The quarter disk in the first quadrant bounded by x^2+y^2=4

[tj Cho]

Find the coordinate of center of mass.
Given: The quarter disk in the first quadrant bounded by x^2+y^2=4
I tried to solve this problem but can't figure out how to do it.
so y integration limit is: 0 <= y <= sqrt(4-x^2))
x limit of integration: 0 <= x <= 2

and then after the dy integral I got something really messy
integral from 0 to 2(sqrt(4-x^2)+x^2sqrt(4-x^2)+(sqrt(4-x^2)^3/3)dx)
Where did I do wrong?

 

[Svein]

You could look at it another way: The quarter disk is given by R=2, φ∈[0, π/2].

 

[HallsofIvy]

The area of the quarter circle with radius 2 is, of course, $(1/4)\pi (4)= \pi$. Because of symmetry the x and y coordinates of the centroid (I would not say "center of mass" since we are not given a density) are the same:
$$\overline{x}= \overline{y}= \frac{\int xdA}{A}= \frac{\int ydA}{A}$$.

Personally, I would do the integration in polar coordinates, as Svein suggests:
$$\int xdA= \int_{r= 0^2}\int_{\theta= 0}^{\pi/2} (r cos(\theta) rd\theta dr= \left(\int_0^2 r^2dr\right)\left(\int_0^{\pi/2} cos(\theta)d\theta\right)$$

But since you specifically ask about the integration in xy- coordinates,
$$\int x dA= \int_0^2\int_0^{\sqrt{4- x^2}} x dydx= \int_0^2x\sqrt{4- x^2}dx$$
and now use the substitution $u= 4- x^2$.

I can't say what you did wrong because you don't show what you did!