Why doesn't this way of calculating the moment of inertia of a sphere work? Instead of the usual approach of integrating a bunch of discs, I do it differently. I integrate by subdividing the sphere into a bunch of concentric spheres containing mass on their surface.
The integral becomes:
$\int_0^R r^2 \cdot (4\pi r^2) \cdot \rho \space dr$ which ends up equaling $\frac{3}{5}MR^2$
Inside the integral I'm taking the squared distance from the axis times the surface area of the sphere times the density. The right answer is of course $\frac{2}{5}MR^2$, but why does my approach fail? Conceptually, why does this method of breaking the sphere into smaller shells not work?
 A: A little conceptual tweak and your approach works:
The moment of inertia integral is given as  $$I= \int (x^2 +y^2) dm $$
Here I find the inertia of the sphere about the $z$ axis, but that is same as finding inertia of sphere about any other axis. We can add and subtract $\int z^2 dm$:
$$ I= \int (x^2 + y^2 +z^2) dm - \int z^2 dm = \frac{3}{5} mr^2 - \int z^2 dm$$
The above simplification used the fact that $x^2 +y^2 +z^2= r^2$, i.e: the spherical radius.
Due to the symmetery of sphere, we may argue that $\int x^2 dm = \int y^2 dm = \int z^2 dm$, hence $\int z^2 dm = \frac{I}{2}$:
$$I = \frac35 mr^2 - \frac{I}{2}$$
Or,
$$ I = \frac{2}{5} mr^2$$
QED
A: The moment of inertia is defined relative to an axis, not a point. Therefore the distance you need is from a mass element in the sphere to the $z$-axis, not to the origin. This is why the disk method is used -- it takes advantage of the symmetry. Your expression uses the distance $r$ to the origin, whereas the distance to the $z$ axis in spherical coordinates is $r\sin \theta$ (where $\theta$ is the angle between the $z$ axis and the point of interest).
The correct expression for a spherical object with arbitrarily varying density in spherical coordinates is
\begin{eqnarray}
I &=& \int_0^R {\rm d} r r^2 \int_0^{\pi} {\rm d\theta} \sin \theta \int_0^{2\pi}{\rm d}\phi \rho(r,\theta,\phi)\Delta(r,\theta,\phi)^2,
\end{eqnarray}
where $\rho(r,\theta,\phi)$ is the density and $\Delta(r,\theta,\phi)=r\sin\theta$ is the distance from a mass element at $r,\theta,\phi$ to the $z$ axis.
For completeness we can work this out for a constant density sphere, $\rho(r,\theta,\phi)=3 M/(4\pi R^3)$ (note that the integrand doesn't depend on $\phi$, so the $\phi$ integral is just $2\pi$)
\begin{eqnarray}
I &=& 2\pi \frac{3M}{4\pi R^3} \int_0^R {\rm d} r r^2 \int_0^{\pi} {\rm d\theta} \sin \theta\ \left(r \sin\theta\right)^2 \\
&=& \frac{3 M}{2 R^3} \int_0^R {\rm d} r r^4 \int_0^\pi {\rm d}\theta \sin^3 \theta \\
&=& \frac{3 M}{2 R^3} \times \frac{R^5}{5} \times \frac{4}{3} \\
&=& \frac{2}{5} MR^2
\end{eqnarray}
A: Andrew's answer explains what's wrong with your reasoning: the first $r^2$ should be replaced by just $x^2+y^2$, since the moment is based on the distance to the $xy$-axis, not the distance to the origin. But there is a slick trick we can use taking advantage of the symmetry of the sphere: since it doesn't matter whether we use the $xy$-, $xz$-, or $yz$-axis, we should get the same result for all three:
$$I=\int_0^R (x^2+y^2)(4\pi r^2)\rho\,dr=\int_0^R (x^2+z^2)(4\pi r^2)\rho\,dr=\int_0^R (y^2+z^2)(4\pi r^2)\rho\,dr.$$
If we add these three expressions we get
$$3I=\int_0^R (2x^2+2y^2+2z^2)(4\pi r^2)\rho\,dr=\int_0^R 2r^2(4\pi r^2)\rho\,dr.$$
So this explains why the correct result is precisely $2/3$ of what you got. (This same trick also explains the $2/3$ factor in the moment of inertia of a spherical shell.)
A: That's a good, well-stated question and the premise is indeed correct: your independent approach has failed. It fails because you're using $r$ for the distance of the mass of a spherical shell of radius $r$ from the axis. But that's the distance from the center, the formula should actually use the distance from the closest part of the axis to that particular mass.
And unfortunately, your spherical shell is not of a fixed distance from the axis. The parts at the "poles" are very close to the axis, and the parts at the equator really are the maximum distance $r$. So you can't really put in any number here, you'll have to slice your sphere up into pieces that are of a uniform distance from your axis (or pieces like disks with moments of inertia that you already know).
A: Since $V=\int_0^R (4\pi r^2) \space dr$ produces the correct formula for the volume of a sphere, the problem is elsewhere. Namely in the $r^2$ factor inside the volume integral.
The problem is that MMOI is measured by the perpendicular distance of a particle to an axis. So the $r^2$ factor inside the integral is incorrect, as it used the radial distance and not the perpendicular distance.
The correct way to integrate is to consider spherical coordinates and use
$$ V = \int_0^R \int_{-\pi/2}^{\pi/2} \int_0^{2 \pi} ( r^2 \cos \psi) \space {\rm d}\varphi \space {\rm d}\psi \space {\rm d} r = \frac{4}{3} \pi R^3$$
and since $m = \rho V$ the MMOI tensor is
$$ \mathrm{I} = \frac{m}{V} \int_0^R \int_{-\pi/2}^{\pi/2}   2 \pi \begin{bmatrix} 
  \frac{ r^4 \cos \psi (\sin^2 \psi+1)}{2} & & \\ & r^4 \cos^3 \psi & \\ & & \frac{ r^4 \cos \psi (\sin^2 \psi+1)}{2} \end{bmatrix} (2 \pi r^2 \cos \psi)\space {\rm d}\psi \space {\rm d} r  $$
$$ \mathrm{I} = \frac{m}{V} \begin{bmatrix}
 \frac{8}{15} \pi R^5 & & \\ & \frac{8}{15} \pi R^5 & \\ & & \frac{8}{15} \pi R^5 \end{bmatrix} = \begin{bmatrix}
 \frac{2}{5} m R^2 & & \\ & \frac{2}{5} m R^2 & \\ & & \frac{2}{5} m R^2 \end{bmatrix}$$

The definition of the MMOI tensor from the volume integral is
$$ I = \frac{m}{V} \int \begin{bmatrix} y^2+z^2 & -x y & -x z \\ -x y & x^2+z^2 & -y z \\ -x z & -y z & x^2+y^2 \end{bmatrix} {\rm d}V $$
and the position in spherical coordinates is
$$ \pmatrix{x \\ y \\ z} = \pmatrix{ r \cos \psi \cos \varphi \\ r \sin \psi \\ r \cos \psi \sin \varphi} $$
