What is Moment of Inertia of a cube about an axis passing through its centre and perpendicular to one of its faces? Give a solid cube of uniform mass distribution and total mass $M$. I want Moment of Inertia of this cube about an axis passing through its centre and perpendicular to one of its faces where a is the lenght of side of cube?
I tried taking a cuboidal volume element of length $a$, breadth $dy$ and height $a$ and integrating it over entire length of cube but I reach with an answer that moment of inertia $I = Ma$ but I know $I = M\frac{a^2}{6}$. What am I doing wrong?
 A: I'm not entirely clear on what you've already tried, but you're probably better off using a slice of a cylindrical arc as your volume element instead. Remember that the "radius" of the cube will vary depending on direction - what you've derived is the moment of inertia of a cylindrical tube.
You should end up with an integral of the form $\int\int\int_0^{r(\theta)}\cdots dz dr d\theta$.
A: An element of mass dM for the cube will be $dM = \rho * dV$, where dV is a volume element of dimension dx * dy * dz
Let's say the origin of the coordinate system for the cube is its center and the positive x-axis passes through the center of one face
In cartesian coordinates, the moment of inertia about the x-axis is defined as
$$I_{xx} = \int (y^2+z^2)*dM = \rho \int (y^2 + z^2)\, dxdydz,$$
integrated over the region of the cube, which has sides of length a.
This representation of the MOI is a little more natural than using cylindrical coordinates, especially since it is desired to calculate the MOI for a cube and not a cylinder.
