How I cut my orange - spherical volume integral

I cut my orange in six eatable pieces, following some rules. My orange is a perfect sphere, and there is a cylindrical volume down through my orange, that is not eatable.

In the diagram, the orange with radius, $$R$$ is shown as seen from the top. The circle with radius, $$r$$ is the non-eatable center. All lines are vertical cuts to be made. The red area is the final waste. Many of the cuts are tangential to the non-eatable center to minimize waste. The pieces are labeled $$1\ldots6$$. and the angles for each pieces are labeled $$\alpha_{1\ldots7}$$.

I want to optimize my cuts, so that the volume of the pieces are similar. I denote the volume of piece $$n$$ as $$V_n$$. It can be done, so that $$V_1=V_2=V_3$$ and $$V_4=V_5=V_6$$.

The first half: The volume of $$V_1+V_2+V_3$$ can easily be calculated from the spherical cap formula: $$(V_1+V_2+V_3) = \pi/3(R-r)^2(2R+r)$$

To calculate $$\alpha_1$$, all I need is the volume, $$V_1$$ of the skewed spherical wedge as a function of $$\alpha_1$$. How do I set up this integral?

The second half: Making $$V_4=V_5=V_6$$ is a lot more complicated, but I would like to know how to make the volume integral that allows me to calculate volumes like $$V_4$$ and $$V_5$$.

The final cut: My intuition says, that making the final cut, where the waste is separated from piece 6. is best done by making $$\alpha_6 = \alpha_7$$. Does that really minimize the waste?

• Is the vertex of 2. assumed to be at the midpoint of the tangent line, or do we need to prove that? Dec 26, 2022 at 13:11
• @TheSimpliFire: Yes - the vertex of 1. 2. and 3. is at the midpoint of the tangent line. Dec 26, 2022 at 14:08
• The points 5, 6 and the top intersect are colinear Dec 27, 2022 at 16:36

For practical purposes assume that $$r\ne0,R$$.

Proposition 1. Suppose we cut in a straight line between 1. and 2. at angle $$\alpha_1$$. Then $$V_1(\alpha_1)=\frac\pi6(R-r)^2(2R+r)+\frac13q(\alpha_1;R,r)$$ where \begin{align}q(\alpha_1;R,r)&=(3R^2r\cos\alpha_1-r^3\cos^3\alpha_1)\arctan\frac{\sqrt{R^2-r^2}}{r\sin\alpha_1}\\&\quad-2R^3\arctan\frac{\sqrt{R^2-r^2}}{R\tan\alpha_1}-\frac12r^2\sqrt{R^2-r^2}\sin2\alpha_1.\end{align}

This formula has intuitive interpretations. Geometrically, as $$\alpha_1\to0$$, the cut approaches the tangent line so we should expect $$V_1\to0$$. Indeed, we have \begin{align}q(+\infty;R,r)&=(3R^2r-r^3-2R^3)\frac\pi2=-\frac\pi2(R-r)^2(2R+r).\end{align} Also, when $$\alpha_1=\pi/2$$, we have $$V_1=(V_1+V_2+V_3)/2=\pi(R-r)^2(2R+r)/6$$ using the spherical cap formula; by direct computation, $$q(\pi/2;R,r)=0$$ and we are left with the same result.

Proposition 2. Rather unintuitively, $$V_1=V_2=V_3$$ is impossible when $$\alpha_1=\alpha_2=\alpha_3$$.

I have asked on MathOverflow whether this can be extended to $$n>3$$ partitions; that is, whether the only spherical cap whose $$n$$ wedges have the same angles and volumes is the hemisphere.

Remark 3. Note that the integrals for $$V_4$$ and $$V_5$$ can be set up analogously, since it involves cutting by two planes. Since $$V_6$$ involves cutting by three planes, we cannot directly apply the same method. I haven't worked out the bounds yet.

Question 4. If the spherical cap is divided into $$n$$ wedges at an angle of $$\pi/n$$, is $$V_1=V_2=\cdots=V_n$$ also impossible for $$n>3$$?

