Center of Gravity derivation question I need a sanity check, please.
When determining the center-of-gravity of a lamina described by $f(x)$, we know that by definition,
$\bar{x} M = \sum_{i=1}^{N} m_i \tilde{x_i}$
where $\tilde{x_i}$ is the location of the centroid of strip located at $x_i$
Assuming uniform density and thickness, this becomes
$\bar{x} A = \sum_{i=1}^{N} A_i \tilde{x_i}$
or
$\bar{x} = \frac{\sum_{i=1}^{N} A_i \tilde{x_i}}{\sum_{i=1}^{N} A_i}$
Let us consider a stip with width $\Delta x$ at a distance $x_i$ from y-axis
The area of the strip is $\Delta x \times f(x_i)$
The centroid of the strip, $\tilde{x_i}$ will be at a distance $x_i + \Delta x/2$ from y axis
Hence
$\bar{x} = \frac{\sum_{i=1}^{N} (x_i + \Delta x/2)(\Delta x  f(x_i)) }{\sum_{i=1}^{N}\Delta x  f(x_i)} = \frac{\sum_{i=1}^{N} (x_i \Delta x f(x_i)) + ( \frac{\Delta x^2}{2}   f(x_i)) }{\sum_{i=1}^{N}\Delta x  f(x_i)}$
Now, to arrive at the famous equation.
$\bar{x} = \frac{\sum_{i=1}^{N} x_i f(x_i)}{\sum_{i=1}^{N} f(x_i)}$
the term
$\frac{\Delta x^2}{2}   f(x_i)$ is eliminated by some means.
Is this reasoning correct?
Is it because it is safe to assume $\Delta x^2$ is negligible?
Thanks so much

 A: Are you trying to re-invent the integral? When you go from discrete to continuous, you make the intervals infinitely small and you have to sum over a infinite number of "slices", broadly speaking retaining only the terms $\mathscr{O}(\Delta x)$.
x-coordinate of $G$.
In the continuous limit, the distance of the center of mass from the $y$-axis becomes
$M x_G = \displaystyle \int_A \rho x dA = \int_{x=0}^{x_1} \int_{y=0}^{p(x)} x \rho dx dy = \int_{x_0}^{x_1} \underbrace{x}_{\text{distance of the cog}\\ \text{from x-axis}} \underbrace{ p(x) dx}_{dm(x)} $.
With the function of the parabola $p(x) = - 4 A \left( \left(\frac{x}{x_1}\right)^2 - \frac{x}{x_1} \right)$, the integral becomes
$M x_G = -4 \rho A x_1^2 \left( \dfrac{1}{4} - \dfrac{1}{3} \right) = \dfrac{1}{3} \rho A x_1^2$. Since the mass reads:
$M = \displaystyle \int_A \rho dA = \int_{x=0}^{x_1} \int_{y=0}^{p(x)} {\rho \,dx dy} = \int_{x=0}^{x_1} \underbrace{\rho p(x) \,dx}_{=dm(x)} = \frac{2}{3} \rho A x_1$,
so that the position of the center of mass is $x_G = \dfrac{1}{2}x_1$.
y-coordinate of $G$.
In the continuous limit, the distance of the center of mass of the parabolic lamina from the $x$-axis is given by
$M y_G = \displaystyle \int_A \rho y dA = \int_{x=0}^{x_1} \int_{y=0}^{p(x)} \rho y \,dx dy = \int_{x=0}^{x_1} \rho \dfrac{p^2(x)}{2} dx = \int_{x=0}^{x_1}  \underbrace{\dfrac{p(x)}{2}}_{=y_G(x) \\ \text{ COG of the strip}} \underbrace{\rho p(x) dx}_{= dm(x) \text{ mass os the strip}} $.
where the mass is given by $M = \displaystyle \int_A \rho dA = \int_{x=0}^{x_1} \int_{y=0}^{p(x)} {\rho \,dx dy} = \int_{x=0}^{x_1} \underbrace{\rho p(x) \,dx}_{=dm(x)}$.
Now, with the equation of the parabola $p(x) = -4\frac{A}{x_1^2}x(x-x_1) = -4 A \left( \left(\frac{x}{x_1}\right)^2 - \frac{x}{x_1} \right)$, being $A$ the maximum value, and assuming uniform density $\rho$:

*

*the mass reads $M = \frac{2}{3} \rho A x_1$,

*the distance of the center of mass from the $x$-axis reads $y_G M = \dfrac{8}{30} \rho A^2 x_1 $ and thus $y_G = \dfrac{2}{5} A$
A: to obtain the center of mass use those equations
$$y_{CM}=\frac{\iint\,y\,dx\,dy}{A}\\
y_{CM}=\frac{1}{A}\int_{x_1}^{x_2}\left(\int_0^{f(x)}y\,dy\right)\,dx$$
$$x_{CM}=\frac{\iint\,y\,dx\,dy}{A}\\
x_{CM}=\frac{1}{A}\int_{x_1}^{x_2}\left(\int_0^{f(x)}dy\right)\,x\,dx$$
where $~A~$ is the area under the curve $~f(x)~$
$$A=\iint\,dx\,dy=\int_{x_1}^{x_2}\,f(x)\,dx$$
$\Rightarrow$
$$y_{CM}=\frac 12\frac{\int_{x_1}^{x_2}f(x)^2\,dx}{\int_{x_1}^{x_2}\,f(x)\,dx}$$
$$x_{CM}=\frac{\int_{x_1}^{x_2}f(x)\,x\,dx}{\int_{x_1}^{x_2}\,f(x)\,dx}$$
with this  function
$$f(x)=a\,(x-x^2)$$
and for example with $~a=2~,x_2=0.4~,x_1=0.2~$ you obtain the results
$~x_{CM}=0.306~,y_{CM}=0.209$


Remark
of course you can obtain analytical solution but I didn't put it , because it is too long
the Parable equation is:
$$f(x)=a\,x^2+b\,x+c$$
for your parable you obtain 3 equations for the 3 unknowns $~a,b,c$
$$f(0)=0\\
f(x_f)=0\\
\frac{df}{dx}\Bigg|_{x=\frac{x_f}{2}}=0$$
$\Rightarrow$
$$a\mapsto \pm a\quad, b=-a\,x_f\quad ,c=0$$
with $~x_f=1,f(x)=a\,(x-x^2)$
