Moment of inertia of a circle

Question

A wire has the shape of the circle $x^2+y^2=a^2$. Determine the moment of inertia about a diameter if the density at $(x,y)$ is $|x|+|y|$

Thank you

Answer 1

Consider a small segment of the wire, going from $\theta$ to $\theta +d\theta$. The length of the small segment is $a \,d\theta$. The density varies, but is approximately $a|\cos\theta|+a|\sin \theta|$.

Take a particular diameter, say with rectangular equation $y=(\tan\phi) x$, or better, $x\sin \phi -y\cos\phi=0$. The perpendicular distance from $(a\cos\theta,a\sin\theta)$ to this diameter is $|a\cos\theta\sin\phi -a\sin\theta\cos\phi|$. So for the moment of inertia, we need to find $$\int_0^{2\pi} \left(a|\cos\theta|+a|\sin \theta|\right)\left(a\cos\theta\sin\phi -a\sin\theta\cos\phi\right)^2a\,d\theta.$$ The integration is doable, but not easy. Special cases such as $\phi=0$ or $\phi=\pi/4$ will not be too hard.

Remark: The perpendicular distance from a point $(p,q)$ to the line with equation $ax+by+c=0$ is $$\frac{|ap+bq+c|}{\sqrt{a^2+b^2}}.$$ There is a reasonably good discussion of the formula in Wikipedia.