Iterated Integrals and Fubini's Theorem

Core Idea

A double integral ∬ f(x,y) dA can be computed as an iterated integral ∫∫ f(x,y) dy dx. By Fubini's theorem, the order can be swapped if f is continuous; rewriting bounds requires understanding the region carefully.

Explainer

You know from antiderivatives that a single integral ∫ₐᵇ f(x) dx computes the signed area under a curve on an interval. Now extend the question: what is the signed volume under a surface z = f(x, y) over a 2D region R in the xy-plane? The double integral ∬_R f(x, y) dA answers this, but computing it directly requires a strategy. The iterated integral reduces the 2D problem to two sequential 1D integrations, each of which you already know how to perform.

The idea is to slice the volume into thin cross-sections. Fix a value of x and integrate f(x, y) over y — this gives A(x), the area of the cross-sectional slice at that x. Then integrate A(x) over x to accumulate all slices into a total volume. Written explicitly: ∫ₐᵇ ( ∫_c^d f(x,y) dy ) dx. The inner integral treats x as a constant and integrates in y; the outer integral then integrates the result in x. The parentheses are usually omitted, and you simply evaluate the innermost integral first, working outward.

Fubini's theorem guarantees that when f is continuous on a rectangle [a,b] × [c,d], the order of integration can be reversed without changing the answer:

∫ₐᵇ ∫_c^d f(x,y) dy dx = ∫_c^d ∫ₐᵇ f(x,y) dx dy

This is powerful because one order may be algebraically much simpler than the other. If ∫ f(x,y) dy is hard to compute, try reversing the order and integrating in x first. The theorem guarantees you get the same answer, so algebraic convenience guides the choice.

For non-rectangular regions, the bounds of the inner integral become functions of the outer variable. Integrating over the triangular region where 0 ≤ x ≤ 1 and 0 ≤ y ≤ x gives ∫₀¹ ∫₀ˣ f(x,y) dy dx — y runs from 0 to x, a bound that depends on the current value of x. To reverse the order, you must re-describe the same region from y's perspective: y ranges from 0 to 1, and for each y, x ranges from y to 1. The reversed integral is ∫₀¹ ∫_y¹ f(x,y) dx dy. Always sketch the region first, label the corners and curves, and read the bounds directly from the sketch — this is the single most reliable technique for setting up iterated integrals correctly.