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I'm trying to calculate the height of a spherical cap on a sphere where you know the volume of the cap and the radius of the sphere.
I was able to find the equation for the volume of the cap with radius $r$ and height $h$ which is
$$V=\frac{\pi h^2}{3}(3r-h)$$
I feel really dumb but for the life of me I can't seem to rewrite the equation in such a way to find $h$ when given a known volume and radius.
On
If you don't have to solve for the height algebraically, you could solve for the height by graphing the equation and finding the x-intercepts. This can be done on a graphing calculator or online at Desmos Graphing Calculator.
Graph the equation $$ \frac{\pi}{3}h^2\ (3r - h) - V $$ and locate the x-intercepts
On
Given $V$ in order to find $h$ we have to find roots of the cubic equation.
EDIT1:
It is more convenient to express given volume as a fraction$f$ of the full sphere$ \;4\pi r^3/3$
$$\dfrac{\pi h^2(3r-h)}{3}=f \dfrac43 \pi r^3 \;\;\rightarrow 4fr^3+h^3-3r h^2=0$$
The following is given by Mathematica for $h$ value.
Solve[-3 r h^2 + h^3 + 4 f r^3 == 0, h]
$$ \left(r+\frac{1}{2} \left(1+i \sqrt{3}\right) \sqrt[3]{2 \sqrt{f^2 r^6-f r^6}-2 f r^3+r^3}+\frac{\left(1-i \sqrt{3}\right) r^2}{2 \sqrt[3]{2 \sqrt{f^2 r^6-f r^6}-2 f r^3+r^3}}\right) $$
in which the imaginary part cancels on evaluation.
Else if you want a graphic solution read off $h$ for given $ (V,r)$ or vice-versa as shown:
On
By solving the volume formula $[V=\frac{\pi h^2}{3}(3r-h)]$ for h to obtain the cubic equation $[\frac{\pi}{3}h^3-\pi r h^2+V=0]$, converting to the depressed cubic form of the equation $[t^3+pt+q]$, using the trigonometric solution $\Bigg[\Bigg(t_k=2\sqrt{-\frac{p}{3}}\cos\bigg[\frac{1}{3}\cos^{-1}\bigg(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\bigg)-\frac{2\pi k}{3}\bigg]\Bigg)$ for $k=0,1,2\Bigg]$ to solve for the root where k=1, converting back to the original form of the cubic equation, and simplifying, the following solution can be obtained for h in terms of V and r.
$$ h=r[\sqrt{3}\sin(\frac{1}{3}cos^{-1}A)-\cos(\frac{1}{3}\cos^{-1}A)+1] $$ $$ A=1-\frac{3V}{2\pi r^3}=1-\frac{2V}{V_{sphere}} $$
You have to solve the cubic $$ \frac{\pi}{3}h^3-\pi r h^2+V=0 $$
that always have a real solution for $h$, and, in general, can be solved with Cardano's formula.