Center of Mass for a Lamina. Why is the value inside the integral $x$ and not $x/2$?

Question

TLDR

Why is the center of mass formula for a lamina equal to $A_1$ but not $A_2$? (Specifically the $0.5$ in the $x$-coordinate integral).

$$ A_1=\left(\frac{\int_a^bxf(x)dx}{\int_a^bf(x)dx}, \frac{\int_a^b 0.5 f(x)^2dx}{\int_a^bf(x)dx}\right)$$

$$ A_2=\left(\frac{\int_a^b0.5xf(x)dx}{\int_a^bf(x)dx}, \frac{\int_a^b 0.5 f(x)^2dx}{\int_a^bf(x)dx}\right)$$

Longer Explanation and Question

The center of mass of $n$ point masses in the $xy$-plane is point $P$ where $P$ is defined as:

$$ P = \left(\frac{\sum_{i=1}^n m_ix_i}{\sum_{i=1}^n m_i}, \frac{\sum_{i=1}^n m_iy_i}{\sum_{i=1}^n m_i} \right) .$$

When applying this to functions we could find the center of mass for an area under a curve bounded by two endpoints. For a given function $f(x)$ bounded by points $a$ and $b$, we can chop up the function into smaller and smaller rectangles and average out their center of masses. Point $Q$ represents the center of mass within this interval and it is defined as:

$$ Q = \left(\frac{\int_a^bxf(x)dx}{\int_a^bf(x)dx}, \frac{\int_a^b 0.5 f(x)^2dx}{\int_a^bf(x)dx} \right) . $$

When taking the center of mass for the $y$-coordinate, it makes sense that we would be summing up the halfway height of every rectangle. However, we do not sum up the halfway width of every rectangle. The center of mass of a rectangle lies directly in the middle of the $dx$ slice we use to integrate the region. However, the formula doesn't include this, and it is wrong.

I cannot find an explanation as to why it is wrong. As the widths of our rectangles approach zero, I guess the center of mass can be approximated as being $x$ instead of $\frac{x}{2}$. We could just as easily factor out the $\frac{1}{2}$ value.

Answer 1

In the $x$-dimension, the midpoint of your infinitesimal is not $x / 2$. Given a slice between $x_1$ and $x_2$ with width $\Delta x = x_2 - x_1$, the midpoint is

$$ \frac{x_1 + x_2}{2} $$ $$ = \frac{x_1 + x_1 + \Delta x}{2} $$ $$ = x_1 + \frac{\Delta x}{2} . $$

In the limit as $x_2 \to x_1$, we have $\Delta x \to 0$, and

$$ x_1 + \frac{\Delta x}{2} \to x_1 .$$

Answer 2

The actual mass of the lamina is not distributed along a line but is distributed over a region of the plane. To work out the equivalent of $$\sum_{i=1}^n m_i x_i $$ for the area between the graph of $y = f(x)$ and the line $y = 0$ between the lines $x = a$ and $x = b$, where $f(x)$ is a positive function over $[a,b]$, you can integrate a uniform mass density (density $1$ is fine) over the entire region. You can do this with a double integral like this:

$$ \int_a^b \int_0^{f(x)} 1x\,dy\,dx. $$

The integrals $\int_a^b \int_0^{f(x)}$ correspond to $\sum_{i=1}^n,$ the density $1$ corresponds to $m_i,$ and the variable $x$ in the integrand corresponds to $x_i.$

The equivalent integral for $$ \sum_{i=1}^n m_iy_i $$ is quite similar, except instead of $x_i \sim x$ we have $y_i \sim y$: $$ \int_a^b \int_0^{f(x)} 1y\,dy\,dx. $$

The denominator is similar, but without multiplication by either $x$ or $y$; $$ \sum_{i=1}^n m_i $$ becomes $$ \int_a^b \int_0^{f(x)} 1\,dy\,dx. $$

Where things start to look less symmetric is when we start to do the integration. Note that $$ \int_0^{f(x)} 1x\,dy = x \int_0^{f(x)} 1\,dy = x f(x), $$ but $$ \int_0^{f(x)} 1y\,dy = \int_0^{f(x)} y\,dy = \frac12 (f(x))^2 $$ and $$ \int_0^{f(x)} 1\,dy = f(x). $$

These first steps of integration work out this way for any positive $f(x)$, so it is possible to write the formula for $Q$ that you have shown. But you could also have written the same solution as $$ Q = \left(\frac{\int_a^b\int_0^{f(x)} x\,dy\,dx}{\int_a^b\int_0^{f(x)} 1\,dy\,dx}, \frac{\int_a^b\int_0^{f(x)} y\,dy\,dx}{\int_a^b\int_0^{f(x)} 1\,dy\,dx} \right). $$