Find center of mass and coordinates using double integrals?

[monnapomona]

Homework Statement

Find the mass and center of mass of the lamina that occupies the region D and has the given density function ρ.
D is bounded by the parabolas y = x^2 and x = y^2; ρ(x, y) = 23√x

Homework Equations

m = $\int$$\int$$_{D}$ ρ(x, y) dA
x-bar = $\int$$\int$$_{D}$ x*ρ(x, y) dA
y-bar = $\int$$\int$$_{D}$ y*ρ(x, y) dA

The Attempt at a Solution

I integrated 23√x using order dydx with limits: x^2 ≤ y ≤ √x and 0 ≤ x ≤ 1.
m = 69/14.

The problem I'm having is with the coordinates. I first got (x-bar,y-bar) as (14/62, 28/1265) but that was wrong in my online assignment. I used the same limits of integration and the above equations to find x-bar and y-bar. I don't know where I'm going wrong with the coordinates... Am I suppose to be using different limits?

 

[Dick]

Homework Statement

Find the mass and center of mass of the lamina that occupies the region D and has the given density function ρ.
D is bounded by the parabolas y = x^2 and x = y^2; ρ(x, y) = 23

Homework Equations

m = $\int$$\int$$_{D}$ ρ(x, y) dA
x-bar = $\int$$\int$$_{D}$ x*ρ(x, y) dA
y-bar = $\int$$\int$$_{D}$ y*ρ(x, y) dA

The Attempt at a Solution

I integrated 23√x using order dydx with limits: x^2 ≤ y ≤ √x and 0 ≤ x ≤ 1.
m = 69/14.

The problem I'm having is with the coordinates. I first got (x-bar,y-bar) as (14/62, 28/1265) but that was wrong in my online assignment. I used the same limits of integration and the above equations to find x-bar and y-bar. I don't know where I'm going wrong with the coordinates... Am I suppose to be using different limits?

If you are trying to compute m you just integrate 23. Why are you integrating sqrt(x) as well? m isn't 69/14. Can you spell out what you are doing in a little more detail?

 

[SteamKing]

Your equations for x-bar and y-bar are incorrect. What you have are the expressions for the moments of D about the x and y axes. In order to find x-bar and y-bar, you must divide the moment values by the mass of the region D.

 

[monnapomona]

Your equations for x-bar and y-bar are incorrect. What you have are the expressions for the moments of D about the x and y axes. In order to find x-bar and y-bar, you must divide the moment values by the mass of the region D.

Oh my mistake, I used:

y-bar = 1/m $\int$$\int$$_{D}$ y*ρ(x, y) dA
x-bar = 1/m $\int$$\int$$_{D}$ x*ρ(x, y) dA

to find the coordinates.

 

[HallsofIvy]

Oh my mistake, I used:

y-bar = 1/m $\int$$\int$$_{D}$ y*ρ(x, y) dA
x-bar = 1/m $\int$$\int$$_{D}$ x*ρ(x, y) dA

to find the coordinates.

And in your case, $\rho(x,y)= 23$, a constant. So
$$\overline{x}= \frac{23}{m}\int\int x dxdy$$
$$\overline{y}= \frac{23}{m}\int\int y dxdy$$
with
$$m= 23\int\int dxdy$$

You are given "D is bounded by the parabolas y = x^2 and x = y^2". Those, of course, intersect at x= 0 and at x= 1. What will the limits of integration be?

 

[monnapomona]

If you are trying to compute m you just integrate 23. Why are you integrating sqrt(x) as well? m isn't 69/14. Can you spell out what you are doing in a little more detail?

Sorry for the confusion, the question was asking to integrate 23√x. I edited my first post.