Derive, making use of an integral, the formula for the moment of inertia of a sphere about its diameter.

Answer

`0.4 MR^2`

Explanation

Divide the sphere into thin disks perpendicular to the axis of rotation (Fig.) The differential of mass is
`dm=(Mpir^2dz)/(4//3piR^3)=(3M)/(4R^3)(R^3-z^2)dz`
Since `r^2=R^2-z^2` . The moment of inertia of such a disk is
`dI=(r^2dm)/2=(3M)/(8R^2)(R^2-z^2)dz`
The moment of inertia of the sphere is
`I = 2 int_0^R(3M)/(8R^3)(R^4-2R^2z^2+z^4)dz`=
`=(3M)/(4R^3)[R^4int_0^Rdz-2R^2int_0^(R)z^2dz+int_0^Rz^4dz]=`
`=(3M)/(4R^3)[R^4z-2R^(2)(z^3)/2+z^5/5]_0^R=`
`=(3M)/(4R^3)(R^5-(2R^5)/(3)+R^5/5)=2/5MR^2`