# Thread: Line integral over a circle, how to do it?

1. ## Line integral over a circle, how to do it?

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

Let me do a simple integration of a ring when the axis is in the center of the circle. Then
$dm=\frac{M}{2\pi R}ds$

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

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

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

This was quite easy. Now suppose I put the axis at the bottom of the circle. Then
$y=\sqrt{R^{2}-x^{2}}+R$

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

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

Now how do I define ds in a good way to calculate this integral. Is it as simple as letting x run from -R to R? No it cant be!

2. ## Re: Line integral over a circle, how to do it?

Ok, I solved the problem, so maybe i should post the rest in the calculus forum. Anyway this is what i did: $ds=\sqrt{1+(\frac{dy}{dx})^{2}}$

3. ## Re: Line integral over a circle, how to do it?

Originally Posted by fysikbengt
Ok, I solved the problem, so maybe i should post the rest in the calculus forum. Anyway this is what i did: $ds=\sqrt{1+(\frac{dy}{dx})^{2}}$
I assume you mean $ds= \sqrt{1+ \left(\frac{dy}{dx}\right)^2}dx$

Another thing you could do is use polar coordinates. $x= cos(\theta)$, $y= sin(\theta)$. Then $ds= \sqrt{\left(\frac{dx}{d\theta}\right)^2+ \left(\frac{dy}{d\theta}\right)^2}= d\theta$.

4. ## Re: Line integral over a circle, how to do it?

Originally Posted by HallsofIvy
I assume you mean $ds= \sqrt{1+ \left(\frac{dy}{dx}\right)^2}dx$

Another thing you could do is use polar coordinates. $x= cos(\theta)$, $y= sin(\theta)$. Then $ds= \sqrt{\left(\frac{dx}{d\theta}\right)^2+ \left(\frac{dy}{d\theta}\right)^2}= d\theta$.
Yes that sounds more clever. I had to change variables anyway. After a few hours I managed to solve it only to find there is a theorem (the parallell axis theorem) which made the whole calculation obsolete. But I dont regret anything.