Center of Mass
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 2.6: “Moments and Centers of Mass” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/2-6-moments-and-centers-of-mass |
| Textbook used in class | Stewart, Calculus, Section 8.3: “Applications to Physics and Engineering” |
Opening Scenario
A thin metal plate (lamina) of variable thickness needs to be balanced on a pin. Where should the pin be placed so the plate does not tip? The balance point is the center of mass. If the plate has uniform density, the balance point is the centroid -- the geometric average of its shape. If density varies, the heavy parts pull the balance point toward them.
The center of mass extends a familiar idea from discrete masses on a seesaw (the pivot point that balances the seesaw) to continuous distributions of mass (a plate, a beam, or a region bounded by curves).
Quick Reference
1D system. Masses $m_1, \ldots, m_n$ at positions $x_1, \ldots, x_n$. Total mass $m = \sum m_i$. Center of mass: $$\bar{x} = \frac{\sum m_i x_i}{m} = \frac{\text{moment about origin}}{\text{total mass}}.$$
Continuous lamina with uniform density $\rho$ over a region bounded by $f(x) \geq g(x)$ on $[a,b]$: $$m = \rho \int_a^b [f(x) - g(x)]\,dx, \quad M_y = \rho \int_a^b x[f(x)-g(x)]\,dx, \quad M_x = \frac{\rho}{2}\int_a^b [f(x)^2 - g(x)^2]\,dx.$$
Centroid (center of mass for uniform density): $$\bar{x} = \frac{M_y}{m}, \qquad \bar{y} = \frac{M_x}{m}.$$
Since $\rho$ appears in both numerator and denominator of $\bar{x}$ and $\bar{y}$, it cancels: the centroid depends only on the shape, not the density.
Key Concepts
1. From Discrete to Continuous
A seesaw balances when $\sum m_i(x_i - \bar{x}) = 0$, i.e., $\bar{x} = \dfrac{\sum m_i x_i}{\sum m_i}$.
For a lamina with area density $\rho(x,y)$, replace the sum with an integral. On a flat plate with uniform density $\rho$, the mass of a thin vertical strip at $x$ of width $dx$ and height $f(x) - g(x)$ is $dm = \rho[f(x)-g(x)]\,dx$.
Integrating: $m = \int dm = \rho\int_a^b [f(x)-g(x)]\,dx = \rho A$ where $A$ is the area.
2. The Two Moments
The moment about the $y$-axis, $M_y$, measures the tendency to rotate around the $y$-axis: $$M_y = \rho \int_a^b x[f(x)-g(x)]\,dx.$$
The moment about the $x$-axis, $M_x$, measures tendency to rotate around the $x$-axis. The height of the center of each strip is $\dfrac{f(x)+g(x)}{2}$: $$M_x = \rho \int_a^b \frac{f(x)+g(x)}{2}\cdot [f(x)-g(x)]\,dx = \frac{\rho}{2}\int_a^b [f(x)^2-g(x)^2]\,dx.$$
3. Centroid for Uniform Density
When $\rho$ is constant, it cancels: $$\bar{x} = \frac{M_y}{m} = \frac{\int_a^b x[f(x)-g(x)]\,dx}{\int_a^b [f(x)-g(x)]\,dx} = \frac{\int_a^b x[f(x)-g(x)]\,dx}{A}.$$
$$\bar{y} = \frac{M_x}{m} = \frac{\frac{1}{2}\int_a^b [f(x)^2-g(x)^2]\,dx}{A}.$$
The centroid $(\bar{x}, \bar{y})$ is the geometric center of the region. It need not lie inside the region (e.g., for a ring shape).
Worked Example
Find the centroid of the region bounded by $y = \sqrt{x}$ and $y = x$ on $[0,1]$.
Area: $$A = \int_0^1 (\sqrt{x} - x)\,dx = \left[\frac{2}{3}x^{3/2} - \frac{x^2}{2}\right]_0^1 = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}.$$
$\bar{x}$: $$\bar{x} = \frac{1}{A}\int_0^1 x(\sqrt{x}-x)\,dx = 6\int_0^1 (x^{3/2}-x^2)\,dx = 6\left[\frac{2}{5}x^{5/2}-\frac{x^3}{3}\right]_0^1 = 6\left(\frac{2}{5}-\frac{1}{3}\right) = 6\cdot\frac{1}{15} = \frac{2}{5}.$$
$\bar{y}$: $$\bar{y} = \frac{1}{A}\cdot\frac{1}{2}\int_0^1 (x - x^2)\,dx = 3\int_0^1 (x-x^2)\,dx = 3\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = 3\left(\frac{1}{2}-\frac{1}{3}\right) = 3\cdot\frac{1}{6} = \frac{1}{2}.$$
Boxed answer: Centroid $= \left(\dfrac{2}{5},\, \dfrac{1}{2}\right)$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using $M_x = \rho\int x[f-g]\,dx$ for the moment about $x$-axis | Mixing up which moment uses $x$ and which uses $y$ | $M_y = \rho\int x[f-g]\,dx$ (moment about $y$); $M_x = \frac{\rho}{2}\int[f^2-g^2]\,dx$ (moment about $x$) |
| Dividing by $A$ but not canceling $\rho$ first | Keeping $\rho$ in $\bar{x}$ formula when density is uniform | For uniform density, $\bar{x} = M_y/m = M_y/(\rho A)$; $\rho$ cancels |
| Using $f(x)-g(x)$ in $M_x$ instead of $[f^2-g^2]/2$ | $M_x = \rho\int_a^b y[f(x)-g(x)]\,dx$ where $y = f-g$ | The center of the strip is $(f+g)/2$, not $f-g$; use $M_x = \frac{\rho}{2}\int[f^2-g^2]\,dx$ |
Common Misconceptions
the centroid of a region is the midpoint of its bounding box, that is, the point halfway between the left and right edges and halfway between the top and bottom.
