Let $f : [0,1] \to [0,\infty)$ be a strictly increasing continuous function. Let $R$ be the region bounded by $x = 0$, $x = 1$, $y = 0$, and $y = f(x)$. Let $x_1$ be the $x$-coordinate of the centroid of the planar region $R$, and let $x_2$ be the $x$-coordinate of the centroid of the solid generated by rotating $R$ around the $x$-axis. Prove that $x_1 < x_2$.
Domain: $x \in [0,1]$ | Region: $R = \{(x,y): 0 \le y \le f(x)\}$ | Goal: Show $x_1 < x_2$
We work with a strictly increasing, continuous function $f : [0,1] \to [0,\infty)$ and the region $$R = \{(x,y) \mid 0 \le x \le 1,\; 0 \le y \le f(x)\}.$$ Rotating $R$ around the $x$-axis produces a solid of revolution. Our goal is to prove that the $x$-coordinate of the solid's centroid ($x_2$) lies strictly to the right of the planar region's centroid ($x_1$).
The centroid formulas from calculus give us: $$x_1 = \frac{\int_0^1 x f(x)\,dx}{\int_0^1 f(x)\,dx}, \qquad x_2 = \frac{\int_0^1 x f(x)^2\,dx}{\int_0^1 f(x)^2\,dx}.$$
Both are weighted averages of the position $x$, but with different weights:
Key insight: Since $f$ is strictly increasing, larger values of $x$ have larger values of $f(x)$. Squaring $f(x)$ amplifies this effect—large values become disproportionately larger. Therefore, $x_2$ places relatively more weight toward the right end of the interval than $x_1$ does.
To make this rigorous, we prove: $$\frac{\int_0^1 x f(x)\,dx}{\int_0^1 f(x)\,dx} < \frac{\int_0^1 x f(x)^2\,dx}{\int_0^1 f(x)^2\,dx}.$$
Cross-multiplying (all denominators are positive): $$\left(\int_0^1 x f(x)\,dx\right)\left(\int_0^1 f(x)^2\,dx\right) < \left(\int_0^1 x f(x)^2\,dx\right)\left(\int_0^1 f(x)\,dx\right).$$
Let $I$ be the left side minus the right side. Expanding as double integrals and simplifying (details in the symmetrization approach), we get: $$I = \frac{1}{2}\int_0^1\int_0^1 f(x)f(y)(x-y)(f(y)-f(x))\,dy\,dx.$$
Since $f$ is strictly increasing, whenever $x \neq y$:
We interpret the centroids as expectations under different probability measures. Define densities: $$p_1(x) = \frac{f(x)}{\int_0^1 f(t)\,dt}, \qquad p_2(x) = \frac{f(x)^2}{\int_0^1 f(t)^2\,dt}.$$
Then $x_1 = \mathbb{E}_{p_1}[X]$ and $x_2 = \mathbb{E}_{p_2}[X]$. The transformation $p_1 \to p_2$ represents a multiplicative reweighting by $f(x)$, which is increasing.
The Chebyshev correlation inequality states that if $u, v : [0,1] \to \mathbb{R}$ are both increasing (or both decreasing), then: $$\int_0^1 u(x)v(x)\,dx \cdot \int_0^1 1\,dx \ge \int_0^1 u(x)\,dx \cdot \int_0^1 v(x)\,dx.$$
However, we need something more refined. Consider the likelihood ratio: $$\frac{p_2(x)}{p_1(x)} = \frac{f(x)^2 / \int_0^1 f^2}{f(x) / \int_0^1 f} = \frac{f(x) \cdot \int_0^1 f(t)\,dt}{\int_0^1 f(t)^2\,dt}.$$
Since only the numerator depends on $x$ and $f$ is strictly increasing, this ratio is strictly increasing in $x$.
Monotone Likelihood Ratio Property: When $p_2/p_1$ is increasing, we say $p_2$ dominates $p_1$ in the monotone likelihood ratio order. This is a strong form of stochastic dominance.
Theorem: If $p_2$ dominates $p_1$ in MLR order, then for any increasing function $g$: $$\mathbb{E}_{p_1}[g(X)] \le \mathbb{E}_{p_2}[g(X)].$$
Taking $g(x) = x$ (which is increasing), we immediately get: $$x_1 = \mathbb{E}_{p_1}[X] < \mathbb{E}_{p_2}[X] = x_2.$$
The strict inequality follows because $f$ is strictly increasing on a set of positive measure, making the likelihood ratio strictly increasing almost everywhere.
Let $\mathcal{M}^+([0,1])$ denote the cone of finite positive Borel measures on $[0,1]$. The barycenter functional is: $$\text{bar} : \mathcal{M}^+([0,1]) \to [0,1], \quad \text{bar}(\mu) = \frac{\int_0^1 x\,\mu(dx)}{\mu([0,1])}.$$
Define $\mu(dx) = f(x)\,dx$ and $\nu(dx) = f(x)\,\mu(dx) = f(x)^2\,dx$. Then $x_1 = \text{bar}(\mu)$ and $x_2 = \text{bar}(\nu)$.
The transformation $\mu \mapsto \nu$ is a monotone multiplicative tilt by the increasing function $f$.
Theorem: Let $\mu \in \mathcal{M}^+([0,1])$ and $w : [0,1] \to (0,\infty)$ be strictly increasing. Define $\nu(dx) = w(x)\mu(dx)$. Then: $$\text{bar}(\nu) > \text{bar}(\mu).$$
Proof. We compute: $$\text{bar}(\nu) - \text{bar}(\mu) = \frac{\int x w(x)\,\mu(dx)}{\int w(x)\,\mu(dx)} - \frac{\int x\,\mu(dx)}{\mu([0,1])}.$$
Cross-multiplying and rearranging: $$\left(\int x w(x)\,\mu(dx)\right) \cdot \mu([0,1]) - \left(\int x\,\mu(dx)\right) \cdot \left(\int w(x)\,\mu(dx)\right).$$
Writing as a double integral over $[0,1]^2$ with respect to the product measure $\mu \otimes \mu$: $$\int\int [x w(x) - y w(y)]\,\mu(dx)\mu(dy).$$
Symmetrizing: $$\frac{1}{2}\int\int [x w(x) - y w(y) - x w(y) + y w(x)]\,\mu(dx)\mu(dy)$$ $$= \frac{1}{2}\int\int (x-y)(w(x)-w(y))\,\mu(dx)\mu(dy).$$
Since $w$ is strictly increasing, $(x-y)$ and $(w(x)-w(y))$ have the same sign for all $x \neq y$, making the integrand strictly positive off the diagonal. Therefore $\text{bar}(\nu) > \text{bar}(\mu)$. ∎
Corollary: In our setting, $w(x) = f(x)$ is strictly increasing, so $x_2 = \text{bar}(\nu) > \text{bar}(\mu) = x_1$.
This framework generalizes to arbitrary monotone transformations on measure spaces and connects to:
Imagine you have a rod of varying thickness along its length, getting thicker as you move from left to right. The center of mass (balance point) of this rod is at position $x_1$.
Now imagine spinning this rod around its length to create a solid object (like a vase). The parts that were thicker create larger circular cross-sections.
Here's the key: when you square the thickness to calculate circular areas, the thick parts (on the right) become disproportionately heavier compared to the thin parts (on the left). This shifts the center of mass further to the right, to position $x_2$.
Why? If one part is twice as thick as another, its circular cross-section has four times the area (since area = $\pi r^2$). This amplification effect always pushes the solid's center of mass rightward compared to the original rod.
Mathematical essence: Squaring an increasing function makes it "more increasing," concentrating more relative weight where the function is large.