Question

Find the centroid of the area bounded by

y² = x³, y = 2x

Answer 1

First, we find the points of intersection of the graphs $y^2=x^3$ and y=2x.

$y^2=x^3=(2x)^2=4x^2$ : x=0, y=0 and x=4, y=8.

On the segment $0\leq x\leq 4$ we have $2x\geq x^{3/2}$.

Let $\Omega$ denotes the region $\{x^{3/2}\leq y\leq 2x, 0\leq x\leq 4\}$.

The centroid $C$ of $\Omega$ has coordinates $x_C=I_x/A,\ y_C=I_y/A$, where $I_x=\int\int_{\Omega}xdxdy$, $I_y=\int\int_{\Omega}ydxdy$, $A=\int\int_{\Omega}dxdy$.

Calculate these integrals:

$A=\int\int_{\Omega}dxdy=\int\limits_0^4\left(\int\limits_{x^{3/2}}^{2x}dy\right)dx=\int\limits_0^4(2x-x^{3/2})dx=(x^2-2/5x^{5/2})|_0^4 = 16-12.8=3.2$

$I_x=\int\int_{\Omega}xdxdy=\int\limits_0^4\left(\int\limits_{x^{3/2}}^{2x}dy\right)xdx=\int\limits_0^4x(2x-x^{3/2})dx=(\frac{2}{3}x^3-\frac{2}{7}x^{7/2})|_0^4=128/3-256/7=128/21$

$I_y=\int\int_{\Omega}ydxdy=\int\limits_0^4\left(\int\limits_{x^{3/2}}^{2x}ydy\right)dx=\int\limits_0^4\frac{y^2}{2}\left|^{2x}_{x^{3/2}}\right.dx=\int\limits_0^4(2x^2-x^3/2)dx=\left(\frac{2}{3}x^3-\frac{1}{8}x^4\right)\left|_0^4\right.=128/3-32=32/3$

Now we find the coordinates of the centroid:

$x_C=(128/21)/3.2=40/21$

$y_C=(32/3)/3.2=10/3$

Answer. $C=\left(\frac{40}{21},\frac{10}{3}\right)$.