A seesaw balances when $m_1 d_1 = m_2 d_2$: mass times distance from the pivot must match on both sides. Mass times signed distance is called a moment. For a flat plate cut in a curved shape, the balance point (its centroid) comes from integrals, built with the same strips you used for area between curves.
Why it works
For point masses $m_i$ at positions $x_i$, the balance point $\bar{x}$ is where the total moment about it is zero: $\sum m_i (x_i - \bar{x}) = 0$, so
$$\bar{x} = \frac{\sum m_i x_i}{\sum m_i} = \frac{\text{moment}}{\text{mass}}.$$
For a plate of constant density $\rho$ between $y = g(x)$ and $y = f(x)$, cut it into vertical strips. A strip at $x$ has mass $\rho\,(f - g)\,\Delta x$ and balances at its middle, $\left(x, \frac{f + g}{2}\right)$. So the moments are
$$M_y = \rho\int_a^b x\,(f - g)\,dx,$$ $$\begin{aligned} M_x &= \rho\int_a^b \frac{f + g}{2}(f - g)\,dx \\ &= \rho\int_a^b \frac{f^{2} - g^{2}}{2}\,dx. \end{aligned}$$
Dividing by the mass $\rho A$ cancels $\rho$, so the centroid depends only on the shape.
Key idea
For a plate between $y = g(x)$ and $y = f(x)$ with area $A$:
$$\bar{x} = \frac{1}{A}\int_a^b x\left[f(x) - g(x)\right]dx,$$ $$\bar{y} = \frac{1}{A}\int_a^b \frac{1}{2}\left[f(x)^{2} - g(x)^{2}\right]dx.$$
Worked example: Masses on a rod
Masses of 2 kg, 3 kg and 5 kg sit at $x = -1$, $x = 2$ and $x = 4$. Find the center of mass.
- Moment: $2(-1) + 3(2) + 5(4) = 24$.
- Mass: $2 + 3 + 5 = 10$.
- $\bar{x} = \frac{24}{10} = 2.4$.
Answer: $\bar{x} = 2.4$
Worked example: Under a parabola
Find the centroid of the region under $y = x^{2}$ from $x = 0$ to $x = 2$.
- Area: $A = \int_0^2 x^{2}\,dx = \frac{8}{3}$.
- $M_y = \int_0^2 x\cdot x^{2}\,dx = 4$, so $\bar{x} = 4 \div \frac{8}{3} = \frac{3}{2}$.
- $M_x = \int_0^2 \frac{1}{2}(x^{2})^{2}\,dx = \frac{1}{2}\cdot\frac{32}{5} = \frac{16}{5}$, so $\bar{y} = \frac{16}{5}\div\frac{8}{3} = \frac{6}{5}$.
Answer: $\left(\frac{3}{2}, \frac{6}{5}\right)$
Worked example: Using symmetry
Find the centroid of the region between $y = 4 - x^{2}$ and the $x$-axis.
- The region is symmetric about the $y$-axis, so $\bar{x} = 0$ (the moments of the two halves cancel).
- $A = \int_{-2}^{2} (4 - x^{2})\,dx = \frac{32}{3}$.
- $$\begin{aligned} M_x &= \int_{-2}^{2} \frac{1}{2}(4 - x^{2})^{2}\,dx \\ &= \frac{1}{2}\cdot\frac{512}{15} \\ &= \frac{256}{15}, \end{aligned}$$ so $\bar{y} = \frac{256}{15}\div\frac{32}{3} = \frac{8}{5}$.
Answer: $\left(0, \frac{8}{5}\right)$
Common mistake
Using $\int f(x)^{2}\,dx$ for $M_x$ without the $\frac{1}{2}$. A strip's mass sits at its middle height $\frac{f}{2}$, not at its top. Check every answer: the centroid must lie inside the smallest box around the region, on any line of symmetry, and closer to where the region is thicker.
The theorem of Pappus ties this to volumes: revolving a region of area $A$ about an axis it does not cross gives volume $V = 2\pi\bar{r}A$, where $\bar{r}$ is the centroid's distance to the axis. For the region under $y = x^{2}$ on $[0, 2]$ about the $x$-axis: $2\pi\cdot\frac{6}{5}\cdot\frac{8}{3} = \frac{32\pi}{5}$, matching the disk method.