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Suppose I have a half disc and the coordinates axes at the centre of base of the disc. For the given system, I have surface mass density $S$ as $$S=S_0 sin\theta$$($S_0$ being positive constant). I need to get to the center of mass coordinates. Since the half disc is in $x-y$ plane, so $$z_{cm}=0$$Also since $sin(\frac{\pi}{2}+\theta)=sin(\frac{\pi}{2}-\theta)$, so it turns out that the surface mass density of two points symmetrical about the $y$ axis are equal. This means the mass of left and right part of half disc are equal, so $$x_{cm}=0$$Now for the $y$ centre of mass, I have $$y_{cm}=\frac{\int ydm}{\int dm}=\frac{\int y S_0sin\theta dA}{S_0sin\theta dA}$$How do I compute this integral? Do I use polar coodinates or cartesian ones? I do not see a way of expressing $dA$ in terms of $dx$ and $dy$. Does this turn out to be a surface integral?(I've never evaluated one before)

Thanks in advance. I'm sorry if this looks more of a Maths SE question than a Physics SE question. This is indeed, a physics question.

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I think you mean $y$ instead of $y^2$ in the integral. –  Michael Brown Apr 2 '13 at 3:54
@MichaelBrown:Thank you, I've corrected that. –  Ashish Gaurav Apr 2 '13 at 4:07
Also, I would recommend not carrying around your denominator. What should your answer for $\int dm$ be? –  Jerry Schirmer Apr 2 '13 at 4:38
Try breaking disc into small half rings and then integrate . I would be much simpler. $y_{cm-half-ring}=2\pi/R$ –  Mr.ØØ7 Apr 4 '13 at 15:00
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2 Answers 2

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I'd recommend performing the integrals in polar coordinates $(r, \theta)$ on the plane. In these coordinates, the area element is $$ dA = r\,dr\,d\theta $$ If you're having trouble even after this suggestion, comment with your confusion.

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this worked well; what if I had to compute it in cartesian coordinates-I'm not saying I'll derive it, just the area is what I need in future. Is it $dA=dx\times dy$? –  Ashish Gaurav Apr 2 '13 at 9:07
one more thing: does computing wrt $d\theta$ first and then $dr$ also work? –  Ashish Gaurav Apr 2 '13 at 9:15
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$$y_{cm}=\frac{\int ydm}{\int dm}=\frac{\int_0^\pi\int_0^R(rsin\theta)\times (Ssin\theta)\times rdrd\theta}{\int_0^\pi\int_0^R(Ssin\theta)\times rdrd\theta}=\frac{\int_0^\pi\int_0^Rr^2sin^2\theta drd\theta}{\int_0^\pi\int_0^Rrsin\theta drd\theta}$$ $$y_{cm}=\frac{R}{3}\times\frac{\int_0^\pi(1-cos2\theta)d\theta}{\int_0^\pi sin\theta d\theta}=\frac{\pi R}{6}$$ Whoa, I computed a double integral for the first time!

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