<s>[INST] 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
 [/INST] 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.
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