Moment of inertia of the ring through the diameter

I know that the moment of inertia of the ring through the Diameter is $I_{x}=I_{y} = mr^2/2$.

But I cannot get this formula using the integral. Give me some hints how to do this. Thanks a lot!

• Your question is not clear. Please provide a diagram or plot. Oct 19, 2017 at 0:04
• @CyeWaldman The axis of rotation passes through the center of the circle and lies in the plane of this circle Oct 19, 2017 at 0:16

$$I=\int r'^2 dm$$

$$dm={{M}\over{2\pi}}d\theta$$

$$r'=r\cos\theta$$

$$I=\int_0^{2\pi} r^2\cos^2\theta {{M}\over{2\pi}}d\theta$$

$$I={{Mr^2}\over{2\pi}}\int_0^{2\pi} \cos^2\theta d\theta$$

$$I={{Mr^2}\over{2\pi}}[{{\theta}\over{2}}+{{\sin 2\theta}\over{4}}]\biggl|_{\theta=0}^{2\pi}$$

$$I={{Mr^2}\over{2\pi}}[(\pi+0)-(0+0)]$$

$$I={{Mr^2}\over{2}}$$