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?
 A: 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.
A: 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 
$$I=\frac23\,4\pi\int_0^R\rho(r)r^4\mathrm dr.$$
