# Dbl. int. - center of mass

1. Mar 1, 2005

### tandoorichicken

Any hints on how to approach this problem?

A lamina occupies the part of the disk $x^2 + y^2 <= 1$ in the first quadrant. Find the center of mass if the density at any point is proportional to the square of its distance from the origin

Last edited: Mar 1, 2005
2. Mar 1, 2005

### tandoorichicken

And just for future reference, how do you do inequalities in latex?

3. Mar 1, 2005

### stunner5000pt

$$\leq {and} \geq$$ the code reference file is found by clicking on the code and then clicking the link

4. Mar 1, 2005

### Gamma

I don't know what responses you got for your earlier thread. I could not find it. Any way, use the following to find the CM.

$$M r_{cm} = \int r_m dm$$

Since the density depends only on the radius, CM should be along the $\theta = \pi/2$ line.

Choose a small element at $(r, \theta )$ and proceed.