Proof of Proposition 1: Consider a shift in coordinates $$x\to x+r$$ so that the tangent line spanning 1. to 3. is $$x=0$$.

The volume of 1. is therefore bounded by the sphere $$(x+r)^2+y^2+z^2=R^2$$ and the planes $$x=0$$ and $$y=-mx$$ where $$\alpha_1=\pi/2-\arctan m$$. Since $$x=0$$, we have $$-\sqrt{R^2-r^2}\le z\le\sqrt{R^2-r^2}.$$ The plane $$y=-mx$$ meets the sphere at $$(x+r)^2+m^2x^2+z^2=R^2$$ so $$0\le x\le\frac{-r+\sqrt{r^2+(m^2+1)(R^2-r^2-z^2)}}{m^2+1}:=f(z).$$ Finally, by definition we obtain $$-\sqrt{R^2-(x+r)^2-z^2}\le y\le-mx$$ so the integral is thus \begin{align}V_1&=\int_{-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}}\int_0^{f(z)}\left(\sqrt{R^2-(x+r)^2-z^2}-mx\right)\,dx\,dz\\&=\small\frac12\int_{-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}}\left[(x+r)\sqrt{R^2-(x+r)^2-z^2}+(R^2-z^2)\arctan\frac{x+r}{\sqrt{R^2-(x+r)^2-z^2}}-mx^2\right]_0^{f(z)}\,dz.\end{align} Recalling that $$f(z)$$ satisfies $$mf(z)=\sqrt{R^2-(f(z)+r)^2-z^2}$$, the inner integral becomes $$\small mrf(z)+(R^2-z^2)\arctan\frac{f(z)+r}{mf(z)}-r\sqrt{R^2-r^2-z^2}-(R^2-z^2)\arctan\frac r{\sqrt{R^2-r^2-z^2}}.$$ Since this is a function of $$z^2$$, we can write $$V_1=I_1+I_2+I_3+I_4$$ where \begin{align}I_1&=\int_0^{\sqrt{R^2-r^2}}mrf(z)\,dz\\&=-\frac{mr^2\sqrt{R^2-r^2}}{m^2+1}+\frac1{m^2+1}\int_0^{\sqrt{R^2-r^2}}\sqrt{r^2+(m^2+1)(R^2-r^2-z^2)}\,dz\\&=\frac{mr}{2\sqrt{m^2+1}}\left[\left(\frac{r^2}{m^2+1}+R^2-r^2\right)\arctan\frac{\sqrt{(m^2+1)(R^2-r^2)}}r-r\sqrt{\frac{R^2-r^2}{m^2+1}}\right]\\\\I_2&=\int_0^{\sqrt{R^2-r^2}}(R^2-z^2)\arctan\frac{f(z)+r}{mf(z)}\,dz\\&=\frac16\left(\frac{mr^2\sqrt{R^2-r^2}}{m^2+1}+\pi(r^2+2R^2)\sqrt{R^2-r^2}-4R^3\arctan\left(\frac{m\sqrt{R^2-r^2}}R\right)\right.\\&\left.\quad\quad+\frac{mr(m^2r^2+3(m^2+1)R^2)}{(m^2+1)^{3/2}}\arctan\left(\frac{\sqrt{(m^2+1)(R^2-r^2)}}r\right)\right)\\\\I_3&=-\int_0^{\sqrt{R^2-r^2}}r\sqrt{R^2-r^2-z^2}\,dz=-\frac{\pi r}4(R^2-r^2)\\\\I_4&=-\int_0^{\sqrt{R^2-r^2}}(R^2-z^2)\arctan\frac r{\sqrt{R^2-r^2-z^2}}\,dz\\&=\frac\pi{12}\left(4R^3-3R^2r-r^3-2(2R^2+r^2)\sqrt{R^2-r^2}\right).\end{align} This reduces to $$V_1=\frac\pi6(R-r)^2(2R+r)+g(m;R,r)$$ where \begin{align}g(m;R,r)&=\frac{3m(m^2+1)R^2r-m^3r^3}{3(m^2+1)^{3/2}}\arctan\frac{\sqrt{(m^2+1)(R^2-r^2)}}r\\&\quad-\frac23R^3\arctan\frac{m\sqrt{R^2-r^2}}R-\frac{mr^2\sqrt{R^2-r^2}}{3(m^2+1)}\end{align} and substituting $$m=\cot\alpha_1$$ and $$g=3q$$ yields the desired result.

