A circular lamina of radius `a` and centre `O` has a mass per unit area of `kx^(2)`, where `x` is the distance from `O` and `k` is a constant. If the mass of the lamina is `M`, find in terms of `M` and `a`, the moment of inertia of the lamina about an axis through `O` and perpendicular to the lamina.

Explanation

`M=int_(0)^(a)(dM)`
`=int_(0)^(1)(2pix)dX(kx^(2))=(pika^(4))/(2)`
`thereforepik=(2M)/(a^(4))`
now `I=int_(0)^(a)dI=int_(0)^(aa)(dM)X^(2)`
`=int_(0)^(a)(2pikx^(3))x^(2)=((pik)d^(6))/(3)`
Substituting the value of `pik`, we get
`I=(2)/(3)Ma^(2)`