Double Integrals

A weather service wants the total rainfall over an entire county in one storm, not just the depth at a single rain gauge. Rainfall depth f(x, y) varies from point to point across the county's shape — heavier in the hills, lighter near the coast — so "total rainfall" means adding up f over every point of a two-dimensional area, not along a one-dimensional stretch. The same question shows up as the total mass of a metal sheet whose density varies from spot to spot, or the total sales across a city whose population density varies by neighbourhood. All of these are asking for a sum over a region, and that sum is exactly what a double integral computes.

A single integral measures the area under a curve. Lift the idea one dimension: let f(x, y) be a surface floating over a flat region R in the plane. The double integral

\iint_R f(x, y)\, dA

is the signed volume trapped between that surface and the region — the natural sequel to the definite integral, and a first taste of integrating a function of several variables.

From a double Riemann sum to a volume

Just as the definite integral was born from Riemann sums, the double integral is the limit of a double sum. Chop a rectangle R = [a,b] \times [c,d] into a grid of m \times n small tiles, each of area \Delta A = \Delta x\, \Delta y.

Step 1 — build one matchstick. Over the tile with corner (x_i^*, y_j^*), erect a thin box of height f(x_i^*, y_j^*). Its volume is height times base:

\text{box}_{ij} = f(x_i^*, y_j^*)\, \Delta A.

Step 2 — stack every matchstick. Add the boxes over the whole grid to approximate the volume under the surface:

V \approx \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_i^*, y_j^*)\, \Delta A.

Step 3 — refine the grid. Let the tiles shrink (m, n \to \infty). The staircase of boxes settles onto the smooth volume, and the double sum becomes the double integral:

\iint_R f(x, y)\, dA = \lim_{m, n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_i^*, y_j^*)\, \Delta A.

Elegant, but a double limit is no way to compute. The rescue is Fubini's theorem: do one variable at a time.

Fubini's theorem: integrate one variable, then the other

Slice the solid with planes parallel to the y-axis. Each slice at a fixed x has cross-sectional area A(x) = \int_c^d f(x, y)\, dy — an ordinary single integral in which x is held constant. Sweeping the slices across [a, b] recovers the volume, exactly the volumes-by-slicing principle:

\iint_R f\, dA = \int_a^b A(x)\, dx = \int_a^b \!\left( \int_c^d f(x, y)\, dy \right) dx.

Fubini says the order does not matter: slice the other way and you get the same number.

Worked example — \iint_R xy\, dA over [0,1]\times[0,2]

Step 1 — write the iterated integral. Inner variable y runs 0 \to 2, outer variable x runs 0 \to 1:

\iint_R xy\, dA = \int_0^1 \!\left( \int_0^2 xy\, dy \right) dx.

Step 2 — do the inner integral, treating x as a constant. Pull x out and integrate y:

\int_0^2 xy\, dy = x \int_0^2 y\, dy = x \left[ \frac{y^2}{2} \right]_0^2 = x \cdot \frac{4}{2} = 2x.

Step 3 — feed that into the outer integral. The inner slice-area is A(x) = 2x:

\int_0^1 2x\, dx = \left[ x^2 \right]_0^1 = 1.

So \iint_R xy\, dA = 1.

Check — swap the order. Fubini claims it makes no difference which variable goes first. Try inner x, outer y instead:

\int_0^2 \!\left( \int_0^1 xy\, dx \right) dy = \int_0^2 y \left[ \frac{x^2}{2} \right]_0^1 dy = \int_0^2 \frac{y}{2}\, dy = \left[ \frac{y^2}{4} \right]_0^2 = 1.

Same region, same integrand, opposite slicing direction, and the answer lands on 1 both times — exactly what Fubini promises for a rectangle.

Non-rectangular regions: Type I

Real regions are rarely tidy rectangles. A Type I region sits between two curves of x: for a \le x \le b, y runs from a floor g_1(x) to a ceiling g_2(x). The inner limits now depend on x:

\iint_R f\, dA = \int_a^b \!\left( \int_{g_1(x)}^{g_2(x)} f(x, y)\, dy \right) dx.

Worked example — the region under y = x^2 on [0,1], with f = 1 (so we get its area).

Step 1 — read the limits off the region. For each x \in [0,1] the strip runs y: 0 \to x^2:

\text{Area} = \int_0^1 \!\left( \int_0^{x^2} 1\, dy \right) dx.

Step 2 — inner integral. The length of the strip is the ceiling minus the floor:

\int_0^{x^2} 1\, dy = x^2 - 0 = x^2.

Step 3 — outer integral. Sum the strip lengths across x:

\int_0^1 x^2\, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}.

The region has area \tfrac13. (A Type II region flips the roles — y outer, with x running between curves h_1(y) and h_2(y) — and you slice horizontally instead.)

Worked example — a triangular plate under y = x

A metal plate occupies the triangle R with corners (0,0), (2,0) and (2,2) — the region under the line y = x, for 0 \le x \le 2. Its density at (x, y) is xy (heavier towards the far corner). Find the total mass, \iint_R xy\, dA.

