# Moment of inertia of a cube about its diagonal

I want to calculate the moment of inertia of a cube with a side length $a$ and constant density $\rho$ about one of its diagonals.

I am a bit scared by the formula I have obtained for the square of the distance to the diagonal of the cube. Here's how I get it:

Let's define the diagonal as a vector with the start point $(0,0,0)^T$ and end point $(a,a,a)^T$. Thus our diagonal will be $\mathbf{c}=(a,a,a)^T$. A general point $A$ is given by $(x,y,z)^T$ and a point that is lying on the vector $\mathbf{c}$ is given by $P = (ta,ta,ta)^T$. We define the distance vector $$\mathbf{d} = \left(\begin{matrix}ta-x\\ta-y\\ta-z\\\end{matrix}\right)$$

which indeed will be the distance when the dot product $\mathbf{d}\cdot\mathbf{c}$ equals $0$. Calculating the dot product and finding $t={x+y+z \over3a}$we obtain $$\lvert\mathbf{d}\rvert^2=\frac19\left[(y+z-2x)^2+(x+z-2y)^2+(x+y-2z)^2\right]$$

a pretty huge equation for our squared distance. Calculating $$\Theta=\int_V\lvert\mathbf{d}\rvert^2\rho\operatorname{d}\mu(x,y,z)$$ seems to cost much time.

Is it an optimal solution and is it correct?

• That integral does not look so terrible to deal with. An alternative interesting opportunity is to find, through the Huygens-Steiner theorem, the moment of inertia of the (uniform massive) boundary of a unit cube rotating along its diagonal, then integrate this expression. Aug 3, 2014 at 20:34
• Also consider that if you replace $x$ with $u+a/2$ and so on, many terms of your integral just vanish due to symmetry. Aug 3, 2014 at 20:37
• @JackD'Aurizio it's not terrible at all, the integration itself is simple, but since there are a lot of terms, it's a bit time-consuming and not that elegant. As it was an exam task I thought I might have made a mistake when finding the distance. I'm not familiar with the theorem you mentioned, but will investigate it, thanks. I'm still interested if I computed the distance correctly Aug 3, 2014 at 20:39
• The distance computation looks ok. Consider that only the quadratic terms survive, since $$\int_{[-a/2,a/2]^3}uv\,d\mu = 0.$$ Aug 3, 2014 at 20:42
• This gives: $$\Theta = \frac{1}{a^3}\int_{[-a/2,a/2]^3}(u^2+v^2+w^2)\,d\mu = \frac{a^2}{4}.$$ Aug 3, 2014 at 20:45