x^2}$? Since $u$ is 'area' and $r$ is radius, the $\sqrt{4r^2 - u^2}$ term in $g^{-1}$ yields an imaginary value, which is not desired.
My answer is for solving the equation in closed form by only elementary operations and for solving the equation in closed form by only elementary operations with exception of a constant summand.
a)
Because your equation depends on $\arccos(x)$ and $x$ and both are algebraically independent, your equation cannot be solved in closed form by rearranging it only by applying only elementary operations (Elementary functions) that you can derive from the equation.
It is not clear if your equation has solutions that are Elementary numbers.
b)
The method Yuriy S proposed is as follows.
Because it is not possible to solve the equation in the way written in a), we try to calculate the inverse in that way with exception of a constant summand.
All elementary functions (in particular the elementary standard functions) can be represented as compositions of only $\exp$, $\ln$ and/or algebraic functions.
As written in a), $arccos$ disturbs, or $\ln$ because $arccos(x)=i\ln\left(\sqrt{1-x^2}+ix\right)+\frac{\pi}{2}$.
$\arccos$ or $\ln$ can be removed by calculating the first derivative: $\frac{d}{dx}\ln(x)=\frac{1}{x}$.
our strategy:
$g(x)\rightarrow g'(x)$
$g'(x)\rightarrow g'^{-1}(x)$
$\int g'^{-1}(x)dx=g^{-1}(x)+c$
$$\ $$
$$r,x\ge 0$$
$$g(x)=2r^2\arccos\left(\frac{x}{2r}\right)-\frac{x}{2}\sqrt{4r^2-x^2}$$
$$g'(x)=-\sqrt{4r^2-x^2}$$
$$g'(g'^{-1}(x) )=x:$$
$$-\sqrt{4r^2-\left(g'^{-1}(x)\right)^2}=x$$
$$g'^{-1}(x)\colon \left[\sqrt{4\,{r}^{2}-{x}^{2}},-\sqrt{4\,{r}^{2}-{x}^{2}}\right]$$
$c,c_1,c_2$: integration constants
$$\int g'^{-1}(x)dx=g^{-1}(x)+c:$$
$$g^{-1}(x)\colon \left[\frac{1}{2}\,x\sqrt{4\,{r}^{2}-{x}^{2}}+2\,{r}^{2}\arctan\left({\frac{x}{\sqrt{4\,{r}^{2}-{x}^{2}}}}\right)+c_1,-\frac{1}{2}\,x\sqrt{4\,{r}^{2}-{x}^{2}}-2\,{r}^{2}\arctan\left({\frac{x}{\sqrt{4\,{r}^{2}-{x}^{2}}}}\right)+c_2\right]$$
That are the different branches of $g^{-1}$, the partial inverses of $g$.
You have to look for the suitable branch.
Now you have to determine the integration constants $c_1$ and $c_2$. Because $g$ is an elementary function that is not invertible by an elementary function, $g^{-1}$ is not an elementary function, and therefore you cannot calculate the integration constants in closed form in the way written in a). You have to determine them numerically therefore.