Find center of mass given Lamina pictured: https://s3.amazonaws.com/wamapdata/qimages/qtrring.gif

with inner radius of 3 and an outer radius of 7, and a density function $$\rho(x,y) = \frac{x}{x^2+y^2}.$$

My Assumptions: $$0 < \theta < \frac{\pi}{2}, \quad 3 < r < 7.$$

Originally I went with polar coordinates and got $M = 4, M_x = 2, M_y = \pi$. It was incorrect. Can anyone tell me what the answer is and what I did wrong?


We note that we have generically for the mass of a plane object $\Sigma$ we have:


In this case we have that: $x=r\cos(\theta),y=r\sin(\theta)$,where $r\in[3,7]$ and $\theta\in[0,\frac{\pi}{2}]$. And thus:


We also note that:

$$\begin{pmatrix}\overline{x} \\ \overline{y}\end{pmatrix}=\frac{1}{M}\iint_{\Sigma}\vec{r}\rho(\vec{r})\:\mathrm{d}A$$

Using the same bounds as before we get:

$$\begin{pmatrix}\overline{x}\\ \overline{y}\end{pmatrix}=\frac{1}{4}\int_{3}^{7}\int_{0}^{\frac{\pi}{2}}\begin{pmatrix}r\cos(\theta)\cos(\theta) \\ r \sin(\theta)\cos(\theta)\end{pmatrix}\:\mathrm{d}\theta\:\mathrm{d}r=\frac{1}{4}\int_{3}^{7}\begin{pmatrix}\frac{\pi r}{4} \\ \frac{r}{2}\end{pmatrix}\:\mathrm{d}r=\begin{pmatrix}\frac{5\pi}{4} \\ \frac{5}{2}\end{pmatrix}$$

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  • $\begingroup$ He's correct for the mass: $7-3=4$. $\endgroup$ – Mark Fantini Jun 15 '14 at 19:29
  • $\begingroup$ @Fantini Ahaha, my mistake! I've corrected it now, thanks! $\endgroup$ – Thomas Russell Jun 15 '14 at 19:30
  • $\begingroup$ It happens (more often than it should :(). $\endgroup$ – Mark Fantini Jun 15 '14 at 19:32
  • $\begingroup$ @Fantini Tell me about it; mental arithmetic has never been my strong point haha! $\endgroup$ – Thomas Russell Jun 15 '14 at 19:33
  • $\begingroup$ @Shaktal Its good to know that my Mass is correct but I'm not sure how I got my Moments wrong. How did you find those? $\endgroup$ – Danh Jun 15 '14 at 19:37

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