This problem is found on Stewart Calculus Metric Version, 7th Edition. It's quite difficult since you can only use integration along $y$. Here are the curve equations: $x = y^2$ and $x = y + 2$
If I solve it the area the usual way, the values of x are the SQUARE ROOT OF 2 AND -1, thus leading to an imaginary solution. If we solve it along y, the values of y are 2 and -1, but the problem is using the formulas.
Can the following equations be modified in order to solve this problem? And how do we arrive to the answer (which are the coordinates of the centroid $(\frac85, \frac{-1}2)$ OR $(1.6, -0.5)$? Thank you!
Formulas for Area Between Two Curves:
$$ \overline{x}=\frac{1}{A}\int_{a}^{b}xf(x) {\mathrm{d} x} $$ $$ \overline{y}=\frac{1}{A}\int_{a}^{b}\frac{1}{2}[f(x)]^2 {\mathrm{d} x} $$
HINT
$$x_G=\frac{\int ydA}{\int dA}\qquad y_G=\frac{\int xdA}{\int dA}$$
Note that the set up for the area should be
$$A=\int_{-2}^1 [(2-y)-y^2] dy$$