Mass and Density (2D)
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 5.6: “Calculating Centers of Mass and Moments of Inertia” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/5-6-calculating-centers-of-mass-and-moments-of-inertia |
| Textbook used in class | Stewart, Calculus, Section 15.5: “Applications of Double Integrals” (Examples 1, 2) |
Quick Reference
Mass of a lamina with density $\rho(x,y)$ over region $D$: $$m = \iint_D \rho(x,y)\,dA.$$
If $\rho$ is constant, then $m = \rho \cdot \text{Area}(D)$.
Motivation
A thin flat plate (lamina) with variable density $\rho(x,y)$ (in kg/m$^2$) has mass equal to the integral of the density over the plate. This is the direct two-dimensional analogue of the single-variable formula $m = \int_a^b \rho(x)\,dx$ for a rod.
Key Concept
The mass formula $m = \iint_D \rho(x,y)\,dA$ works because the plate can be broken into tiny area elements $dA$; the mass of each element is $\rho(x,y)\,dA$, and the total mass is the limit of the sum.
The setup requires: identifying the region $D$, expressing the density $\rho(x,y)$, and evaluating the double integral.
Worked Example
Find the mass of the lamina occupying the triangular region with vertices $(0,0)$, $(1,0)$, $(0,1)$ if the density is $\rho(x,y) = 1 + x + y$. (Stewart 15.5, Example 1.)
Region: Type I, $0 \leq x \leq 1$, $0 \leq y \leq 1-x$.
$$m = \int_0^1\int_0^{1-x}(1+x+y)\,dy\,dx.$$
Inner integral: $$\int_0^{1-x}(1+x+y)\,dy = \left[(1+x)y + \frac{y^2}{2}\right]_0^{1-x} = (1+x)(1-x)+\frac{(1-x)^2}{2} = (1-x^2)+\frac{(1-x)^2}{2}.$$
Simplify: $(1-x)(1+x) + \frac{(1-x)^2}{2} = (1-x)\!\left(1+x+\frac{1-x}{2}\right) = (1-x)\cdot\frac{3+x}{2}$.
Outer integral: $$m = \int_0^1\frac{(1-x)(3+x)}{2}\,dx = \frac{1}{2}\int_0^1(3+x-3x-x^2)\,dx = \frac{1}{2}\int_0^1(3-2x-x^2)\,dx.$$ $$= \frac{1}{2}\left[3x - x^2 - \frac{x^3}{3}\right]_0^1 = \frac{1}{2}\left(3 - 1 - \frac{1}{3}\right) = \frac{1}{2}\cdot\frac{5}{3} = \frac{5}{6}.$$
The mass is $\boxed{5/6}$ (in appropriate units).
if the density is not constant, the mass equals the density at the center times the area. That formula holds only for constant density. For variable density, you must integrate $\rho(x,y)$ over $D$. The density at the center is not representative of the average density unless the density is uniform.
Leveled Practice
Problem 1. Find the mass of the unit square $[0,1]\times[0,1]$ with density $\rho(x,y) = xy$.
Show answer
$$m = \int_0^1\int_0^1 xy\,dy\,dx = \int_0^1 x\left[\frac{y^2}{2}\right]_0^1 dx = \int_0^1\frac{x}{2}\,dx = \left[\frac{x^2}{4}\right]_0^1 = \frac{1}{4}.$$