# Inertia Tensor of an Ellipsoidal Shell

In this post we derive the inertia tensor for an ellipsoidal shell with negligible thickness, such that all mass is distributed uniformly across the boundary of the ellipsoid. This result is not as easy to find online as the inertia tensors for other common 3D shapes, hence the post. If you're in a hurry, the inertia tensor for an ellipsoidal shell of mass $m$ and semi-axes $a,b,c$ is given in $\eqref{8}$.

## Spherical shell inertia

The inertia of a spherical shell with mass $m$ and radius $r$ is well-known to be $\bm{J}_{\circ}=(2/3)mr^2\bm{I}_3$, where we've denoted the inertia as $\bm{J}$ to avoid confusion with the $3\times 3$ identity matrix $\bm{I}_3$. We also use the subscript $(\cdot)_\circ$ to differentiate the inertia of the shell from that of the uniform solid, which we will denote $\bm{J}_\bullet$. All of the inertia tensors in this post are taken about the centroid of the shape, which coincides with the center of mass.

We will derive $\bm{J}_\circ$ for the sphere here before moving on to the more general ellipsoidal case, because the derivations are similar. We will use a similar but slightly different approach to the derivation here.

Recall that the inertia tensor of a solid sphere of uniform density is $\bm{J}_\bullet=(2/5)mr^2\bm{I}_3$. Substituting in $m=\rho V=\rho(4/3)\pi r^3$, where $\rho$ is the density and $V$ is the volume of the sphere, we get $\bm{J}_\bullet=(8/15)\pi\rho r^5\bm{I}_3$.

To obtain a shell, we will take one solid sphere of radius $r$ and subtract its inertia from that of a slightly larger solid sphere, which has radius $(1+\Delta)r$ with $\Delta\geq 0$, and taking the limit $\Delta\to 0$. This gives us

$\begin{equation}\label{1} \bm{J}_\circ = \lim _{\Delta\to0}\ (8/15)\pi\rho((1+\Delta)^5-1)r^5\bm{I}_3. \end{equation}$

The density of the shell is given by

$\begin{equation}\label{2} \rho = \frac{m}{(4/3)\pi((1+\Delta)^3-1)r^3}, \end{equation}$

where the volume (the denominator) is the difference between the volumes of the outer and inner spheres. Substituting $\eqref{2}$ into $\eqref{1}$, we get

$\begin{equation}\label{3} \bm{J}_\circ = \lim _{\Delta\to0}\ (2/5)m\frac{(1+\Delta)^5-1}{(1+\Delta)^3-1}r^2\bm{I}_3. \end{equation}$

We can easily determine that

$\begin{equation}\label{4} \lim _{\Delta\to0}\ \frac{(1+\Delta)^5-1}{(1+\Delta)^3-1} = \frac{5}{3} \end{equation}$

using the small block of Python code

import sympy
Δ = sympy.symbols("Δ")
lim = ((1 + Δ) ** 5 - 1) / ((1 + Δ) ** 3 - 1)
print(lim.expand().simplify().subs({Δ: 0}))

Substituting $\eqref{4}$ back into $\eqref{3}$, we achieve the expected result for the spherical shell:

$\begin{equation}\label{5} \bm{J}_{\circ} = (2/3)mr^2\bm{I}_3. \end{equation}$

## Ellipsoidal shell inertia

We are now ready to generalize to the inertia of the ellipsoidal shell. In this section of the article we'll now use $\bm{J}_\circ$ and $\bm{J}_\bullet$ to refer to the inertia tensors of the ellipsoidal shell and solid ellipsoid, respectively. We will consider an ellipsoid of mass $m$ and semi-axes $a,b,c$. This ellipsoid consists of the set of all points $\bm{x}\in\mathbb{R}^3$ that satisfy $\bm{x}^T\bm{E}^{-1}\bm{x}\leq 1$, where

$\begin{equation*} \bm{E} = \begin{bmatrix} a^2 & 0 & 0 \\ 0 & b^2 & 0 \\ 0 & 0 & c^2 \end{bmatrix}. \end{equation*}$

The inertia tensor of a solid ellipsoid of uniform density is $\bm{J}_\bullet=(1/5)m\bm{S}$, where

$\begin{equation*} \bm{S} = \begin{bmatrix} b^2+c^2 & 0 & 0 \\ 0 & a^2+c^2 & 0 \\ 0 & 0 & a^2+b^2 \end{bmatrix} = \mathrm{tr}(\bm{E})\bm{I}_3 - \bm{E}, \end{equation*}$

with $\mathrm{tr}(\cdot)$ denoting the matrix trace. Again we substitute in $m=\rho V=\rho(4/3)\pi abc$ to obtain $\bm{J}_\bullet=(4/15)\pi\rho abc\bm{S}$.

In the same way as for the spherical shell, we will obtain an ellipsoidal shell by subtracting an ellipsoid with semi-axes $a,b,c$ from a slightly larger ellipsoid with semi-axes $(1+\Delta)a,(1+\Delta)b,(1+\Delta)c$. Taking the limit $\Delta\to 0$, we get

$\begin{equation}\label{6} \bm{J}_\circ = \lim _{\Delta\to0}\ (4/15)\pi\rho((1+\Delta)^5-1)abc\bm{S}. \end{equation}$

with density

$\begin{equation}\label{7} \rho = \frac{m}{(4/3)\pi((1+\Delta)^3-1)abc}. \end{equation}$

Substituting $\eqref{7}$ into $\eqref{6}$, we get

$\begin{equation*} \bm{J}_\circ = \lim _{\Delta\to0}\ (1/5)m\frac{(1+\Delta)^5-1}{(1+\Delta)^3-1}\bm{S}. \end{equation*}$

Again making use of $\eqref{4}$, we obtain our final result:

$\begin{equation}\label{8} \bm{J}_\circ = (1/3)m\bm{S}. \end{equation}$

## Equivalent inertias

• The spherical shell inertia $\eqref{5}$ is of course a special case of the ellipsoidal shell inertia $\eqref{8}$ when $r=a=b=c$.
• The ellipsoidal shell inertia $\eqref{8}$ is equal to the inertia of a solid cuboid of uniform density with mass $m$ and side lengths $2a,2b,2c$.
• The ellipsoidal shell inertia $\eqref{8}$ is equal to the inertia of a system of six point masses, each with mass $m/6$, with one point each located at the ends of the semi-axes. That is, the point masses are located at $(\pm a,0,0)$, $(0, \pm b, 0)$, and $(0, 0, \pm c)$.