Is the area on both sides of a centroid equal?

Question

I am given the function $f(x)=x$ and I am trying to find the $x$ centroid between $0$ and $1$. I know the that $\bar x $ can be found by the formula $$\frac{\int_0^1xdA}{\int_0^1dA}$$ which then becomes $$\frac{\int_0^1x^2dx}{\int_0^1xdx}$$ Following this formula, we get $\bar x = \frac{2}{3}$. However, the areas on either side of $\bar x$ are not equal. Shouldn't $\bar x$ lie at the center of mass for the function, and therefore make $\int_0^{\bar x}xdx = \int_{\bar x}^1xdx$? Why is this not the case?

Answer 1

Short answer: the "center of mass" is not the center of the mass.

Long answer: the center of mass (called "center of gravity" in some contexts) is a place where various torques balance out, not where various masses balance out. Despite the name, from the mathematical definition there's no reason to think that it should separate mass or area into two equal parts.

For a toy example, consider a tiny unit square centered at $(-2,0)$ and a tiny vertical domino (two unit squares glued together) centered at $(1,0)$. The center of mass will be the origin, but that certainly doesn't cause half the mass to be on each side.

Answer 2

Why should the areas be equal? The way you've set up the problem, you're trying to find the centroid of a right-angle triangle with corners $(0,0)$, $(1,0)$, and $(1,1)$, which is at (using standard results), $(\frac{1}{3}(0+1+1), \frac{1}{3}(0+0+1)) = (\frac{2}{3}, \frac{1}{3})$. This is exactly what you get when you calculate the $x$ and $y$ centroids separately:

$$\bar{x} = \frac{\int_0^1 x\cdot x\,dx}{\int_0^1 x\,dx} = \frac{2}{3},\qquad \bar{y} = \frac{\int_0^1 y\cdot (1-y)\,dy}{\int_0^1 (1-y)\,dy} = \frac{1}{3}.$$

Perhaps you're confused because you're thinking of a one-dimensional bar lying along the $x$ axis and not extending in the $y$ direction. In this case the formula you used doesn't apply. In one dimension, the centroid (or center of mass) is given by

$$\bar{x}_{\text{1D}} = \frac{\int x\cdot dx}{\int dx}.$$

For your problem, this just gives $\bar{x}_{\text{1D}} = 1/2$ as you would expect.