# Find center of mass of the lamina

1. Mar 26, 2012

### MeMoses

1. The problem statement, all variables and given/known data

Find the center of mass of the lamina which occupies if the density at any point is proportional to the distance from the origin.

36 <= x^2+y^2 <= 81, y >= 0

2. Relevant equations

3. The attempt at a solution
Rewrote it in polars to get 6<r<9. The x is clearly 0 as you can see from symmetry, but I can't get y. The total mas should be 45pi/2 takeing the integral of r dr d(theta) with 0<theta<pi and 6<r<9. And to find y I need to take the Momentx/mass, but im not getting the right answer. For momentx I took the double integral of r**2*sin(theta) with the same limits to get 342. Can you see where I'm going wrong. Also density=k*r but that just leaves an extra k on mass and Mx which cancel. Any help would be great

2. Mar 26, 2012

### HallsofIvy

Staff Emeritus
Yes, the x coordinate is 0, by symmetry. The integration, using polar coordinates, will be with r= 6 to 9 and $\theta= 0$ to $\pi$.

I suggest you do the mass integral over again. I get much more then "45/pi/2".

I do get the same thing for Momentx as you did.

Last edited: Mar 26, 2012
3. Mar 26, 2012

### MeMoses

I it turns out I just forgot an r in my integrals, just another careless mistake.