Moments and Center of Mass (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 3, 4) |
Quick Reference
Moments of a lamina with density $\rho(x,y)$ over region $D$: $$M_x = \iint_D y\,\rho(x,y)\,dA \quad \text{(moment about the }x\text{-axis)},$$ $$M_y = \iint_D x\,\rho(x,y)\,dA \quad \text{(moment about the }y\text{-axis)}.$$
Center of mass: $\bar x = M_y/m$, $\bar y = M_x/m$.
Centroid (uniform density $\rho = 1$): $\bar x = \frac{1}{\text{Area}}\iint_D x\,dA$, $\bar y = \frac{1}{\text{Area}}\iint_D y\,dA$.
Motivation
The center of mass $(\bar x, \bar y)$ is the balance point of the lamina: if you place the plate on a pin at that point, it balances. A moment measures the tendency of the mass to rotate about an axis; a large moment about the $x$-axis means the mass is far from the $x$-axis on average.
Note the naming: $M_x$ is the moment about the $x$-axis and involves $y$ (the distance from the $x$-axis). Similarly, $M_y$ involves $x$.
Key Concept
The naming convention $M_x$ for the moment about the $x$-axis seems backward because it uses $y$ (the perpendicular distance from the $x$-axis), but the subscript indicates which AXIS you are taking the moment about, not which coordinate appears in the formula. Keeping this straight prevents a persistent confusion.
Worked Example
Find the center of mass of the lamina from the previous skill: triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$ and density $\rho(x,y) = 1+x+y$. (We found $m = 5/6$.)
Computing $M_y$: $$M_y = \int_0^1\int_0^{1-x} x(1+x+y)\,dy\,dx.$$
Inner: $x\!\int_0^{1-x}(1+x+y)\,dy = x\cdot\frac{(1-x)(3+x)}{2}$ (from the mass calculation).
$$M_y = \frac{1}{2}\int_0^1 x(1-x)(3+x)\,dx = \frac{1}{2}\int_0^1(3x+x^2-3x^2-x^3)\,dx$$ $$= \frac{1}{2}\int_0^1(3x-2x^2-x^3)\,dx = \frac{1}{2}\left[\frac{3x^2}{2}-\frac{2x^3}{3}-\frac{x^4}{4}\right]_0^1 = \frac{1}{2}\left(\frac{3}{2}-\frac{2}{3}-\frac{1}{4}\right).$$
$= \frac{1}{2}\cdot\frac{18-8-3}{12} = \frac{7}{24}.$
Computing $M_x$: by symmetry in $x$ and $y$ of the region and the density $1+x+y$, $M_x = 7/24$ as well.
Center of mass: $$\bar x = \frac{M_y}{m} = \frac{7/24}{5/6} = \frac{7}{24}\cdot\frac{6}{5} = \frac{7}{20}, \quad \bar y = \frac{7}{20}.$$
The center of mass is $(7/20, 7/20) = (0.35, 0.35)$, displaced toward the high-density corner $(0,0)$ relative to the centroid of the triangle (which is $(1/3, 1/3)$).
$M_x = \iint_D x\,\rho\,dA$. The moment about the $x$-axis is $M_x = \iint_D y\,\rho\,dA$ (with $y$, the perpendicular distance from the $x$-axis). The moment about the $y$-axis is $M_y = \iint_D x\,\rho\,dA$ (with $x$). The subscript names the axis; the integrand uses the coordinate perpendicular to that axis.
Leveled Practice
Problem 1. Find the centroid of the semicircular disk $D: x^2+y^2 \leq 1$, $y \geq 0$.
Show answer
By symmetry, $\bar x = 0$.
$\text{Area}(D) = \pi/2$.
$M_x = \iint_D y\,dA$. In polar: $0 \leq r \leq 1$, $0 \leq \theta \leq \pi$.
$M_x = \int_0^\pi\int_0^1 r\sin\theta\cdot r\,dr\,d\theta = \int_0^\pi\sin\theta\,d\theta\cdot\int_0^1 r^2\,dr = 2\cdot\frac{1}{3} = \frac{2}{3}$.
$\bar y = M_x/\text{Area} = \frac{2/3}{\pi/2} = \frac{4}{3\pi}$.
Centroid: $(0, 4/(3\pi))$.