First post here. I'm having serious trouble understanding how the Moment $Mx$ is solved for in a typical Centre of Mass problem.
So, many people online, are teaching methods to solve for $Mx$ that are completely different from each other.
And these different methods give different results!
Some solve the 'Moment' $Mx$ as
$$Mx = \int \frac{y}{2} [ f(x) ]^2 - [ g(x) ]^2~\mathrm{d}x$$
Something like this : https://www.youtube.com/watch?v=SWu_i-19Rn0
And some others solve the moment $Mx$ as
$$Mx = \iint y \rho(x ,y)~\mathrm{d}x\mathrm{d}y$$
Like this : https://www.youtube.com/watch?v=yJDRwKGfKDs
Which is the correct one?
Or are they both related to two different things?
The center of mass $(x,y)_{\rm CoM}$ of a suitable region $R$ with mass $M$ has coordinates equal to the average values of $x$ and $y$ over $R$.
$$x_{\rm CoM} = \frac{\displaystyle \iint_R x\rho\,dA}{\displaystyle \iint_R\rho\,dA}, y_{\rm CoM} = \frac{\displaystyle \iint_R y\,\rho\,dA}{\displaystyle\iint_R\rho\,dA}$$
The numerators are the moments of $R$ about the $y$-axis ($M_y$) and $x$-axis ($M_x$), respectively.
If $R$ is suitable enough to express it as the set
$$R = \{(x,y) \mid a\le x\le b \text{ and } g(x)\le y\le f(x)\}$$
then
$$M_x = \int_a^b \int_{g(x)}^{f(x)} y\rho(x,y)\,dy\,dx$$
Suppose density is constant, $\rho(x,y)=\rho$. Then
$$M_x = \rho \int_a^b \int_{g(x)}^{f(x)} y\,dy\,dx = \rho\int_a^b \frac{y^2}2\bigg|_{y=g(x)}^{y=f(x)}\,dx = \frac\rho2 \int_a^b \bigg(f(x)^2 - g(x)^2\bigg)\,dx$$