Math Core

A double integral adds up a quantity spread over a region. Choose what is being spread (mass, charge, probability, surface area) and the same integral computes something new. This lesson collects the most important interpretations and shows how each one turns into $\iint_D (\text{something})\,dA$ .

Mass and density

Picture a thin flat plate, a lamina, occupying a region $D$ in the plane. If its density varies from point to point, described by $\rho(x, y)$ in mass per unit area, then a tiny piece of area $\Delta A$ near $(x, y)$ has mass about $\rho(x, y)\,\Delta A$ . Adding and taking the limit:

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

The same formula gives total electric charge from a charge density $\sigma(x, y)$ , or total population from a population density. When $\rho = 1$ , mass equals area.

Moments and center of mass

The moment of the lamina about an axis measures its tendency to rotate about that axis. A small piece at $(x, y)$ is at distance $y$ from the $x$ -axis and distance $x$ from the $y$ -axis, so

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

Notice the switch: $M_x$ (moment about the $x$ -axis) uses $y$ , the distance from that axis.

Center of mass

The center of mass $(\bar{x}, \bar{y})$ of a lamina with density $\rho$ on $D$ is

\bar{x} = \frac{M_y}{m} = \frac{1}{m}\iint_D x\,\rho\,dA, \qquad \bar{y} = \frac{M_x}{m} = \frac{1}{m}\iint_D y\,\rho\,dA.

It is the balance point: the lamina would balance on a pin placed there. When $\rho$ is constant, the center of mass is called the centroid and depends only on the shape.

In words, $\bar{x}$ is the density-weighted average of $x$ over the region. That is the same idea as average value, with $\rho$ as the weight.

Worked example: A plate that gets heavier to the right

A lamina occupies the rectangle $[0, 2] \times [0, 3]$ with density $\rho(x, y) = x$ . Find its mass and center of mass.

Solution. Every integrand here is separable on a rectangle.

m = \int_0^3 \int_0^2 x\,dx\,dy = 2 \cdot 3 = 6.

M_y = \int_0^3 \int_0^2 x^2\,dx\,dy = \frac{8}{3} \cdot 3 = 8, \qquad M_x = \int_0^3 \int_0^2 xy\,dx\,dy = 2 \cdot \frac{9}{2} = 9.

So $\bar{x} = \dfrac{8}{6} = \dfrac{4}{3}$ and $\bar{y} = \dfrac{9}{6} = \dfrac{3}{2}$ . The balance point is right of center ( $\bar{x} > 1$ ) because the plate is denser on the right, but it is centered vertically because the density does not depend on $y$ .

A uniform half-disk of radius 1. Its centroid lies on the axis of symmetry at height 4/(3π) ≈ 0.42.Open in grapher →

Worked example: Centroid of a half-disk

Find the centroid of the upper half of the disk $x^2 + y^2 \le a^2$ .

Solution. Take $\rho = 1$ . By symmetry across the $y$ -axis, $\bar{x} = 0$ . The area is $m = \dfrac{\pi a^2}{2}$ . Using polar coordinates,

M_x = \iint_D y\,dA = \int_0^{\pi} \int_0^a (r\sin\theta)\,r\,dr\,d\theta = \frac{a^3}{3}\int_0^{\pi}\sin\theta\,d\theta = \frac{2a^3}{3}.

So $\bar{y} = \dfrac{2a^3/3}{\pi a^2/2} = \dfrac{4a}{3\pi}$ , about $0.42a$ .

Tip

Use symmetry before integrating. If both the region and the density are symmetric across a line, the center of mass lies on that line, so one coordinate comes free.

Moments of inertia

The moment of inertia (second moment) measures resistance to rotation. It weights each piece of mass by the square of its distance from the axis:

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

$I_0$ is the moment of inertia about the origin (the polar moment). Mass far from the axis contributes much more than mass close to it, which is why a figure skater spins faster when pulling in their arms.

Probability

A pair of continuous random variables $X$ and $Y$ has a joint density function $f(x, y)$ when $f \ge 0$ , $\iint_{\mathbb{R}^2} f\,dA = 1$ , and

P\big((X, Y) \in D\big) = \iint_D f(x, y)\,dA.

Probability plays the role of mass, with total mass $1$ . The expected values $E[X] = \iint x f\,dA$ and $E[Y] = \iint y f\,dA$ are exactly the center of mass of that probability distribution.

The event X < Y is the part of the first quadrant above the line y = x.Open in grapher →

Worked example: Which happens first?

Two components have lifetimes $X$ and $Y$ (in years) with joint density $f(x, y) = 6e^{-2x - 3y}$ for $x, y \ge 0$ , and $0$ otherwise. Find $P(X < Y)$ .

Solution. The event is the region $0 \le x$ , $y > x$ . Integrate $y$ from $x$ to $\infty$ , then $x$ from $0$ to $\infty$ :

\int_x^{\infty} 6e^{-2x - 3y}\,dy = 6e^{-2x} \cdot \frac{e^{-3x}}{3} = 2e^{-5x},

P(X < Y) = \int_0^{\infty} 2e^{-5x}\,dx = \frac{2}{5}.

Surface area

The part of the surface $z = f(x, y)$ lying over a region $D$ has area

A = \iint_D \sqrt{1 + f_x^2 + f_y^2}\,dA.

The reason: over a tiny rectangle of area $\Delta A$ , the surface is nearly its tangent plane, and the tilted patch of tangent plane has area $\sqrt{1 + f_x^2 + f_y^2}\,\Delta A$ . The steeper the surface, the larger the factor. For a horizontal plane the factor is $1$ .

Worked example: Area of a paraboloid cap

Find the area of the part of $z = x^2 + y^2$ that lies over the disk $x^2 + y^2 \le 1$ .

Solution. $f_x = 2x$ and $f_y = 2y$ , so the factor is $\sqrt{1 + 4x^2 + 4y^2} = \sqrt{1 + 4r^2}$ . In polar,

A = \int_0^{2\pi} \int_0^1 \sqrt{1 + 4r^2}\,r\,dr\,d\theta = 2\pi \cdot \frac{1}{12}\left(1 + 4r^2\right)^{3/2}\Big|_0^1 = \frac{\pi\left(5\sqrt{5} - 1\right)}{6}.

Common mistake

Keep the moments straight: $M_x$ uses the factor $y$ and $I_x$ uses $y^2$ , because the distance from the $x$ -axis is $|y|$ . And always divide a moment by the mass, not the area, unless the density is $1$ .