Rotation 3

Lets take an elemental disc with $x$ radius from the center of the half disc and width $dx$

Then see, If the Mass is $M$ , Then Mass of Elemental disc is = $dM$ = $\dfrac{2M x dx}{\pi(R_2^2-R_1^2)}$

So now see ,

We can find out the moment of inertia of the elemental disc like

$dI = dM . x^2$

So $dI=\dfrac{2M x^3 dx}{\pi(R_2^2-R_1^2)}$

So Integrating All Such Elemental Disc's Moment of Inertia will give us our required answer

So

$\large{\int_{R_1}^{R_2}{ dI}=\int_{R_1}^{R_2}\dfrac{2M x^3 dx}{\pi(R_2^2-R_1^2)}}$

So Now Integrating we'll get

$\large{\int_{R_1}^{R_2}{ dI}= I =\int_{R_1}^{R_2}\dfrac{2M x^3 dx}{\pi(R_2^2-R_1^2)}= \dfrac{2M}{\pi{R_2^2-R_1^2}} \int_{R_1}^{R_2}{x^3 .dx}}$

$\dfrac{2M}{\pi{R_2^2-R_1^2}} \int_{R_1}^{R_2}{x^3 .dx} = \boxed{\dfrac{M(R_2^2+R_1^2)}{2}}$