Sketch proof of Proposition 2: $$V_1=V_2=V_3$$ forces $$q(\alpha_1;R,r)=-\frac\pi6(R-r)^2(2R+r).$$ If the angles are equal, $$\alpha_i=\pi/3$$. We can eliminate both $$R$$ and $$r$$ by substituting $$s=\sqrt{(1-r^2/R^2)/3}$$ which gives $$\small2\arctan s+\frac{3s}4(1-3s^2)-\frac{11+3s^2}8\sqrt{1-3s^2}\arctan\frac{2s}{\sqrt{1-3s^2}}=\frac\pi6(1-\sqrt{1-3s^2})^2(2+\sqrt{1-3s^2})$$ over the range $$s\in[0,1/\sqrt3]$$. The case $$s=0$$ means $$r=R$$ (impossible) and the case $$s=1/\sqrt3$$ means $$r=0$$ (also impossible). In this answer, @IosifPinelis shows that the equation has no solution in $$(0,1/\sqrt3)$$ which proves the lemma.

• Thanks for your answer. It will take me some days to try to understand it (some of it). If 2*R=10cm and 2*r=1cm, can I use your answer to calculate \alpha_1 to nearest degree? Dec 27, 2022 at 17:05
• @hpekristiansen Yes, the answer is $58$ degrees. Substituting $R=5$ and $r=0.5$ into $q(\alpha;R,r)=-\pi(R-r)^2(2R+r)/6$ gives $$\left(\frac{75}2\cos\alpha_1-\frac1{64}\cos^3\alpha_1\right)\arctan\frac{\sqrt{99}}{\sin\alpha_1}-250\arctan\frac{\sqrt{99}}{10\tan\alpha_1}-\frac{\sqrt{99}}{16}\sin2\alpha_1+\frac{567}{16}\pi=0$$ The Mathematica command FindRoot[(37.5Cos[x]-Cos[x]^3/64)ArcTan[Sqrt[99]/Sin[x]]-250ArcTan[0.1Sqrt[99]Cot[x]]-Sqrt[99]Sin[2x]/16+567Pi/16,{x,1}] gives $$\alpha_1=1.00619\cdots\text{radians}=57.65\cdots^\circ$$ Dec 27, 2022 at 17:57
• For others/completeness: wolframalpha.com/… Jan 9, 2023 at 6:57

Too long for comment but may give some useful ideas for setting up the integrals in spherical coordinates (cylindrical may be slightly simpler):

Make the first cut into the sphere $$x^2+y^2+z^2=R^2 \iff \rho=R$$ along the plane $$x=r \iff \rho=r\sec(\theta)\csc(\phi)$$. Letting $$\theta$$ vary over a fixed symmetric interval, solve for $$\phi$$ when the plane and sphere meet to get

$$r\sec(\theta)\csc(\phi)=R \implies \sin(\phi) = \frac rR \sec(\theta)$$

In the plane $$z=0$$, when $$\phi=\frac\pi2$$, we have

$$r\sec(\theta)=R \implies \cos(\theta) = \frac rR$$

Hence the volume of the first half is given by the integral(s)

