Mass, Center of Mass & Moments — Compute mass, center of mass, and moments (intro) - Multivariable Calculus

SECTION 1

Historical Context & Motivation

Long before calculus existed, ancient engineers needed to know where to place a support so a beam wouldn't tip over. The Greeks called this special balancing point the center of gravity, and finding it accurately was crucial for building stable structures, designing ships, and constructing levers. As mathematics advanced, scholars realized that the same question — 'Where does an object balance?' — could be answered precisely using integration, especially when the object's density varies from point to point.

~250 BC • Archimedes & the Lever

Archimedes formalized the law of the lever and showed how to locate the center of gravity of simple shapes like triangles and parabolic segments using geometric reasoning.

1600s • Newton & Leibniz Invent Calculus

With the tools of integration, mathematicians could extend Archimedes' ideas to objects with continuously varying density, turning balance problems into integrals.

1700s • Euler & Mechanics

Leonhard Euler applied moments and center-of-mass calculations systematically to rigid-body mechanics, laying the foundation for modern engineering analysis.

1800s • Multiple Integrals Formalized

Mathematicians like Riemann and Lebesgue rigorously defined double and triple integrals, making it possible to compute mass and center of mass for any region with a known density function.

The central question this lesson addresses is straightforward: if you have a flat plate (called a lamina) whose density might change from one point to another, how do you compute its total mass, and where exactly would you place a single support so it balances perfectly? Double integrals give us the answer.

SECTION 2

Core Principles & Definitions

Before diving into formulas, let's establish the key ideas. We work with a thin, flat plate — a lamina — that occupies some region R in the xy-plane. The plate can have a density function ρ(x, y) that tells us how much mass is packed into each tiny piece of area. When density is constant, the problem simplifies; when density varies, double integrals are essential.

Mass (m)

The total amount of matter in the lamina. Computed by integrating the density function ρ(x, y) over the entire region R.

Moment

A measure of how mass is distributed relative to an axis. The moment about the x-axis (M_x) uses the y-coordinate as a 'lever arm,' and the moment about the y-axis (M_y) uses the x-coordinate.

Center of Mass (x̄, ȳ)

The single point where you could balance the entire lamina on a pin. Found by dividing each moment by the total mass: x̄ = M_y / m and ȳ = M_x / m.

Density Function ρ(x, y)

Describes mass per unit area at each point. Units are typically kg/m² or g/cm². A constant ρ means uniform (homogeneous) density.

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KEY TAKEAWAY

Think of a lamina like a cookie that might be thicker (denser) in some spots. If the cookie is perfectly uniform, its balance point is at the geometric center. But if one side has extra chocolate chips (more mass), the balance point shifts toward the heavier side. The center of mass is that shifted balance point, and moments are the mathematical tool that tells us exactly how far to shift.

SECTION 3

Visual Explanation

The diagram below shows a lamina (shaded region) in the xy-plane with varying density. Notice how the density is represented by color intensity — darker areas are denser. The center of mass is marked and sits closer to the denser region, not at the geometric center of the shape.

The elliptical lamina has density increasing from left to right. The center of mass (pink dot) is shifted right of the geometric center (cyan dot) because more mass is concentrated on the right side.

This visual captures the core idea: when density is not uniform, the balance point (center of mass) does not coincide with the geometric center. The moments M_x and M_y quantify exactly how much the mass distribution is 'weighted' toward one side, and dividing by total mass yields the precise balance coordinates.

SECTION 4

Mathematical Framework

All three quantities — mass, moments, and center of mass — are computed using double integrals over the region R. The density function ρ(x, y) plays the starring role in every formula. Let's walk through each one.

TOTAL MASS

m = ∬_R ρ(x, y) dA

m = total mass of the lamina; ρ(x, y) = density at point (x, y); dA = differential area element (dx dy or dy dx); R = region occupied by the lamina.

