How to get thickness of a spherical shell based on radius and volume?

Question

I have a problem converting an equation. I want to flip an independent to dependent.

Volume of a spherical shell is:

$$V= \frac{4}{3}\pi\bigl[(R+t)^3-(R-t)^3\bigl]$$

Where R is radius, t is half of shell's thickness. It is important to me that it has a form of ± t rather than the most common $V= \frac{4}{3}\pi\bigl[R^3-(R-2t)^3\bigl]$

How to turn the equation to have the following form?: $$t(V,R) = ...$$

I know it must be simple but no matter what I do, I end up with R somehow tangled to t. Please, help. I am getting mad...

Answer 1

Mathematica gives:

$$t(V,R)=\frac{2\cdot(2\pi)^{1/3}R^2}{\left(-3V+\sqrt{256\pi^2R^6+9V^2}\right)^{1/3}}-\frac{\left(-3V+\sqrt{256\pi^2R^6+9V^2}\right)^{1/3}}{2\cdot(2\pi)^{1/3}}$$

Answer 2

Starting from $$V= \frac{4}{3}\pi\bigl[(R+t)^3-(R-t)^3\bigl]=\frac{4}{3}\pi\left(6 R^2 t+2 t^3 \right)$$ and intoducing $$x=\frac t R \qquad\text{and}\qquad b=\frac{3 V}{8 \pi R^3}$$ we end with a cubic equation $$x^3+3x-b=0\tag 1$$ Now, let $x=y-\frac 1y$ to get $$y^6-b y^3-1=0 $$ which is a quadratic in $y^3$ making the analytical real solution of$(1)$ to be $$y^3=\frac{b+\sqrt{b^2+4}}{2}\qquad \implies \qquad\color{red}{x=\sqrt[3]{\frac{b+\sqrt{b^2+4}}{2}}-\sqrt[3]{\frac{2}{b+\sqrt{b^2+4}}}}\tag 2$$

Assuming that $t$ is small, you can also write $$t=\sum_{i=1}^n a_i\, b^{2i-1} \tag 3$$ which will give by identification $$a_1=\frac{1}{3}\qquad a_2=-\frac{1}{81}\qquad a_3=\frac{1}{729}\qquad a_4=-\frac{4}{19683}\qquad a_5=\frac{55}{1594323}\tag 4$$ and so on.

This is in fact the Taylor expansion of $(2)$ built around $b=0$.

In order to check, we can use $(1)$ for given value of $x$ to deduce $b$ and from $b$ recompute $x$ from the expansion. Below are the results $$\left( \begin{array}{ccc} x_{given} & b_{calc} & x_{calc} \\ 0.00 & 0.000000 & 0.000000 \\ 0.05 & 0.150125 & 0.050000 \\ 0.10 & 0.301000 & 0.100000 \\ 0.15 & 0.453375 & 0.150000 \\ 0.20 & 0.608000 & 0.200000 \\ 0.25 & 0.765625 & 0.250000 \\ 0.30 & 0.927000 & 0.300002 \\ 0.35 & 1.092880 & 0.350014 \\ 0.40 & 1.264000 & 0.400064 \\ 0.45 & 1.441130 & 0.450253 \\ 0.50 & 1.625000 & 0.500879 \end{array} \right)$$ which seems to be a quite good approximation.

Edit

Nicer would be to use the hyperbolic solution for one real root. This would give $$x=2 \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{b}{2}\right)\right)\tag 5$$ that is to say $$\color{red}{t=2 R \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{3 V}{16 \pi R^3}\right)\right)}\tag 6$$ which is the exact formula you are looking for.