Slice an ellipsoid into equally thick slices for maximal surface After seeing a colleague slicing a nearly ellipsoid piece of ginger for his cup of tea into almost equally thick slices to get more surface area (so the tea would suck out the ginger taste better), i wondered which way of cutting would provide the most surface area.
Alongside the longer axis, the shorter axis or another skewed axis?
Does it also depend on the slices' thickness?
Looking forward to optimize the ginger tea making process.
 A: The limiting case
At least if the slices are very thin relative to the narrowest axis of the ellipsoid, it doesn't matter: Choose some slicing direction and call the axis perpendicular to the slicing planes the $x$-direction. Then, denote the thickness of the slices by $\Delta x$, and denote the cross-sectional area of the ginger ellipsoid at the value $x$ by $A(x)$. Note that the surface area contributed by the outside of the ellipsoid doesn't depend on the slicing, so it's enough to compare the total of the surface areas exposed by cutting.
The slice at the coordinate $x_i$ exposes an area of $2 A(x_i)$, and adding these contributions gives a total of $$E = \sum_i 2 A(x_i).$$ Now, for fine enough slicing (small enough $\Delta x$) we can approximate the volume $V$ in terms of these cross-sectional areas:
$$V = \int dV = \int A(x) \,dx \approx \sum_i A(x_i) \Delta x;$$
rearranging gives
$$E \approx \frac{2 V}{\Delta x},$$
but this doesn't depend on the direction of slicing.
Asymptotics Of course, one can only slice ginger so finely, and so practically one might be interested in the best one can do with a finite number of slices. Perhaps some inequality manipulation can handle this, but it's likely to be ugly. Instead, we can use some asymptotic analysis to show that, in the limit, it's marginally better to take slices perpendicular to the longest direction. (Some numerical experimentation suggests this holds even for very small numbers of slices.
For convenience, let's choose units so that the slice width is $1$, and for convenience suppose that the semi-axes $p, q, r$ of the ellipsoid are all multiples of the slice width (i.e., integers in our units).
If we take slices orthogonal to the $p$ direction, some elementary geometry shows that area of the cross section at $x$ is
$$\pi q r \sqrt{1 - \left(\frac{x}{p}\right)^2},$$
so the total area exposed by slicing (again, at the integer values $x = -(p - 1), \ldots, p - 1$ is
$$E = \sum_{x = -p}^p 2 A(x) = 2 \pi q r \sum_{x = -p}^p \sqrt{1 - \left(\frac{x}{p}\right)^2}.$$
Expanding the argument $\sqrt{1 - \left(\frac{x}{p}\right)^2}$ of the summation in a Taylor series based at $0$ gives
$$\sqrt{1 - \left(\frac{x}{p}\right)^2} = 1 - \sum_{k = 1}^{\infty} \frac{(2 k - 3)!}{4^{k - 1} k! (k - 2)!} \left(\frac{x}{p}\right)^{2k}.$$
In fact, this converges on $[-p , p]$, so substituting in the above expression gives
\begin{align}E &= 2 \pi q r \left[\sum_{x = -p}^p \left(1 - \sum_{k = 1}^{\infty} \frac{(2 k - 3)!}{4^{k - 1} k! (k - 2)!} \left(\frac{x}{p}\right)^{2k}\right)\right] \\ &= 2 \pi q r \left[ 2 p + 1 - \sum_{k = 1}^{\infty} \frac{(2 k - 3)!}{4^{k - 1} k! (k - 2)!} \cdot \frac{1}{p^{2k}} \sum_{x = -p}^p x^{2k}\right]. \end{align}
The inner sum can be rewritten as
$$\sum_{x = -p}^p x^{2k} = 1 + 2 \sum_{x = 1}^p x^{2k};$$
expanding the first few terms of Faulhaber's formula gives that
$$\sum_{x = 1}^p x^{2k} = \frac{1}{2k + 1} p^{2k + 1} + \frac{1}{2} p^{2k} + \frac{k}{6} p^{2k - 1} + O(p^{2 k - 2}),$$ so
\begin{align}
\frac{1}{p^{2k}} \sum_{x = -p}^p x^{2k}
&= \frac{1}{p^{2k}} \left[1 + 2 \left(\frac{1}{2k + 1} p^{2k + 1} + \frac{1}{2} p^{2k} + \frac{k}{6} p^{2k - 1} + O(p^{2 k - 2})\right)\right] \\
&= \frac{2}{2 k + 1} p + 1 + \frac{k}{2 p} + O\left(\frac{1}{p^2}\right)
\end{align}
Substituting again gives
$$E = 2 \pi q r \left[ 2 p + 1 - \sum_{k = 1}^{\infty} \frac{(2 k - 3)!}{4^{k - 1} k! (k - 2)!} \left[\frac{2}{2 k + 1} p + 1 + \frac{k}{2 p} + O\left(\frac{1}{p^2}\right)\right]\right]$$
Using that the value of $\sqrt{1 - x^2}$ at $x = 1$ is $0$ gives (after a little work) that the constant terms in $p$ cancel. So, the argument in the above leading case gives that the linear coefficient of $E$ is $$2V = \frac{8 \pi}{3},$$ and that the next nonzero term is some positive multiple of
$-\frac{1}{p}$.
So, the direction of slicing that maximizes $E$ for large numbers of slices is the one that minimizes $\frac{1}{p}$, that is, the one with maximal $p$. In other words, it is (asymptotically) most efficient to slice in the direction perpendicular to the long axis.
(Disclaimer: The careful reader will note that exchanging the order of summation isn't justifiable, at least not past the constant term in $p$; we're only analyzing the sign of the $\frac{1}{p}$ term of the Laurent series, and to determine this we don't need the information that this reordering discards.)

What about an average slice thickness of let's say a quarter of the narrowest axis?

In this case, even for the "coarsest" example (so that the ellipsoid has ratio 6:4:4), slicing in the long direction gives a surface area of $\sim 150.58$ units (of square slice thickness), which is only marginally ($\sim 9.65\%$) better than the surface area $\sim 137.33$ given by slicing in the narrow direction.
