MoI of a Sphere using Spherical Coordinates

[Ackbach]

Homework Statement

Calculate the moment of inertia of a uniformly distributed sphere about an axis through its center.

Homework Equations

I know that
$$I= \frac{2}{5} M R^{2},$$
where $M$ is the mass and $R$ is the radius of the sphere. However, for some reason,
when I do this integration in spherical coordinates, I do not get this result.

The Attempt at a Solution

The density is
$$ \rho= \frac{M}{V}= \frac{3M}{4 \pi R^{3}}.$$
Then the moment of inertia is
$$I= \int r^{2} \,dm= \rho \iiint_{V}r^{2} \,dV
= \rho \int_{0}^{2 \pi} \int_{0}^{ \pi} \int_{0}^{R} r^{4} \sin(\theta) \, dr \, d\theta \, d\phi
=4 \pi \rho \frac{R^{5}}{5}
$$
$$=4 \pi \frac{3M}{4 \pi R^{3}} \frac{R^{5}}{5}= \frac{3}{5} MR^{2}.$$
Where is my error?

I know, I know: all the derivations of this result I can find split the sphere up into
disks. I understand those derivations: I want to know where this one is wrong.

 

[WannabeNewton]

The $r$ in the moment of inertia integral is the perpendicular distance from the axis of rotation to a given mass element $dm$. The $r$ in spherical coordinates is the distance to $dm$ from the center of the 2-sphere. The two only agree for mass elements that lie on the equatorial plane.

 

[WannabeNewton]

If you really want to use spherical coordinates, then note that the perpendicular distance from the axis of rotation to a given mass element will be given by $r_{\perp } = r\sin\theta$ hence the moment of inertial integral becomes $I = \rho \int _{S^{2}}r^{2}\sin^{2}\theta dV = \frac{3}{10}MR^{2} \int_{0}^{\pi}\sin^{3}\theta d\theta = \frac{3}{10}\frac{4}{3}MR^{2} = \frac{2}{5}MR^{2}$.