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.

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.