To find the total mass, you integrate the density over every tiny piece of the region. If ρ is constant, this simplifies to ρ × (area of R). When ρ varies, the integral 'adds up' different densities at each point.

MOMENT ABOUT THE x-AXIS

Mₓ = ∬_R y · ρ(x, y) dA

M_x measures how mass is distributed relative to the x-axis. The factor y acts as a 'lever arm' — points farther from the x-axis contribute more.

MOMENT ABOUT THE y-AXIS

Mᵧ = ∬_R x · ρ(x, y) dA

M_y measures how mass is distributed relative to the y-axis. Here x is the lever arm — points farther right or left from the y-axis contribute more.

CENTER OF MASS

x̄ = Mᵧ / m , ȳ = Mₓ / m

x̄ = x-coordinate of the center of mass; ȳ = y-coordinate of the center of mass. Notice the 'cross' pattern: x̄ uses M_y and ȳ uses M_x.

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Watch the Subscripts!

A common mistake is mixing up which moment goes with which coordinate. Remember: x̄ = M_y / m (moment about the y-axis gives the x-coordinate). The subscripts 'swap.' Think of it this way: the moment about an axis tells you how far to go in the perpendicular direction.

SECTION 5

Setting Up the Integrals — Step by Step

The hardest part of these problems is usually setting up the limits of integration correctly. Let's look at a concrete example: a triangular lamina with a density function, and see how the integral gets built from the region's geometry.

A triangular region R with vertices at (0, 0), (4, 0), and (0, 3). The hypotenuse has equation y = 3 − (3/4)x. For a fixed x-value (pink dashed line), y ranges from 0 up to the hypotenuse. The outer integral sweeps x from 0 to 4.

The process always follows the same pattern. First, sketch the region R and identify its boundaries. Second, decide whether to integrate with respect to y first (vertical strips) or x first (horizontal strips). Third, express the limits of the inner integral in terms of the outer variable. Finally, write the integrand as ρ(x, y) for mass, y · ρ(x, y) for M_x, or x · ρ(x, y) for M_y.

Sketch the region R and label all boundary curves.
Choose order of integration (dy dx or dx dy) — pick whichever makes limits simpler.
Write limits — inner limits depend on the outer variable; outer limits are constants.
Place the correct integrand — ρ for mass, y·ρ for M_x, x·ρ for M_y.
Evaluate the inner integral first, then the outer integral.

SECTION 6

Worked Example

Let's compute the mass and center of mass of a rectangular lamina occupying the region 0 ≤ x ≤ 2, 0 ≤ y ≤ 3, with density function ρ(x, y) = 1 + x. This density increases linearly as we move to the right.

Mass & Center of Mass of a Rectangle with ρ(x, y) = 1 + x

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Step 1 — Identify the Region and Density

The region R is a rectangle with 0 ≤ x ≤ 2 and 0 ≤ y ≤ 3. The density is ρ(x, y) = 1 + x. Because the density depends only on x, the lamina gets heavier as we move right.
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Step 2 — Compute the Mass m

m = ∫₀² ∫₀³ (1 + x) dy dx. Evaluate the inner integral first: ∫₀³ (1 + x) dy = (1 + x) × y |₀³ = 3(1 + x). Now the outer integral: m = ∫₀² 3(1 + x) dx = 3[x + x²/2]₀² = 3[2 + 2] = 3 × 4 = 12.

m = 12
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Step 3 — Compute the Moment Mᵧ (about the y-axis)

M_y = ∫₀² ∫₀³ x(1 + x) dy dx. Inner integral: ∫₀³ x(1 + x) dy = 3x(1 + x) = 3x + 3x². Outer integral: M_y = ∫₀² (3x + 3x²) dx = 3[x²/2 + x³/3]₀² = 3[2 + 8/3] = 3 × 14/3 = 14.