This is the average-rate-as-arithmetic-mean error. The centroid is a weighted average of position, with each infinitesimal strip weighted by its area. For a region shaped like a triangle with all its area concentrated near one vertex, the centroid is pulled toward that vertex, not toward the geometric center of the bounding rectangle. For the region under $y = x^2$ on $[0,1]$, the centroid is at $\bar{x} = 3/4$, not $1/2$, because the parabola concentrates area toward the right where the strip heights are larger.
the moment about the $y$-axis $M_y$ uses $y$ as the lever arm, and the moment about the $x$-axis $M_x$ uses $x$.
This is the concept-image-conflicts-definition error. The moment about an axis uses the perpendicular distance to that axis as the lever arm. $M_y$ measures the tendency to rotate around the $y$-axis, so the lever arm is the horizontal distance $x$: $M_y = \rho\int_a^b x[f(x)-g(x)]\,dx$. $M_x$ uses the vertical distance $y$, which for a strip centered at height $(f+g)/2$ gives $M_x = \tfrac{\rho}{2}\int_a^b [f(x)^2 - g(x)^2]\,dx$. Swapping $x$ and $y$ in these formulas gives the wrong centroid coordinates.
Leveled Practice
Level 1 -- Single Strip
Problem 1. Find the centroid of the region under $y = 4-x^2$ on $[-2,2]$ (above the $x$-axis).
Show answer
By symmetry: $\bar{x} = 0$.
$A = \displaystyle\int_{-2}^{2}(4-x^2)\,dx = \left[4x-\frac{x^3}{3}\right]_{-2}^2 = \left(8-\frac{8}{3}\right)-\left(-8+\frac{8}{3}\right) = \frac{32}{3}$.
$\bar{y} = \dfrac{1}{A}\cdot\dfrac{1}{2}\displaystyle\int_{-2}^{2}(4-x^2)^2\,dx$.
$(4-x^2)^2 = 16-8x^2+x^4$.
$\displaystyle\int_{-2}^{2}(16-8x^2+x^4)\,dx = 2\displaystyle\int_0^2(16-8x^2+x^4)\,dx = 2\left[16x-\frac{8x^3}{3}+\frac{x^5}{5}\right]_0^2 = 2\left(32-\frac{64}{3}+\frac{32}{5}\right) = 2\cdot\frac{256}{15} = \frac{512}{15}$.
$\bar{y} = \dfrac{3}{32}\cdot\dfrac{1}{2}\cdot\dfrac{512}{15} = \dfrac{3\cdot 256}{32\cdot 15} = \dfrac{8}{5}$.
Boxed answer: $(\bar{x},\bar{y}) = \left(0,\, \dfrac{8}{5}\right)$.
Level 2 -- Region Between Curves
Problem 2. Find the centroid of the triangular region with vertices $(0,0)$, $(2,0)$, and $(0,4)$.
Show answer
The hypotenuse has equation $y = 4 - 2x$, so $f(x) = 4-2x$ for $x \in [0,2]$, $g(x) = 0$.
$A = \displaystyle\int_0^2(4-2x)\,dx = [4x-x^2]_0^2 = 4$.
$\bar{x} = \dfrac{1}{4}\displaystyle\int_0^2 x(4-2x)\,dx = \dfrac{1}{4}\int_0^2(4x-2x^2)\,dx = \dfrac{1}{4}\left[2x^2-\dfrac{2x^3}{3}\right]_0^2 = \dfrac{1}{4}\left(8-\dfrac{16}{3}\right) = \dfrac{1}{4}\cdot\dfrac{8}{3} = \dfrac{2}{3}$.
$\bar{y} = \dfrac{1}{4}\cdot\dfrac{1}{2}\displaystyle\int_0^2(4-2x)^2\,dx = \dfrac{1}{8}\int_0^2(16-16x+4x^2)\,dx = \dfrac{1}{8}\left[16x-8x^2+\dfrac{4x^3}{3}\right]_0^2 = \dfrac{1}{8}\left(32-32+\dfrac{32}{3}\right) = \dfrac{4}{3}$.
Boxed answer: $(\bar{x},\bar{y}) = \left(\dfrac{2}{3},\, \dfrac{4}{3}\right)$.
(Check: centroid of a triangle is at $\frac{1}{3}$ of the way from each vertex to the opposite side: $(\frac{0+2+0}{3}, \frac{0+0+4}{3}) = (\frac{2}{3}, \frac{4}{3})$. Confirmed.)
Mastery Checklist
Mental Model
The center of mass is the balance point. For a discrete system, it is the weighted average of positions. For a continuous region, replace the sum with an integral. The two moments $M_y$ and $M_x$ count how the mass is distributed horizontally and vertically: dividing each moment by the total mass gives the horizontal coordinate $\bar{x}$ and vertical coordinate $\bar{y}$ of the balance point.
Connections
Looking back
- Definite integral (Chapter 4): The area and moment integrals are standard definite integrals.
- Region between curves (Section 6.1): The area formula $\int[f-g]\,dx$ is extended here to moments.
Looking ahead
- Pappus’s theorem (Section 8.3): Connects the centroid to volumes and surface areas of revolution.