Find the centroid bounded by the curves of $y+x=2$, $y^2=x$ Problem Statement:

Find the centroid of $y+x=2, y^2=x$

I think my main problem is finding the limits of integration. I originally set it at 0 to 1, but that didn't work. 
 A: Draw a picture.  The points of intersection are at $(1,1)$ and $(4,-2)$.  The integral over the region $R$ between these curves is then
$$\int_R dx dy = \int_0^1 dx \, \int_{-\sqrt{x}}^{\sqrt{x}} dy + \int_1^4 dx \, \int_{-\sqrt{x}}^{2-x} dy$$
The centroid $x$ coordinate is
$$\bar{x} = \frac{\displaystyle \int_0^1 dx \, x \, \int_{-\sqrt{x}}^{\sqrt{x}} dy + \int_1^4 dx \, x\, \int_{-\sqrt{x}}^{2-x} dy}{\displaystyle \int_0^1 dx \, \int_{-\sqrt{x}}^{\sqrt{x}} dy + \int_1^4 dx \, \int_{-\sqrt{x}}^{2-x} dy}$$
The centroid $y$ coordinate is
$$\bar{y} = \frac{\displaystyle \int_0^1 dx  \, \int_{-\sqrt{x}}^{\sqrt{x}} dy\, y + \int_1^4 dx \, \int_{-\sqrt{x}}^{2-x} dy \, y}{\displaystyle \int_0^1 dx \, \int_{-\sqrt{x}}^{\sqrt{x}} dy + \int_1^4 dx \, \int_{-\sqrt{x}}^{2-x} dy}$$
Upon evaluating the integrals, I get for the centroid
$$(\bar{x},\bar{y}) = \left(\frac{8}{5},-\frac12\right)$$
A: Hint: To find the limits of integration, solve for x in both equations, set the expressions for x equal to each other, and then find the values of y that satisfy this equation.
