12.1 Calculation of mass, moments, and centers of mass

Mass and Density

Mass Density Function and Total Mass

The mass density function $\rho(x,y)$ tells you the mass per unit area at each point $(x,y)$ in a planar region. Think of it as a map of how "heavy" each tiny piece of the region is. When density varies across the region, you need a double integral to add up all those tiny contributions.

The total mass $M$ of a region $R$ is:

$M = \iint_R \rho(x,y) \, dA$

If the density is constant (uniform), this collapses to $M = \rho \cdot A$, where $A$ is the area of the region.

Example: A rectangular plate measuring $2 \, \text{m} \times 3 \, \text{m}$ has constant density $\rho = 5 \, \text{kg/m}^2$. Then $M = 5 \times (2 \times 3) = 30 \, \text{kg}$.

For variable density, you have to actually evaluate the integral. Suppose a square plate over $0 \le x \le 1$, $0 \le y \le 1$ has density $\rho(x,y) = 3x + 2y$. You'd compute:

$M = \int_0^1 \int_0^1 (3x + 2y) \, dy \, dx$

Evaluate the inner integral first with respect to $y$, then the outer with respect to $x$.

Planar Regions and Setting Up Double Integrals

A planar region is the two-dimensional domain in the $xy$-plane over which you integrate. The shape of this region determines your limits of integration, and choosing the right coordinate system can simplify things dramatically.

• For rectangular regions, Cartesian coordinates with constant limits work directly.
• For circular or radial regions, polar coordinates are almost always the better choice. The area element transforms as $dA = r \, dr \, d\theta$.

Example (polar): For a circular disk of radius $R$ centered at the origin:

$\iint_R f(r,\theta) \, dA = \int_0^{2\pi} \int_0^R f(r,\theta) \, r \, dr \, d\theta$

Don't forget the extra factor of $r$ in polar coordinates. Leaving it out is one of the most common mistakes in these problems.

Moments

First Moments

First moments measure how mass is distributed relative to an axis. They're the building blocks for finding the center of mass.

For a region with density $\rho(x,y)$:

• First moment about the $x$-axis: $M_x = \iint_R y \, \rho(x,y) \, dA$
• First moment about the $y$-axis: $M_y = \iint_R x \, \rho(x,y) \, dA$

Notice the pattern: $M_x$ uses $y$ (the distance from the $x$-axis), and $M_y$ uses $x$ (the distance from the $y$-axis). This notation trips people up constantly. The subscript tells you which axis you're taking the moment about, but the integrand uses the other variable because that's the perpendicular distance to that axis.

Second Moments (Moments of Inertia)

The second moment, or moment of inertia, measures an object's resistance to rotational acceleration about a given axis. Larger moment of inertia means harder to spin.

• About the $x$-axis: $I_x = \iint_R y^2 \, \rho(x,y) \, dA$
• About the $y$-axis: $I_y = \iint_R x^2 \, \rho(x,y) \, dA$
• About the $z$-axis (origin): $I_z = \iint_R (x^2 + y^2) \, \rho(x,y) \, dA$

The $z$-axis version, also called the polar moment of inertia, uses $x^2 + y^2$ because that's the squared distance from the origin. Notice that $I_z = I_x + I_y$, which is the perpendicular axis theorem for planar objects.

Example: A thin circular disk of radius $R$ with constant density $\rho$ has moment of inertia about its center:

$I_z = \frac{1}{2} \rho \pi R^4$

To see where this comes from, set it up in polar coordinates:

$I_z = \int_0^{2\pi} \int_0^R r^2 \cdot \rho \cdot r \, dr \, d\theta = \rho \int_0^{2\pi} d\theta \int_0^R r^3 \, dr = \rho \cdot 2\pi \cdot \frac{R^4}{4} = \frac{1}{2}\rho\pi R^4$

Since the total mass is $M = \rho \pi R^2$, this can also be written as $I_z = \frac{1}{2}MR^2$, which you may recognize from physics.

Center of Mass

Center of Mass and Centroid

The center of mass $(\bar{x}, \bar{y})$ is the point where the region would balance perfectly if placed on a pin. It's computed from the first moments and total mass:

$\bar{x} = \frac{M_y}{M}, \quad \bar{y} = \frac{M_x}{M}$

Again, watch the subscripts: $\bar{x}$ uses $M_y$ (not $M_x$), and $\bar{y}$ uses $M_x$ (not $M_y$).

When density is constant, it cancels from the numerator and denominator, and the center of mass becomes the centroid, which depends only on geometry:

$\bar{x} = \frac{1}{A} \iint_R x \, dA, \quad \bar{y} = \frac{1}{A} \iint_R y \, dA$

where $A$ is the area of the region.

Calculating Center of Mass: Step-by-Step

For a region $R$ with variable density $\rho(x,y)$:

1. Sketch the region and determine appropriate limits of integration (Cartesian or polar).
2. Compute total mass: $M = \iint_R \rho(x,y) \, dA$
3. Compute the first moment about the $y$-axis: $M_y = \iint_R x \, \rho(x,y) \, dA$
4. Compute the first moment about the $x$-axis: $M_x = \iint_R y \, \rho(x,y) \, dA$
5. Divide to get the center of mass: $\bar{x} = \frac{M_y}{M}, \quad \bar{y} = \frac{M_x}{M}$

You're evaluating three separate double integrals over the same region. Often you can reuse intermediate calculations, so look for shared factors.

Example: Consider the upper half of a disk of radius $R$ (a semicircular region, $y \ge 0$) with constant density $\rho$. By symmetry, $\bar{x} = 0$. For $\bar{y}$, you compute:

$M = \rho \cdot \frac{\pi R^2}{2}, \quad M_x = \int_0^{\pi} \int_0^R r\sin\theta \cdot \rho \cdot r \, dr \, d\theta = \frac{2\rho R^3}{3}$

So $\bar{y} = \frac{M_x}{M} = \frac{4R}{3\pi}$. The center of mass sits above the geometric midpoint because more area is concentrated near the base.