<s>[INST] Equal slicing of my spherical cake Recently I baked a spherical cake ($3$ cm radius) and invited over a few friends, $6$ of them, for dinner. When done with main course, I thought of serving this spherical cake and to avoid uninvited disagreements over the size of the shares, I took my egg slicer with parallel wedges(and designed to cut $6$ slices at a go; my slicer has $5$ wedges) and placed my spherical cake right in the exact middle of it before pressing uniformly upon the slicer.
What should the relative placement of my wedges of the slicer be like so that I am able to equally distribute the cake among my $6$ friends?The set-up of the egg slicer looks something like this :

 [/INST] Let me recast Lovsovs' answer in another perspective.
Consider at first to split an hemi-sphere into $n$ parts of equal volume.
Let designate as $h(k,n)$ the relative position of the $k$-th cut along the radius of the emisphere starting from the base circle 
(corresponding to $k=0$).
$$
0 \leqslant h(k,n) \leqslant 1\quad \left| {\;0 \leqslant k \leqslant n} \right.
$$
Thus we shall have
$$
\frac{2}
{3}\pi R^{\,3} \frac{k}
{n} = \pi \int_{y\; = \;0}^{h\,R} {\left( {R^{\,2}  - y^{\,2} } \right)dy}  = \pi \left( {R^{\,2} R\,h - \frac{1}
{3}R^{\,3} h^{\,3} } \right)
$$
that means
$$
h^{\,3}  - 3h + 2\frac{k}{n} = 0
$$
We will proceed and solve such depressed cubic equation
according to the method indicated in this work by Alessandra Cauli, refer also to the answer to this post.
Putting apart the case $k=0$, for $\;1 \leqslant k \leqslant n$ we define
$$
u = \sqrt[{3\,}]{{ - \frac{q}
{2} + \sqrt {\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}}} }}
$$
where $p$ and $q$ are respectively the coefficients of $h^1$ and $h^0$.
The 2nd radical is
$$
\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}} = \left( {\frac{k}
{n}} \right)^{\,2}  - 1 =  - \frac{{n^{\,2}  - k^{\,2} }}
{{n^{\,2} }} \leqslant 0
$$
which being non-positive tells us that there are three real solutions.
Completing the calculation for $u$
$$
\begin{gathered}
  u = \sqrt[{3\,}]{{ - \frac{k}
{n} + i\frac{1}
{n}\sqrt {n^{\,2}  - k^{\,2} } }} = \frac{1}
{{\sqrt[{3\,}]{n}}}\sqrt[{3\,}]{{n\,e^{\,i\,\alpha } }} =  \hfill \\
   = e^{\,\,i\,\alpha \,/\,3} \quad \left| \begin{gathered}
  \;1 \leqslant k \leqslant n \hfill \\
  \;\alpha  = \arctan _{\text{4Q}} \left( { - k,\sqrt {n^{\,2}  - k^{\,2} } } \right) = \pi  - \arctan \left( {\sqrt {\left( {\frac{n}
{k}} \right)^{\,2}  - 1} } \right) =  \hfill \\
  \;\;\;\; = \pi  - \beta  \hfill \\ 
\end{gathered}  \right. \hfill \\ 
\end{gathered} 
$$
We take then
$$
v =  - \frac{p}
{{3\,u}} = \frac{1}
{u}\quad \quad \omega  = e^{\,i\,\frac{{2\pi }}
{3}}  = 1/2\left( { - 1 + i\sqrt 3 } \right)
$$
and arrive to obtain the three solutions as 
$$
\begin{gathered}
  1 \leqslant k \leqslant n\quad 0 \leqslant \beta  = \arctan \left( {\sqrt {\left( {\frac{n}
{k}} \right)^{\,2}  - 1} } \right) < \frac{\pi }
{2} \hfill \\
  \left\{ \begin{gathered}
  h_{\,1}  = e^{\,i\,\alpha /3}  + e^{\, - \,i\,\alpha /3}  = 2\cos \left( {\frac{\pi }
{3} - \frac{\beta }
{3}} \right) \hfill \\
  h_{\,2}  = e^{\,i\,\alpha /3 + 2\pi /3}  + e^{\, - \,i\,\alpha /3 - 2\pi /3}  =  - 2\cos \left( {\frac{\beta }
{3}} \right) \hfill \\
  h_{\,3}  = e^{\,i\,\alpha /3 - 2\pi /3}  + e^{\, - \,i\,\alpha /3 + 2\pi /3}  = 2\cos \left( {\frac{\pi }
{3} + \frac{\beta }
{3}} \right) \hfill \\ 
\end{gathered}  \right. \hfill \\ 
\end{gathered} 
$$
of which we can easily determine that only $h_3$ 
respect the physical conditions of our problem.
Passing now to split the entire sphere, imagine it consisting of the two halves, placed base-to-base, one on the positive  and one on the negative $h$ axis.
So, when $n$ is even we will place the cuts at $k = 0,\; \pm 2, \cdots ,\; \pm \left( {n - 2} \right)$,
while for $n$ odd at $k =  \pm 1,\; \pm 3, \cdots ,\; \pm \left( {n - 2} \right)$.
In conclusion, always as a ratio to the radius, the cutting blades shall be positioned at:
$$
\left\{ \begin{gathered}
  2 \leqslant n\quad 0 \leqslant k \leqslant \left\lfloor {\frac{n}
{2}} \right\rfloor  - 1 \hfill \\
  \beta \left( {k,n} \right) = \arctan \left( {\sqrt {\left( {\frac{n}
{k}} \right)^{\,2}  - 1} } \right) \hfill \\
  h_{\,T} (k,n) =  \pm 2\cos \left( {\frac{\pi }
{3} + \frac{{\beta (2k + \bmod (n,2),\,n)}}
{3}} \right) \hfill \\ 
\end{gathered}  \right.
$$
(if you can accept that$\pm 0=0$).
</s>