Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

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.

share|cite|improve this question

migrated from Apr 2 '13 at 5:37

This question came from our site for active researchers, academics and students of physics.

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. $y_{cm-half-ring}=2\pi/R$ – ABC Apr 4 '13 at 15:00
up vote 1 down vote accepted

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.

share|cite|improve this answer
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

$$y_{cm}=\frac{\int ydm}{\int dm}=\frac{\int_0^\pi\int_0^R(rsin\theta)\times (Ssin\theta)\times rdrd\theta}{\int_0^\pi\int_0^R(Ssin\theta)\times rdrd\theta}=\frac{\int_0^\pi\int_0^Rr^2sin^2\theta drd\theta}{\int_0^\pi\int_0^Rrsin\theta drd\theta}$$ $$y_{cm}=\frac{R}{3}\times\frac{\int_0^\pi(1-cos2\theta)d\theta}{\int_0^\pi sin\theta d\theta}=\frac{\pi R}{6}$$ Whoa, I computed a double integral for the first time!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.