Moment of Inertia of a wire A wire has the shape of the circle x^2 + y^2 = a^2. Determine its
mass and moment of inertia about a diameter if the density at (x, y)
is |x| + |y|.
I found that the mass is 4a^2 (don't know if it's the answer), but couldn't find the moment of inertia, wich answer is 4a^4.
Thanks!
 A: Consider an element of circle whose mass is $$\rho a\delta\theta$$ and therefore whose moment of inertia about an axis perpendicular to the plane of the circle and through the centre of the circle is $$\rho a^3 \delta\theta$$
$$=a^3(|x|+|y|)\delta\theta$$
Therefore the moment of inertia about this axis is $$a^3\int_0^{2\pi}(|x|+|y|)d\theta$$
$$=a^4\int_0^{2\pi}(|\cos\theta|+|\sin\theta|)d\theta$$
$$=8a^4$$
Then, applying symmetry in $x$ and $y$ and the perpendicular axes theorem, the required moment of inertia about the diameter is $$\frac 12 8a^4=4a^4$$
A: The mass of the wire is given by the absolute line integral of the density: $$\begin{align}
M=\int_C\rho\;ds=\int_C\left|x\right|+\left|y\right|ds&=\int_0^{2\pi}(\left|a\cos\theta\right|+\left|a\sin\theta\right|)\;a\;d\theta \\
&=a^2\int_0^{2\pi}\left|\cos\theta\right|+\left|\sin\theta\right|d\theta=8a^2.\end{align}$$  
The moment of inertia about a diameter can be found by taking advantage of a symmetry.  
Consider two small sections $\rho\;ds$ of the wire, one at an angle of $\theta$ from the chosen diameter, the other at $\theta+\frac\pi 2$. Since $\rho(\theta)=\rho(\theta+\frac\pi 2)$, their net contribution to the moment of inertia is $(a\cos\theta)^2\rho(\theta)\;ds+(a\sin\theta)^2\rho(\theta)\;ds=a^2\rho(\theta)\;ds$, i.e., the same as if half of their combined mass were concentrated at the “equator.” Therefore, the total moment of inertia about a diameter of the wire loop is $$I_\text{diam.}=\frac12Ma^2=4a^4.$$One can, of course, get the same result by computing $I_x=\int_C\rho y^2 ds$ directly and making a similar appeal to symmetry to show that this equals the moment of inertia about any other diameter.
