Calculating moment of inertia if a solid ball, using infinitesimally thick spheres Consider a Sphere of mass $M$, and volume $V = \frac{4}{3}π(r)^3$ and uniform density $p$, If I want to get its moment of inertia around an axis running through its centroid then I shall Integrate $dm r^2$.
And If I cut the Sphere into infinitesimally small volumes of spheres I get $dV = 4πr^2 dr$, $dm = p dV$. 
So I get to integrate  $p.4πr^4 dr$ from zero to the whole radius if the Sphere, now the final result is $\frac{3MR^2}{5}$ and that's $1.5$ times the real moment if inertia. So what did I do wrong?
 A: The moment of inertia of an object about the $z$-axis is
$$\int r^2 dm = \int x^2 + y^2 dm.$$
However, for the spherical shell, you used
$$\int x^2 + y^2 + z^2 dm.$$
By symmetry, all three of these terms contribute equally, and you only want two of them, so your answer should be $2/3$ as big. This gives $(2/3)(3/5)MR^2 = (2/5)MR^2$, the correct answer.
A: The volume element you are choosing has symetry related to the center of the sphere, i.e., a point instead of a line. The axis around which the sphere will rotate is a line that passes by its center. This will allow you visualize the correct differential of volume you have to choose.
A: You cannot do $ dI =r^2 dm$ for a hollow sphere. In fact, a hollow sphere of radius $R$ and mass  $M$ has $ I = \frac{2MR^2}{3}$ Link to Wiki entry on MOI
You can use this expression,   to derive a solid sphere's MOI, using integral sum over infinitesimal hollow spheres of radius $0$ to $R$. Here are the steps....
$\rho = \frac{M}{ \frac{4}{3} \pi R^3}$
and
$dm = \rho  \;4 \pi r^2 dr  =  \frac{3M}{   R^3}   \;  r^2 dr   $
*(using the formula for hollow sphere)*

$I = \int_0^R \frac{2r^2}{3} dm $   
$I = \int_0^R \frac{2r^2}{3} \frac{3M}{   R^3}   \;  r^2 dr$
$I = \frac{2M}{R^3}  \int_0^R   r^4     dr$
$I = \frac{2M}{R^3}  \frac{R^5}{5}      = \frac{2MR^2}{5}$
