Applications of Double Integrals

Mass, moments, center of mass, and moment of inertia of a lamina, via double integrals.

The unifying idea. Every application here is the same move: take a physical quantity spread over a 2D region and add up its little bits with a double integral. Mass, moments, charge — all the same shape. The only thing that changes is what you put inside the integral. Get that, and the formulas stop being a list to memorize and become one idea with variations.

A flat plate (a lamina) has density ρ(x, y) — mass per unit area, which can vary from point to point. Its total mass:

Read it as: density times a little bit of area is a little bit of mass; add them all up. If ρ is constant, this is just ρ times the area.

m = \iint_D \rho(x,y)\, dA

Mass of a lamina

$\rho(x,y)$ Density: mass per unit area, may vary across the plate.
$dA$ An infinitesimal patch of area.
$m$ Total mass of the lamina.

A moment measures how mass is distributed relative to an axis — mass weighted by distance from that axis.

The moment about the x-axis weights by y (distance from the x-axis is the y-coordinate); the moment about the y-axis weights by x:

M_x = \iint_D y\, \rho(x,y)\, dA \qquad M_y = \iint_D x\, \rho(x,y)\, dA

Moments about the x- and y-axis

$M_x$ Moment about the x-axis — weights mass by distance from that axis (i.e. by y).
$M_y$ Moment about the y-axis — weights mass by distance from that axis (i.e. by x).

The subscript trap, worth burning in: M_x — moment about the x-axis — uses y inside. M_y uses x. The subscript names the axis; the integrand is the other variable. Students swap these constantly.

Center of mass — the balance point: divide each moment by the total mass.

\bar{x} = \dfrac{M_y}{m} \qquad \bar{y} = \dfrac{M_x}{m}

Center of mass

$\bar{x}, \bar{y}$ Coordinates of the center of mass — the balance point of the plate.

Again crossed: x̄ (a horizontal position) comes from M_y. The x-coordinate of the balance point depends on the moment about the y-axis. If you keep M_x = ∬ y ρ dA straight, the rest follows.

Moment of inertia (second moment) — like a moment, but weighted by distance squared. It measures resistance to rotation:

Weight by distance-squared instead of distance:

I_x = \iint_D y^2\, \rho(x,y)\, dA \qquad I_y = \iint_D x^2\, \rho(x,y)\, dA \qquad I_0 = I_x + I_y

Moments of inertia

$I_x, I_y$ Moments of inertia about the x- and y-axis — weight mass by distance squared; resistance to rotation.
$I_0$ Polar moment of inertia, about the origin — the sum I_x + I_y.

I_0 (the polar moment of inertia, about the origin) is just the sum, because x² + y² is distance-squared from the origin.

How to solve one of these

1.
Identify which quantity is asked. Mass, a moment, center of mass, or moment of inertia — that alone tells you what goes inside the integral.
2.
Write the integrand as density times the weight. No weight means mass. Weight by $y$ or $x$ means a first moment. Weight by $y^2$ or $x^2$ means a moment of inertia.
3.
For center of mass, compute the mass and both moments first. Find $m$ , $M_x$ , and $M_y$ , then divide — it's three integrals, then arithmetic. Don't try to shortcut it.
4.
Set up the region as in Type I, Type II, or polar form. Applications don't change how you integrate over the region $D$ , only what you integrate.

m = \iint_D \rho(x,y)\, dA

Density times a little bit of area, added up over the whole plate, gives the total mass.