# Centroid of an Area Between Two Curves by Calculus

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:

Formulas for the Centroid:

$$\overline{x}=\frac{1}{A}\int_{a}^{b}xf(x) {\mathrm{d} x}$$ $$\overline{y}=\frac{1}{A}\int_{a}^{b}\frac{1}{2}[f(x)]^2 {\mathrm{d} x}$$

• Welcome to MSE. Please type your questions (using MathJax) instead of posting links to pictures. – José Carlos Santos Mar 4 '18 at 13:02
• note that the equation for the straight line should be x+y=2 and not x=y+2 – gimusi Mar 4 '18 at 13:45
• That was the given problem in the book. – Kenneth Ligutom Mar 5 '18 at 12:19
• @KennethLigutom Please remember that you can choose an aswer among the given is the OP is solved, more details here meta.stackexchange.com/questions/5234/… – gimusi Mar 9 '18 at 23:00

HINT

• make a plot of the functions and find the intersection points
• then compute

$$x_G=\frac{\int ydA}{\int dA}\qquad y_G=\frac{\int xdA}{\int dA}$$

Note that the set up for the area should be

$$A=\int_{-2}^1 [(2-y)-y^2] dy$$

• 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. – Kenneth Ligutom Mar 4 '18 at 13:03
• @KennethLigutom could please show your work for the set up of the integrals? – gimusi Mar 4 '18 at 13:29