\begin{align*} V_1+V_2+V_3&=\int_{-\cos^{-1}\left(\frac rR\right)}^{\cos^{-1}\left(\frac rR\right)} \int_{\sin^{-1}\left(\frac rR \sec(\theta)\right)}^{\pi - \sin^{-1}\left(\frac rR \sec(\theta)\right)} \int_{r\sec(\theta)\csc(\phi)}^R \rho^2 \sin(\phi)\,d\rho\,d\phi\,d\theta\\[1ex] &= \frac13 \int_{-\cos^{-1}\left(\frac rR\right)}^{\cos^{-1}\left(\frac rR\right)} \int_{\sin^{-1}\left(\frac rR \sec(\theta)\right)}^{\pi - \sin^{-1}\left(\frac rR \sec(\theta)\right)} \left(R^3\sin(\phi)-r^3\sec^3(\theta)\csc^2(\phi)\right)\,d\phi\,d\theta \\[1ex] &= \frac23 \int_{-\cos^{-1}\left(\frac rR\right)}^{\cos^{-1}\left(\frac rR\right)} \frac{\left(R^2\cos^2(\theta)-r^2\right)^{\frac32}}{\cos^3(\theta)}\,d\theta \\[1ex] &= \frac{4R^3}3 \int_r^R \frac{\left(\lambda^2-r^2\right)^{\frac32}}{\lambda^3\sqrt{R^2-\lambda^2}} \, d\lambda \end{align*}

which agrees with the spherical cap formula.

To isolate the volume of just the first wedge, it would be convenient to fix the point of tangency of the cut to the core to lie in the planes $$x=r$$ and $$y=0$$, and to know $$\alpha_1$$. Then we can find an equation for the plane representing the cut that forms $$V_1,V_2$$, $$\rho_1 = \frac{r\csc(\phi)}{\cos(\theta)+\tan(\alpha_1)\sin(\theta)}=\frac{r\csc(\phi)}{\sec(\alpha_1)\cos(\theta-\alpha_1)}$$.

The bounds highlighted in red (part of the very first cut) and green represent the lower and upper limits of $$\rho$$ - in particular, the green boundary is $$\rho=\min\left(\rho_1,R\right)$$. The south-southeasternmost ray is $$\theta=-\cos^{-1}\left(\frac rR\right)$$. Let the other two be given by $$\theta=\theta_1$$ and $$\theta=\theta_2$$.

When $$\theta\in\left[-\cos^{-1}\left(\frac rR\right),\theta_1\right]$$, the integral takes the same the limits for $$\phi$$ as the integral for the first half.

When $$\theta\in[\theta_1,\theta_2]$$, we see the cut $$\rho_1$$ meets the sphere for

$$\frac{r\csc(\phi)}{\sec(\alpha_1)\cos(\theta-\alpha_1)} = R \implies \sin(\phi) = \frac{r\cos(\alpha_1)}{R\cos(\theta-\alpha_1)}$$

We find $$\theta_1$$ by solving for $$\theta$$ in the above equation when $$\phi=\frac\pi2$$:

$$\frac{r}{\sec(\alpha_1)\cos(\theta-\alpha_1)} = R \implies \cos(\theta-\alpha_1) = \frac rR \cos(\alpha_1) \\ \implies \theta_1 = \alpha_1-\cos^{-1}\left(\frac rR \cos(\alpha_1)\right)$$

We find $$\theta_2$$ by finding where the first cut meets the cut $$\rho_1$$:

$$r\sec(\theta)\csc(\phi) = \frac{r\csc(\phi)}{\sec(\alpha_1)\cos(\theta-\alpha_1)} \implies \frac{\cos(\theta)}{\cos(\theta-\alpha_1)}=\cos(\alpha_1) \\ \implies \theta_2 = 0$$

Then the integral for $$V_1$$ has limits

$$\int_{-\cos^{-1}\left(\frac rR\right)}^{\alpha_1-\cos^{-1}\left(\frac rR \cos(\alpha_1)\right)} \int_{\sin^{-1}\left(\frac rR \sec(\theta)\right)}^{\pi-\sin^{-1}\left(\frac rR \sec(\theta)\right)} \int_{r\sec(\theta)\csc(\phi)}^R \\ + \int_{\alpha_1-\cos^{-1}\left(\frac rR \cos(\alpha_1)\right)}^0 \int_{\sin^{-1}\left(\frac{r\cos(\alpha_1)}{R\cos(\theta-\alpha_1)}\right)}^{\pi-\sin^{-1}\left(\frac{r\cos(\alpha_1)}{R\cos(\theta-\alpha_1)}\right)} \int_{r\sec(\theta)\csc(\phi)}^{\frac{r\csc(\phi)}{\sec(\alpha_1)\cos(\theta-\alpha_1)}}$$