Mᵧ = 14
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Step 4 — Compute the Moment Mₓ (about the x-axis)

M_x = ∫₀² ∫₀³ y(1 + x) dy dx. Inner integral: ∫₀³ y(1 + x) dy = (1 + x) × y²/2 |₀³ = (9/2)(1 + x). Outer integral: M_x = ∫₀² (9/2)(1 + x) dx = (9/2)[x + x²/2]₀² = (9/2)(4) = 18.

Mₓ = 18
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Step 5 — Find the Center of Mass (x̄, ȳ)

x̄ = M_y / m = 14/12 = 7/6 ≈ 1.167. ȳ = M_x / m = 18/12 = 3/2 = 1.5. The center of mass is at (7/6, 3/2). Notice that x̄ ≈ 1.167 is to the right of the geometric center x = 1, which makes sense because the density increases toward the right.

Center of mass: (7/6, 3/2) ≈ (1.167, 1.5)

✓

SANITY CHECK

Always verify your answer makes physical sense. The geometric center of the rectangle is (1, 1.5). Since the density increases in the x-direction, x̄ should be greater than 1, and ȳ should remain 1.5 (density doesn't depend on y). Our answer (7/6, 3/2) checks out perfectly!

SECTION 7

Uniform vs. Variable Density

Many problems you encounter will specify either uniform (constant) density or variable density. It's important to understand how these two cases compare, because uniform density problems simplify dramatically.

Feature	Uniform Density (ρ = constant)	Variable Density (ρ = f(x, y))
Mass formula	m = ρ × Area(R) — no integration needed for mass if area is known	m = ∬ ρ(x,y) dA — must integrate
Center of mass	Equals the centroid (geometric center) — ρ cancels out of the formula	Shifts toward regions of higher density
Difficulty	Simpler — focus is on geometry of R	Harder — density adds complexity to the integrand
Real-world example	Uniform sheet of aluminum	A metal plate that is thicker on one end

Comparison of uniform vs. variable density problems

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KEY TAKEAWAY

When density is constant, the center of mass is the same as the centroid — a purely geometric quantity. The centroid is found using the same formulas, just with ρ set to 1. This means every centroid problem you've solved in geometry is actually a special case of center-of-mass calculation!

SECTION 8

Connection to Triple Integrals & Advanced Theory

Everything we've done with double integrals for flat plates extends naturally to three-dimensional objects using triple integrals. Instead of a lamina in the xy-plane, you work with a solid occupying a region in 3D space, and the density function becomes ρ(x, y, z).

Quantity	2D (Double Integral)	3D (Triple Integral)
Mass	m = ∬ ρ(x, y) dA	m = ∭ ρ(x, y, z) dV
Moment	M_x = ∬ y·ρ dA	M_xy = ∭ z·ρ dV (and similar)
Center of mass	(x̄, ȳ) — two coordinates	(x̄, ȳ, z̄) — three coordinates
Region	Flat region R in the plane	Solid region E in space

From 2D to 3D: the same ideas, one more dimension

You'll also encounter moments of inertia in more advanced courses, which measure how mass is distributed relative to rotation. Instead of multiplying by x or y (first power), you multiply by x² or y², giving you a 'second moment.' Moments of inertia are essential in physics for analyzing spinning objects, from figure skaters to car engines.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A lamina has uniform density ρ = 5 and is shaped like a perfect square centered at the origin. Without computing any integrals, where is the center of mass? Explain your reasoning.

Show answer

The center of mass is at the origin (0, 0). When density is uniform, the center of mass coincides with the centroid (geometric center). Because a square centered at the origin is symmetric about both axes, the centroid — and therefore the center of mass — must lie at the intersection of those lines of symmetry, which is (0, 0). Symmetry is a powerful shortcut in these problems.

Question 2: BASIC CALCULATION

Find the mass of a lamina occupying the rectangular region 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, with constant density ρ = 3.

Show answer

For constant density, m = ρ × Area(R). The rectangle has area 1 × 2 = 2, so m = 3 × 2 = 6. You can verify with the integral: m = ∫₀¹ ∫₀² 3 dy dx = ∫₀¹ 6 dx = 6. Both approaches confirm the mass is 6 (in whatever units are appropriate for the density).

