Center of Mass Formula: x̄ = ∫x·ρ dA / ∫ρ dA

Center of Mass Formula

The coordinates $(\overline{x}, \overline{y})$ of the center of mass of a plane figure are given by the formulas below. The formulas only apply for figures of uniform (constant) density.

1. Plane region bounded above by $y = f(x)\text{,}$ below by the x-axis, on the left by $x = a\text{,}$ and on the right by $x = b\text{.}$ $\overline{x} = \frac{M_y}{m} = \frac{1}{A}\int_a^b x f(x)\,dx \qquad \overline{y} = \frac{M_x}{m} = \frac{1}{A}\int_a^b \frac{1}{2}\left[f(x)\right]^2 dx$ Note: $M_y$ is the moment about the y-axis, $M_x$ is the moment about the x-axis, $m$ is the mass, and $A$ is the area.

2. Plane region bounded above by $y = f(x)\text{,}$ below by $y = g(x)\text{,}$ on the left by $x = a\text{,}$ and on the right by $x = b\text{.}$ $\overline{x} = \frac{M_y}{m} = \frac{1}{A}\int_a^b x\left[f(x) - g(x)\right]dx \qquad \overline{y} = \frac{M_x}{m} = \frac{1}{A}\int_a^b \frac{1}{2}\left(\left[f(x)\right]^2 - \left[g(x)\right]^2\right)dx$ Note: $M_y$ is the moment about the y-axis, $M_x$ is the moment about the x-axis, $m$ is the mass, and $A$ is the area.

Key Formula

\bar{x} = \frac{\displaystyle\int_a^b x\,\bigl[f(x) - g(x)\bigr]\,dx}{\displaystyle\int_a^b \bigl[f(x) - g(x)\bigr]\,dx}, \qquad \bar{y} = \frac{\displaystyle\int_a^b \frac{1}{2}\bigl[f(x)^2 - g(x)^2\bigr]\,dx}{\displaystyle\int_a^b \bigl[f(x) - g(x)\bigr]\,dx}

Where:

$\bar{x}$ = The x-coordinate of the center of mass (centroid)
$\bar{y}$ = The y-coordinate of the center of mass (centroid)
$f(x)$ = The upper boundary curve of the region
$g(x)$ = The lower boundary curve of the region
$a, b$ = The left and right limits of integration (where the curves intersect or the region begins and ends)
$\int_a^b [f(x) - g(x)]\,dx$ = The total area A of the region, which appears in the denominator of both formulas

Worked Example

Problem: Find the center of mass (centroid) of the region bounded by y = x² and y = 0 from x = 0 to x = 3.

Step 1: Identify the curves and limits. Here f(x) = x² (upper curve), g(x) = 0 (lower curve), a = 0, and b = 3.

f(x) = x^2, \quad g(x) = 0, \quad a = 0, \quad b = 3

Step 2: Compute the area A of the region (the denominator in both formulas).

A = \int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} = 9

Step 3: Compute the x-coordinate of the center of mass using the moment about the y-axis.

\bar{x} = \frac{1}{9}\int_0^3 x \cdot x^2\,dx = \frac{1}{9}\int_0^3 x^3\,dx = \frac{1}{9}\left[\frac{x^4}{4}\right]_0^3 = \frac{1}{9} \cdot \frac{81}{4} = \frac{81}{36} = \frac{9}{4}

Step 4: Compute the y-coordinate of the center of mass using the moment about the x-axis.

\bar{y} = \frac{1}{9}\int_0^3 \frac{1}{2}(x^2)^2\,dx = \frac{1}{18}\int_0^3 x^4\,dx = \frac{1}{18}\left[\frac{x^5}{5}\right]_0^3 = \frac{1}{18} \cdot \frac{243}{5} = \frac{243}{90} = \frac{27}{10}

Step 5: State the center of mass.

\left(\bar{x},\,\bar{y}\right) = \left(\frac{9}{4},\,\frac{27}{10}\right) = (2.25,\, 2.7)

Answer: The center of mass is at (9/4, 27/10), or equivalently (2.25, 2.7).

Another Example

This example involves a region between two curves rather than a region between a single curve and the x-axis, requiring use of both f(x) and g(x) in the formulas.

