Is a double integral required to find the moment of Inertia of a non-uniform sphere?

Consider some ball of given radius $R$, with a mass density function that depends on the radial variable, $\rho=\rho(r)$ where $r$ is the distance from the center of the sphere. What is the moment of inertia of the ball with respect to an axis through its center?

From the definition of the moment of intertia, it seems that we need to use discs and a double integral here, which is weird because the lecturer insists we don't need to know any multivariable calculus in order to solve this question. Is there a way to approach this problem without using multiple integrals?

• You can use a single integral if you consider multiple shells, like an onion. Commented Oct 23, 2014 at 4:42
• Commented Oct 23, 2014 at 4:46

As you can see, $\rho$ only depends on a single variable: $r$. Thus, it should be intuitive that one can do this problem by integrating only over the variable $r$.

To see what you are supposed to do, consider what happens if you fix $r$: You obtain a spherical shell (as was pointed out in the comments). The moment of inertia of a spherical shell is quite easy to calculate. Once you've done that, use the idea that a ball can be viewed as the infinite sum of thin spherical shells of varying radius $r$ (this is where the integration comes in). You should be able to obtain an answer without performing a double integral.

• The problem I forsaw with spherical shells is that the axis is a straight line through a diameter, thus while every point on the spherical shell will have the same density, it won't have the same distance from the axis.
– MT_
Commented Oct 23, 2014 at 5:41
• @user92774 - What's the moment of inertia of a spherical shell about an axis through the center of the shell? Commented Oct 23, 2014 at 9:58
• @user92774 sorry, are you asked to compute the moment of inertia around an axis that is not through the center of the ball? I didn't get this impression from you original post. Please let me know!
– Danu
Commented Oct 23, 2014 at 11:48
• ah, no, I got it now. I was confused because I was wondering how to deal with the non-uniform density that varied with distance from the center of mass (point) while moment of inertia varied with distance from the axis (line) going through the diameter. But then I realized I can apply the $2/3 MR^2$ formula to account for the latter and I got the right answer. Thanks!
– MT_
Commented Oct 23, 2014 at 18:55

This solution uses multiple integrals but you don't need to compute them. The final computation is a single integral.

The ball has spherical symmetry so its momentum of inertia is the same with respect to the $x$-axis, the $y$-axis and the $z$-axis $I=I_x=I_y=I_z$ with $$I_z=\iiint \rho(r)\left(x^2+y^2\right)\mathrm dx\mathrm dy\mathrm dz$$ now sum up the three momenta $$I_x+I_y+I_z=3I=\iiint \rho(r)\left(x^2+y^2+x^2+z^2+y^2+z^2\right)\mathrm dx\mathrm dy\mathrm dz\\=\iiint\rho(r)\,2r^2\mathrm dx\mathrm dy\mathrm dz$$ as a conclusion the result is $$I=\frac23\,4\pi\int_0^R\rho(r)r^4\mathrm dr.$$