multivariable calculus center of mass, check work.?

Question

Find the center of mass of the lamina which occupies the portion of the circle x^2 + y^2 <= 1, which is in the first quadrant and has density of rho(x,y)= x*y.

first, i calculated the mass: m= [0,1]int [0,1]int x*y dxdy
the mass i got was (1/4).
second i calculated x-bar (x coordinate):
4*[0,1]int[0,1]int x*x*y dxdy = x coordinate
x-coordinate = (4/6)=(2/3)
third, i calculated the y bar ( y coordinate):
4*[0,1]int[0,1]int y*x*y dxdy = y coordinate
y-coordinate = (4/6)=(2/3)

so the center of mass is (2/3,2/3). Is that correct? I just feel awkward seeing both coordinates equal. Is that what should happen in a circle or is it because the density function was just too simple and perfect?

Answer 1

Actually, your work is incorrect. By taking limits of integration 0
The best way to do this is to do the integration in polar coordinates. We have:

m = [0, π/2]∫[0, 1]∫ r cos θ r sin θ r dr dθ (don't forget the extra r when switching to polar)
= 1/2 [0, π/2]∫[0, 1]∫ r³ sin (2θ) dr dθ
= 1/8 [0, π/2]∫ sin (2θ) dθ
= -1/16 cos (2θ)|[0, π/2]
= 1/8

Now, the average x-coordinate, once again doing the integration in polar coordinates:

x-bar = 8 [0, π/2]∫[0, 1]∫ r cos θ r cos θ r sin θ r dr dθ
= 8 [0, π/2]∫[0, 1]∫ r⁴ cos² θ sin θ dr dθ
= 8/5 [0, π/2]∫cos² θ sin θ dθ
Let u=cos θ, du=-sin θ dθ, note that θ=0 ⇒ u=1 and θ=π/2 ⇒ u=0:
= -8/5 [1, 0]∫u² du
= 8/15

We could also calculate the average y-coordinate, but this is also 8/15. Why? We know that both the region over which we are integrating and the mass distribution are symmetric about the line y=x. Thus, if we were to exchange all the y's with x's and vice versa, we would have exactly the same problem. So the one aspect of your calculation that surprised you, namely that the x and y-coordinates of the center of mass are the same, had darn well better be true, otherwise you would know already that you did at least one of them wrong.

So we have that the center of mass is (8/15, 8/15). And we are done.

Answer 2

You wrote mass: m= [0,1]int [0,1]int x*y dxdy. It is wrong. You would calculate the mass of a square lamina with side 1.
Actually m = [0,1]int [0,sqrt(1-x^2)]int x*y dxdy = [0,1]int xdx [0,sqrt(1-x^2)]int ydy = 1/8
then calculate x-bar (x coordinate) = [0,1]int [0,sqrt(1-x^2)]int x*x*y dxdy / m = (1/15)/m = (1/15)/(1/8) = 8/15
similarly (or by symmetry) y coordinate = 8/15
so the center of mass is at (8/15, 8/15)