Problem: Find the centroid of the region enclosed between y = x and y = x² (which intersect at x = 0 and x = 1).

Step 1: Identify the curves. On [0, 1], y = x is above y = x², so f(x) = x and g(x) = x².

f(x) = x, \quad g(x) = x^2, \quad a = 0, \quad b = 1

Step 2: Compute the area of the region.

A = \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}

Step 3: Compute the x-coordinate of the centroid.

\bar{x} = \frac{1}{1/6}\int_0^1 x(x - x^2)\,dx = 6\int_0^1 (x^2 - x^3)\,dx = 6\left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^1 = 6\left(\frac{1}{3} - \frac{1}{4}\right) = 6 \cdot \frac{1}{12} = \frac{1}{2}

Step 4: Compute the y-coordinate of the centroid using the formula with f(x)² − g(x)².

\bar{y} = 6 \cdot \frac{1}{2}\int_0^1 (x^2 - x^4)\,dx = 3\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = 3\left(\frac{1}{3} - \frac{1}{5}\right) = 3 \cdot \frac{2}{15} = \frac{2}{5}

Answer: The centroid of the region between y = x and y = x² is (1/2, 2/5).

Frequently Asked Questions

What is the difference between center of mass and centroid?

For a plane figure with uniform (constant) density, the center of mass and the centroid are the same point. The term 'centroid' is purely geometric and refers to the average position of all points in the shape. 'Center of mass' accounts for density; when density is uniform, the two concepts coincide. If density varies across the figure, you must use the full center-of-mass formulas that include a density function.

Why does the y-bar formula use f(x)² − g(x)² instead of f(x) − g(x)?

The y-coordinate formula computes the moment about the x-axis. Each thin vertical strip of width dx has its own center at height [f(x) + g(x)]/2, and its area is [f(x) − g(x)]dx. Multiplying these together gives (1/2)[f(x)² − g(x)²]dx, which is why the squared terms appear. This is not an arbitrary choice — it comes directly from finding the midpoint height of each strip.

When do you use the center of mass formula in calculus?

You use it whenever a problem asks for the centroid or balancing point of a region defined by curves. It appears frequently in second-semester calculus (Calculus II) and is closely related to moments. It also connects to Pappus's theorem, where the centroid is used to find volumes and surface areas of solids of revolution.

Center of Mass (Uniform Density) vs. Center of Mass (Variable Density)

	Center of Mass (Uniform Density)	Center of Mass (Variable Density)
Definition	Balancing point of a region with constant density	Balancing point of a region where density ρ(x, y) varies
Denominator	Area: ∫[f(x) − g(x)] dx	Total mass: ∬ ρ(x, y) dA
x̄ numerator	∫ x [f(x) − g(x)] dx	∬ x ρ(x, y) dA
When to use	Uniform-density regions in Calculus II	Non-uniform density problems, typically in multivariable calculus or physics

Why It Matters

The center of mass formula is a core topic in Calculus II and appears on virtually every AP Calculus BC exam and college calculus final. It provides a concrete application of integration: finding the exact balance point of an irregular shape. Understanding it also unlocks Pappus's theorem, which lets you compute volumes of revolution using just the centroid and the area — a powerful shortcut in both math and engineering.

Common Mistakes

Mistake: Using f(x) − g(x) in the ȳ numerator instead of (1/2)[f(x)² − g(x)²].

Correction: The ȳ formula requires the squared terms because it accounts for the midpoint height of each vertical strip. Always use (1/2)[f(x)² − g(x)²] in the numerator for ȳ.

Mistake: Forgetting to divide by the area in the final step.

Correction: The integrals in the numerators give moments, not coordinates. You must divide by the total area A to convert a moment into a coordinate. If your answer seems unreasonably large, check whether you divided by A.

Related Terms

Centroid — Same as center of mass for uniform density
Moment — Numerator integrals compute moments about axes
Area between Curves — Denominator of the center of mass formulas
Area under a Curve — Special case when g(x) = 0
Coordinates — The center of mass is expressed as coordinates
Plane Figure — The 2D region the formula applies to
Uniform — Constant density assumption required by these formulas
Formula — General term for the mathematical expressions used