Question 3: INTERMEDIATE

A lamina occupies the region bounded by y = 0, x = 0, and y = 4 − 2x (a triangle). The density is ρ(x, y) = xy. Set up and evaluate the integral for the total mass.

Show answer

The triangle has vertices at (0, 0), (2, 0), and (0, 4). Using dy dx order: m = ∫₀² ∫₀^(4−2x) xy dy dx. Inner integral: ∫₀^(4−2x) xy dy = x · y²/2 |₀^(4−2x) = x(4 − 2x)²/2 = x(16 − 16x + 4x²)/2 = 8x − 8x² + 2x³. Outer integral: m = ∫₀² (8x − 8x² + 2x³) dx = [4x² − 8x³/3 + x⁴/2]₀² = 16 − 64/3 + 8 = 24 − 64/3 = 8/3 ≈ 2.667. The mass is 8/3.

Question 4: APPLIED

A rectangular metal plate has dimensions 0 ≤ x ≤ 3, 0 ≤ y ≤ 2 (meters). Its density is ρ(x, y) = 2 + y kg/m² (denser toward the top). Find the center of mass. An engineer needs to know where to place a support so the plate balances.

Show answer

First compute mass: m = ∫₀³ ∫₀² (2 + y) dy dx = ∫₀³ [2y + y²/2]₀² dx = ∫₀³ 6 dx = 18 kg. For M_y: ∫₀³ ∫₀² x(2 + y) dy dx = ∫₀³ 6x dx = 3x² |₀³ = 27. So x̄ = 27/18 = 3/2 = 1.5 m. For M_x: ∫₀³ ∫₀² y(2 + y) dy dx = ∫₀³ [y² + y³/3]₀² dx = ∫₀³ (4 + 8/3) dx = ∫₀³ 20/3 dx = 20. So ȳ = 20/18 = 10/9 ≈ 1.111 m. The support goes at (1.5, 10/9). Notice x̄ = 1.5 is the geometric midpoint (density doesn't depend on x), while ȳ ≈ 1.111 > 1 because density increases with y, pulling the balance point upward.

Question 5: CRITICAL THINKING

Suppose a lamina has density ρ(x, y) = k − x² − y² over the disk x² + y² ≤ k, where k > 0 is a constant. Without evaluating the integrals fully, explain why the center of mass must be at the origin. Then describe what would happen to the center of mass if the density were changed to ρ(x, y) = k − x² − y² + x.

Show answer

For ρ = k − x² − y², both the region (a disk centered at the origin) and the density function are symmetric about the x-axis and y-axis. In the M_y integral, the integrand x · ρ(x, y) is an odd function of x integrated over a symmetric domain, so it equals zero, giving x̄ = 0. The same argument applies for ȳ by symmetry in y. For the modified density ρ = k − x² − y² + x, the extra term '+x' breaks the x-symmetry. Now the M_y integral includes ∬ x² dA > 0, which is nonzero, pushing x̄ to a positive value. Meanwhile, ȳ remains 0 because the density is still symmetric in y. The center of mass shifts to the right of the origin.

SUMMARY

Summary & Review

Summary

The mass of a lamina with density ρ(x, y) is computed as m = ∬ ρ(x, y) dA. The moments M_x and M_y measure how mass is distributed relative to the x- and y-axes, respectively, by weighting the density with y or x as a lever arm. The center of mass (x̄, ȳ) = (M_y/m, M_x/m) is the balance point of the lamina.

When density is uniform, the center of mass equals the centroid — a purely geometric quantity. When density varies, the center of mass shifts toward the denser regions. Use symmetry as a shortcut whenever both the region and density function share symmetry about an axis. These 2D concepts extend to 3D via triple integrals, and lead to more advanced topics like moments of inertia in physics and engineering.