# 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 label $1\ldots 6$. 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 volumen 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 volumen 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 vaste?