I am trying to compute the distance AB as in the following schema representing the intersection of two spheres, knowing the value of the green volume.
I have based my solution on the sphere cap formula from wikipedia, using this notation:
$V_g$: green volume
$a$ : the distance AB, which is the variable I am trying to solve for
$V_d$: cap volume corresponding to AH
$V_s$ : cap volume corresponding to AG
$h_s$ : the distance AG
$h_d$ : the distance AH
$r$ the radius of the small circle: CD
$R$ the radius of the big circle: OE
We obtain the following equations:
$V_g = V_d - V_s$
$ V_d = \frac{\pi}{6} * h_d * (3*a^2 + h_d^2)$
$ V_s = \frac{\pi}{6}* h_s * (3*a^2 + h_s^2) $
However, $h_s$ and $h_d$ also depends on a as:
$ h_s = R - R*cos(sin_{-1}(a))$ by applying trigonometry to the triangle ABO
$ h_d = r - \sqrt{r^2 - a^2}$ by applying trigonometry to the triangle CAB
Which leads me to this equation:
$ V_g * \frac{6}{\pi} = h_d*(3*a^2 + h_d^{2}) - h_s*(3*a^2+h_s^{2}) $
$h_s$ and $h_d$ being dependent on $a$, I struggle to solve the inverse function like $a = f(V_g)$, is it possible to solve this kind of equation?
Let $R=OB,r=CB$ be the radii of the spheres, $d=OC$ the distance between centers.
Let $b=AC$, then Pythagoras' in right triangles $\triangle ACB, \triangle AOB$ gives:
$$ \require{cancel} \begin{align} AB^2 = CB^2-CA^2 = OB^2-OA^2 \implies r^2 - b^2 &= R^2 - (d+b)^2 \\ r^2 - \cancel{b^2} &= R^2 - d^2 - 2bd - \cancel{b^2} \\ b &= \frac{R^2-r^2 - d^2}{2d} \tag{1} \end{align} $$
The heights of the caps in the two spheres are $AH = r - b\,$, $AG = R - d - b\,$, so the green volume is:
$$ V_g = \frac{\pi AH^2}{3}\left(3 r - AH\right) - \frac{\pi AG^2}{3}\left(3R - AG\right) \tag{2} $$
Equation $(2)$ is an algebraic equation in $R, r, d$ which can be solved for one unknown if the other two are given. Once $d$ is determined, $b$ follows from $(1)$, then $a = \sqrt{r^2 - b^2}$.