Step 1 — read the limits off the picture, not off the integrand. Sketch the triangle first. For a fixed x \in [0, 2], a vertical strip runs from the floor y = 0 up to the ceiling y = x — that ceiling is the whole reason the inner limit is a function of x rather than a number. The outer variable x then sweeps the constants 0 to 2:

\iint_R xy\, dA = \int_0^2 \!\left( \int_0^{x} xy\, dy \right) dx.

Step 2 — inner integral, x frozen.

\int_0^{x} xy\, dy = x \left[ \frac{y^2}{2} \right]_0^{x} = x \cdot \frac{x^2}{2} = \frac{x^3}{2}.

Step 3 — outer integral.

\int_0^2 \frac{x^3}{2}\, dx = \frac{1}{2} \left[ \frac{x^4}{4} \right]_0^2 = \frac{1}{2} \cdot \frac{16}{4} = 2.

The plate's total mass is 2. Notice the discipline that made this work: the inner limits (0 and x) are the region's floor and ceiling, and may depend on the outer variable; the outer limits (0 and 2) must always come out as plain constants. Mixing that up — say, leaving an x in the outer limits — is a sign the region was misread, not that the answer is unusual.

Let f be continuous on the region R. Then the double integral equals either iterated integral, and the order may be swapped:

Rectangle R = [a,b]\times[c,d]: \iint_R f\, dA = \int_a^b \!\int_c^d f\, dy\, dx = \int_c^d \!\int_a^b f\, dx\, dy.
Type I (g_1(x) \le y \le g_2(x)): \displaystyle \iint_R f\, dA = \int_a^b \!\int_{g_1(x)}^{g_2(x)} f\, dy\, dx.
Type II (h_1(y) \le x \le h_2(y)): \displaystyle \iint_R f\, dA = \int_c^d \!\int_{h_1(y)}^{h_2(y)} f\, dx\, dy.
Special case f \equiv 1: \iint_R 1\, dA = \operatorname{area}(R).

Fubini's promise — both orders give the same answer — is a practical superpower, because one order can be elementary while the other is impossible. The classic is

\int_0^1 \!\int_x^1 e^{y^2}\, dy\, dx.

The inner integral \int e^{y^2}\, dy has no elementary antiderivative — dead on arrival. But the region \{ 0 \le x \le 1,\ x \le y \le 1 \} is the triangle below the line y = x's mirror; redescribe it as Type II (0 \le y \le 1, 0 \le x \le y) and swap:

\int_0^1 \!\int_0^y e^{y^2}\, dx\, dy = \int_0^1 y\, e^{y^2}\, dy = \left[ \tfrac12 e^{y^2} \right]_0^1 = \tfrac12 (e - 1).

Now the inner integral is trivial (x is absent, so it just contributes a factor y) and the substitution u = y^2 finishes it. Same region, same volume — only the order changed. Reading the region correctly is the whole game.

The single most common iterated-integral mistake is copying a piece of the integrand into the limits, or vice versa. Keep these straight:

Limits are a description of R, full stop. They answer "where does the region start and end?" — never "what is f doing?" Draw the region before writing a single limit; if you can't sketch it, you can't set up the integral yet.
Inner limits may depend on the outer variable; outer limits may never depend on anything. In \int_a^b \!\int_{g_1(x)}^{g_2(x)} f\, dy\, dx, g_1(x) and g_2(x) are allowed — expected — to carry an x, but a and b must be plain numbers by the time you integrate. An x left over in the outer limits after the inner integral is done is a sure sign of a setup error.
Swapping order over a non-rectangular region is not just swapping dx and dy. It means re-sketching R and reading off brand-new limits for the other slicing direction — the Type I floor/ceiling of one order has nothing to do with the Type II left/right of the other, as the e^{y^2} example above shows: the new limits came from looking at the triangle again, not from any algebra on the old ones.

Take a black-and-white photograph as a grid of pixel brightnesses — a discrete stand-in for a function f(x, y) over a rectangle. Want the photo's total brightness? You could add up brightness row by row (getting one number per row, then adding those numbers), or column by column (one number per column, then adding those). Either way, you visit every pixel exactly once and add it in — so both totals must come out identical.

That is Fubini's theorem with the limit not yet taken: "sum rows-then-columns" is the discrete cousin of \int_a^b\!\int_c^d f\, dy\, dx, and "sum columns-then-rows" is \int_c^d\!\int_a^b f\, dx\, dy. Shrink the pixels to points and the two Riemann-style sums become the two iterated integrals — and Fubini is exactly the statement that a picture's total brightness doesn't care which way you scanned it.

See the slice sweep the region

Fubini works because the inner integral is the length of a strip and the outer integral sweeps the strip across the region. Drag the slider to move the representative vertical strip; its endpoints ride the floor and ceiling of R, which is exactly what the inner limits encode.

The volume, in three dimensions

A double integral is the volume trapped between a surface and its base region. Here the shaded floor is the region of integration, and the roof above it is z = f(x, y) — the double integral \iint_R f\, dA is exactly the volume of the solid between them. Drag the figure to view the hill and its footprint from any angle.