Suppose I want to evaluate the moment of inertia of a circle (or ring or hoop) The moment of inertia is simply this integral $\displaystyle I=\int \rho^{2}dm$, where $\displaystyle \rho$ is the distance from the axis or pivot andmis the mass.

Let me do a simple integration of a ring when the axis is in the center of the circle. Then

$\displaystyle dm=\frac{M}{2\pi R}ds$

$\displaystyle y=\sqrt{R^{2}-x^{2}}$

$\displaystyle y=-\sqrt{R^{2}-x^{2}}$

$\displaystyle I=\int \rho^{2}dm=\int (x^{2}+y^{2})dm=\int_{lower}R^{2}\frac{M}{2\pi R}ds+\int_{upper}R^{2}\frac{M}{2\pi R}ds=R^{2}\frac{M}{2\pi R}\pi R+R^{2}\frac{M}{2\pi R}\pi R=MR^{2}$

This was quite easy. Now suppose I put the axis at the bottom of the circle. Then

$\displaystyle y=\sqrt{R^{2}-x^{2}}+R$

$\displaystyle y=-\sqrt{R^{2}-x^{2}}+R$

$\displaystyle I=\int \rho^{2}dm=\int (x^{2}+y^{2})dm=\int_{lower}(R^{2}-x^{2}+R^{2}-2R\sqrt{R^{2}-x^{2}}+x^{2})\frac{M}{2\pi R}ds+\int_{upper}(R^{2}-x^{2}+R^{2}+2R\sqrt{R^{2}-x^{2}}+x^{2})\frac{M}{2\pi R}ds=\frac{M}{\pi}\int_{lower}(R-\sqrt{R^{2}-x^{2}})ds+\frac{M}{\pi}\int_{upper}(R+\sqrt{R^{2}-x^{2}})ds$

Now how do I definedsin a good way to calculate this integral. Is it as simple as lettingxrun from-RtoR? No it cant be!