# Calculating Mass and Center of Mass

## Homework Statement

Let D be the upper half of the disc x2+y2 ≤1(that is, the part of the disc with y ≥ 0). Suppose the lamina Ω fills the region D, and has density given by ρ(x, y) = |x|.

(a) Calculate the mass of the lamina.
(b) Find the coordinates of the center of mass of the lamina.

## The Attempt at a Solution

So, first, I'm going to rewrite this |x| as y=-x when for when x≤0 and y=x for when x≥0. Therefore, the quarter-circle given by the boundary in the first quadrant allows 0≤x≤1 and 0≤y≤1.

I integrate ∫(0≤x≤1)∫(0≤y≤1) x dydx and get 1/2. Observing that everything is symmetrical, I simply multiply by 2 to find the total mass, which I find to be 1.

Again, by symmetry, I reason that the center of mass must lie on the y-axis, so I will let the x-coordinate of the center of mass equal 0. To find the y-coordinate, I integrate

∫(0≤x≤1)∫(0≤y≤1) xy dydx + ∫(-1≤x≤0)∫(0≤y≤1) -xy dxdy (???)

and get 1/2. Therefore, the center of mass exists at (0,1/2).

It looks like you're integrating over a rectangular region in your solution, but the region given is the upper half of a disk. Also, polar coordinates seem like they would be more friendly here.

I was thinking polar would be easier, too. But I didn't know how to convert y = |x| into polar coordinates. But, yes, clearly my bounds are not correct.

well your bounds should be pretty easy, theta goes from 0 to pi and r goes from 0 to 1. Your question of rho = |x| comes down to knowing x = r cos(theta) and |r cos(theta)| = r |cos(theta)|, which is just a question of knowing where cos(theta) is positive and where it is negative.

Okay. So, first, why have you replaced y with rho in y = |x|? Does rho = rsin(theta)?

But anyway, let's see if I can integrate this properly. I think I should have the sum of two integrals so that I can deal with the bounds independently. So, cos(theta) is non-negative in 0≤theta≤pi/2. I think one of my integrals should then be

∫(0≤theta≤pi/2)∫(0≤r≤1) r^2 * cos(theta) dr dtheta.

This gives me 1/3. My other integral is

∫(pi/2≤theta≤pi)∫(0≤r≤1) r^2 * -1 * cos(theta) dr dtheta.

This, unexpectedly, also gives 1/3. Therefore mass is 2/3 (?)

i did this because in the problem statement you said rho = |x|. But yes, this looks to be correct.

Are you referring to Ω? Isn't that omega?

$$\rho (x,y) = |x|$$

Oh, right. I'm just being stupid.