A double integral adds up "something per unit area" over a region. If that something is density, you get mass; weight it by position and you get the balance point. You did this in one dimension with thin rods; now the objects are thin plates (laminas), and the same integrals measure probability for two random variables.
Key idea
For a plate on $D$ with density $\rho(x, y)$ (mass per unit area):
$$m = \iint_D \rho\,dA, \qquad \bar{x} = \frac{1}{m}\iint_D x\rho\,dA, \qquad \bar{y} = \frac{1}{m}\iint_D y\rho\,dA.$$
Moments of inertia: $I_x = \iint_D y^2\rho\,dA$, $I_y = \iint_D x^2\rho\,dA$. For a joint density $f(x, y)$, $P((X, Y) \in E) = \iint_E f\,dA$, and $\iint f\,dA = 1$ over all outcomes.
Why it works
Chop the plate into small pieces. A piece of area $\Delta A$ near $(x, y)$ has mass about $\rho(x, y)\,\Delta A$; adding the pieces gives the mass. The balance point is a weighted average of positions, each piece weighted by its mass, just as the average of numbers weights each by how often it occurs: $\bar{x} = \frac{\sum x\,\Delta m}{\sum \Delta m}$. Probability works the same way, with "probability per unit area" in place of mass per unit area.
Worked examples
Worked example: Mass and center of mass
A plate covers the triangle with vertices $(0, 0)$, $(1, 0)$, $(0, 1)$, with density $\rho = y$. Find its mass and center of mass.
- $$m = \int_0^1\int_0^{1 - x} y\,dy\,dx = \int_0^1 \frac{(1 - x)^2}{2}dx = \frac{1}{6}.$$
- $\iint x\rho\,dA = \int_0^1 x\frac{(1 - x)^2}{2}dx = \frac{1}{24}$, so $\bar{x} = \frac{1/24}{1/6} = \frac{1}{4}$.
- $$\iint y\rho\,dA = \int_0^1\int_0^{1 - x} y^2\,dy\,dx = \int_0^1 \frac{(1 - x)^3}{3}dx = \frac{1}{12},$$ so $\bar{y} = \frac{1}{2}$.
- The plate is heavier near the top, so $\bar{y} = \frac{1}{2}$ is above the centroid of the triangle ($\frac{1}{3}$).
Answer: $m = \frac{1}{6}$, center $\left(\frac{1}{4}, \frac{1}{2}\right)$
Worked example: A probability
Waiting times $X$ and $Y$ (minutes) have joint density $f(x, y) = 6e^{-2x - 3y}$ for $x, y \ge 0$. Find $P(X \le 1, Y \le 1)$.
- $$P = \int_0^1\int_0^1 6e^{-2x}e^{-3y}\,dy\,dx = \left(\int_0^1 2e^{-2x}dx\right)\left(\int_0^1 3e^{-3y}dy\right).$$
- $= \left(1 - e^{-2}\right)\left(1 - e^{-3}\right) \approx 0.82$.
Answer: $\left(1 - e^{-2}\right)\left(1 - e^{-3}\right) \approx 0.82$
Worked example: Making a density
For which $k$ is $f(x, y) = k(x + y)$ a joint density on the unit square?
- Total probability: $$\int_0^1\int_0^1 (x + y)\,dy\,dx = \int_0^1 \left(x + \frac{1}{2}\right)dx = 1.$$
- So $k \cdot 1 = 1$ and $k = 1$.
Answer: $k = 1$
Common mistake
Dividing by the area instead of the mass. With a variable density, $\bar{x} = \frac{\iint x\rho\,dA}{\iint \rho\,dA}$; dividing by the area gives the geometric centroid only when $\rho$ is constant. Check: the center of mass must move toward the heavy part of the plate, as $\bar{y}$ did in the example.