Consider the two regions bounded by the curves $ \ y=\frac{x+1}{2} \ $ and $ \ x=y^3-y^2-4y-1 \ $

Question

Consider the two regions bounded by the curves $ \ y=\frac{x+1}{2} \ $ and $ \ x=y^3-y^2-4y-1 \ $

Find the Centroid of the region above x-axis .

Find the Centroid of the region below x-axis .

Answer:

The curves $ \ y=\frac{x+1}{2} \ $ and $ \ x=y^3-y^2-4y-1 \ $ intersects at $ \ (-1,0), \ (5,3) , \ (-5,-2) \ $

Now,

$ y=\frac{x+1}{2} \ \Rightarrow x=2y-1=f(y) \ , say $

Also let $ \ g(y)=x=y^3-y^2-4y-1 \ $

The area $ \ A_1 \ $ above x-axis is given by

$ A_1=\int_{0}^{3} (f(y)-g(y)) dy \\ \Rightarrow A_1=\int_{0}^{3} (-y^3+y^2+6y) dy \\ \Rightarrow A_1=15.75 \ $

Let $ \ (\bar x, \bar y) \ $ be the centroid of the region above x-axis .

Then ,

$ \bar x=\frac{1}{A_1} \int_0^3 y(g(y)-f(y)) dy=1.629 \ $

But I am not sure about the work.

Can some help me finding the above centroid.

Answer 1

You found the area of the region above $x$-axis correctly: $$A_1=\int_{0}^{3} (f(y)-g(y)) dy=15.75.$$ The centroid of the region above $x$-axis: $$\bar{y}=\frac{1}{A_1}\int_0^3y\left[f(y)-g(y)\right]dy=1.629; \\ \bar{x}=\frac{1}{A_1}\int_0^3 \frac12\left[f^2(y)-g^2(y)\right]dy=-0.976.$$