Hint: \(\displaystyle I = \int R^2 \ dm\)
Using the hint. I will rotate around the \(\displaystyle z\) axis.
\(\displaystyle I_z = \int R^2 \ dm = \int R^2 \rho \ dV\)
If I take the volume of a thin cylindrical shell, I get:
\(\displaystyle I_z = \int R^2 \rho \ dV = \int R^2 \rho \ 2\pi Rh \ dR = 2\pi \rho h\int R^3 \ dR\)
I am ready to integrate
\(\displaystyle I_z = 2\pi \rho h\int_{R_1}^{R_2} R^3 \ dR = 2\pi \rho h \frac{R^4}{4}\bigg|_{R_1}^{R_2} = 2\pi \rho h\left(\frac{R^4_2}{4} - \frac{R^4_1}{4}\right)\)
\(\displaystyle = \pi \rho h\left(\frac{R^4_2 - R^4_1}{2}\right) = \pi \rho h\frac{\left(R^2_2 + R^2_1\right)\left(R^4_2 - R^2_1\right)}{2}\)
We know that \(\displaystyle \rho\) is the volume density, then
\(\displaystyle \rho = \frac{M}{V} = \frac{M}{h\pi\left(R^2_2 - R^2_1\right)}\)
This gives:
\(\displaystyle I_z = \pi \frac{M}{V} h\frac{\left(R^2_2 + R^2_1\right)\left(R^4_2 - R^2_1\right)}{2} = \pi \frac{M}{h\pi\left(R^2_2 - R^2_1\right)} h\frac{\left(R^2_2 + R^2_1\right)\left(R^4_2 - R^2_1\right)}{2}\)
With a little simplification, finally we get
\(\displaystyle I_z = \textcolor{blue}{\frac{1}{2}M\left(R^2_2 + R^2_1\right)}\)
