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

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.
