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Math Help - Line integral over a circle, how to do it?

  1. #1
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    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!
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  2. #2
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    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}}
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  3. #3
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    Re: Line integral over a circle, how to do it?

    Quote Originally Posted by fysikbengt View Post
    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.
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    Smile Re: Line integral over a circle, how to do it?

    Quote Originally Posted by HallsofIvy View Post
    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.
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