What is the relation between the principal axes (of inertia) and the axes of symmetry of 3D sheet bodies (or 2D shapes)? I suspect that:


*

*if a 2D shape has an odd number of axes of symmetry (e.g. a T-shape (one) or an equilateral triangle (three)), there will be one axis of symmetry coinciding with a principal axis of the corresponding sheet body.

*if a 2D shape has an even or infinite number of axes of symmetry (e.g. an ellipse (two), square (four) or circle (infinite)), there will be two (perpendicular) axes of symmetry that will coincide with principal axes of the corresponding sheet bodies.
Is this intuition correct ? Is there some more formal result that would confirm this ?
 A: If a homogeneous 2D shape has an axis of symmetry, then this axis is also a principal axis of inertia: this is a general result, which can be proved as follows. 
Let's take an orthogonal coordinate system $(x,y)$ in the plane of the flat shape, with the origin at its center of mass. Suppose the shape is symmetric around the $y$-axis: for every point $P=(x,y)$ in the shape, the symmetric point $P'=(-x,y)$ also belongs to the shape. As a consequence, the off-diagonal term in the tensor of inertia (sometimes called a product of inertia) vanishes:
$$
I_{xy}=\iint\rho xy\,dxdy=0,
$$
where $\rho$ is the (constant) surface density.
That means that the $x$-axis and the $y$-axis are principal axes of inertia for the plate.
Can there be another couple of principal axes of inertia, i.e. another coordinate system for which $I_{xy}=0$? It is a well known result in linear algebra that this can happen only if the moments of inertia associated with $x$ and $y$ are equal between them (so-called degenerate case): $I_{xx}=I_{yy}$, where:
$$
I_{xx}=\iint\rho y^2\,dxdy,
\quad\hbox{and}\quad
I_{yy}=\iint\rho x^2\,dxdy.
$$
In that case, ANY two perpendicular axes in the plane (passing through the center of mass) can be taken as principal axes of inertia.
This result has a very interesting consequence: if there are two non-orthogonal axes of symmetry (as in the case of an equilateral triangle), then we must be in the degenerate case and we can take any two perpendicular lines in the plane as principal axes of inertia.
