Dividing a sphere into thirds with two planes

Question

Earlier this week, I asked this question about dividing a circle into thirds, and we were able to use numerical methods to come up with the desired result. Now, I want to extend that question to a $2$-sphere.

Suppose we have a sphere of radius $R$ that is divided into thirds by two parallel planes that are equidistant from the origin, so that the resulting cut leaves three sections of equal volume. What is the distance $d$ between these planes?

I decided to solve by the method shown on this diagram using a "cone" and a "dome":

where $$\frac13 V=V_{\textrm{dome}}=V_{\textrm{cone + dome}}-V_{\textrm{cone}}$$ First, let us solve for $V_{\textrm{cone + dome}}$: $$V_{\textrm{cone + dome}}=V_{\textrm{sphere}}\cdot \frac{A_{\textrm{dome}}}{A_{\textrm{sphere}}}$$ $$=\frac{\frac43 \pi R^3 * 2 \pi R \left(R-d/2\right) }{4 \pi R^2}$$ $$=\frac23 \pi R^2 \left(R-d/2\right)$$ $$=\frac23 \pi R^3 - \frac13 \pi R^2 d$$

Next, let us solve for $V_{\textrm{cone}}$. The radius $R_\textrm{cone}$ of the base of the cone (you have to look at it sideways) is calculated by Pythagorean theorem: $$R_\textrm{cone}^2=R^2-(d/2)^2=R^2-\frac14 d^2$$ $$V_\textrm{cone}=\frac13 \pi R_\textrm{cone}^2 (d/2)=\frac13 \pi \left(R^2-\frac14 d^2\right) (d/2)$$ $$=\frac16 \pi R^2 d -\frac{1}{24}\pi d^3 $$

Thus, we can solve: $$\frac13\left(\frac43 \pi R^3 \right)=\frac13 V=V_{\textrm{cone + dome}}-V_{\textrm{cone}}$$ $$ =\frac23 \pi R^3 - \frac13 \pi R^2 d- \left(\frac16 \pi R^2 d -\frac{1}{24}\pi d^3 \right)$$ $$=\frac23 \pi R^3 - \frac12 \pi R^2 d +\frac{1}{24}\pi d^3$$ Simplifying, we obtain the equality $$0= R^3 - \frac94 R^2 d +\frac{3}{16} d^3$$

I put this in Wolfram Alpha and got:

$$d=-\frac{(1-i\sqrt{3})\sqrt[3]{-R^3+2i\sqrt{2}R^3} }{\sqrt[3]{3}} -\frac{\sqrt[3]{3}(1+i\sqrt{3})R^2}{\sqrt[3]{-R^3+2i\sqrt{2}R^3}}$$ which isn't particularly enlightening. How can I solve for $d$? Can I get a more sensible value? I am particularly interested in the numerical methods approach rather than just a result from a calculator.

Answer 1

Looking up "volume of spherical cap" in many places (for example, here: http://mathworld.wolfram.com/SphericalCap.html) if the radius is $r$ and the height from the end of the sphere is $h$, then the volume is $\frac13\pi h^2(3r-h)$.

Since the volume of the sphere is $\frac43 \pi r^3$, you want the volume of the cap to be one-third of this, or $\frac13\pi h^2(3r-h) =\frac49 \pi r^3 $ or $3 h^2(3r-h) =4 r^3 $. Dividing by $r^3$, this becomes $3(\frac{h}{r})^2(3-\frac{h}{r}) = 4 $. Finally, letting $x = \frac{h}{r}$, this is $3x^2(3-x) = 4$ or $3x^3-9x^2+4 = 0 $.

According to Wolfram Alpha, the root of this is $x\sim 0.77393$, so $d = 2(1-h) \sim 2-1.54786 = .45214 $, which agrees very nicely with your computation.

Answer 2

I let $x=\frac dR$ in Alpha and just plotted it. It gave me the approximate root $0.45125$ Given that, the $x^3$ term is rather small. I then wrote an iterative solution: $x_{i+1}=\frac 49(1+\frac {3x^3}{16})$ It converges rapidly to the correct answer. After five cycles starting with $0.4$ or $0.5$, it has converged to six places. Of course, Newton's would converge even faster, but this was easy to put together in Excel.

Added per request: $$\begin {array} 0.4&0.449777778\\ 0.449777778&0.45202695\\0.45202695&0.452141272\\0.452141272&0.452147113\\ 0.452147113&0.452147411\\0.452147411&0.452147427 \end {array}$$

To make such a spreadsheet, the first column is $x$, the second is $f(x)$ You put the starting value in the first cell of the first column, then each cell in the first column is the value in the cell in the second column in the row above. Copy down and you are done.