# Center of Mass Help

For a question, it asked us to find the center of mass of the region between $y=\sqrt x$ and $y=x^3$. I was wondering if the way that I did it was correct. If not, what is the correct formuala for solving center of mass Center of Mass

One method is to find two lines that "halve the area by mass" and to find the intersection of those two lines. Image of graph with bisecting lines

The green line is at the x value where $$\int_0^x \sqrt{x}-x^3 dx = \int_x^1 \sqrt{x}-x^3 dx$$ By calculating these integrals and setting them to be equal, you can find the $x$-coordinate of the COM.

In the same way, you can find the $y$-coordinate by evaluating the integrals with respect to $y$. $$\int_0^y \sqrt[3]{y}-y^2 dy = \int_y^1 \sqrt[3]{y}-y^2 dy$$

The center of mass would be located by using the following formulas:

$\bar{x}=\frac{1}{A}\int_\limits{a}^bx[f(x)-g(x)]dx$ and $\bar{y}=\frac{1}{A}\int_\limits{a}^b\frac{1}{2}[(f(x))^2-(g(x))^2]dx$

$A=\int_\limits{a}^b[f(x)-g(x)]dx$

$\bar{x}=\frac{\int_\limits{0}^1x(\sqrt{x}-x^3)dx}{\int_\limits{0}^1\sqrt{x}-x^3dx}=\frac{\int_\limits{0}^1({x}^\frac{3}2-x^4)dx}{\int_\limits{0}^1\sqrt{x}-x^3dx}=\frac{[\frac{2}{5}x^\frac{5}{2}-\frac{x^5}{5}]_0^1}{[\frac{2}{3}x^\frac{3}{2}-\frac{x^4}{4}]_0^1}=\frac{12}{25}$

$\bar{y}=\frac{\int_\limits{0}^1\frac{1}{2}[(\sqrt{x})^2-(x^3)^2]dx}{\int_\limits{0}^1\sqrt{x}-x^3dx}=\frac{\int_\limits{0}^1\frac{1}{2}[x-x^6]dx}{\int_\limits{0}^1\sqrt{x}-x^3dx}=\frac{1}{2}\frac{[\frac{1}{2}x^2-\frac{x^7}{7}]_0^1}{[\frac{2}{3}x^\frac{3}{2}-\frac{x^4}{4}]_0^1}=\frac{3}{7}$

Your approach is right just need to do it for the y axis too because its a point which needs an x and y coordinate.