Double and Triple Integrals over General Domains

A double integral ∬_D f(x, y) dA is the limit of sums f · ΔA over a planar region D, and a triple integral ∭_E f dV does the same over a solid E; both are computed as iterated single integrals. In brief: for the rectangle [0, 2] × [0, 3], ∬ xy dA = (∫₀² x dx)(∫₀³ y dy) = 2 · 4.5 = 9. For a general region, the inner limits depend on the outer variable, for instance ∫₀¹ ∫₀^x f dy dx over the triangle 0 ≤ y ≤ x ≤ 1.

The single Riemann integral generalizes naturally: instead of cutting an interval, cut the region into small pieces. This guide explains Fubini’s theorem, how to set limits for type I and type II regions, how to swap the order of integration, volumes, mass and averages, and triple integrals over solids, with a table of standard setups.

What is a double integral?

Cut the region D into small pieces of area ΔA_i, choose a point (x_i, y_i) in each, and form the sum Σ f(x_i, y_i) ΔA_i. When the pieces shrink, the limit is the double integral ∬_D f dA. For f ≥ 0 it equals the volume of the solid under the surface z = f(x, y) above D. For f = 1 it is the area of D.

The integral is linear, additive over disjoint regions and monotone: f ≤ g implies ∬ f ≤ ∬ g. Every continuous function on a bounded region with a piecewise smooth boundary is integrable.

How does Fubini’s theorem work on a rectangle?

If f is continuous on the rectangle R = [a, b] × [c, d], then ∬_R f dA = ∫_a^b (∫_c^d f(x, y) dy) dx = ∫_c^d (∫_a^b f(x, y) dx) dy. The order does not matter. If f(x, y) = g(x)h(y), the integral factors into a product of two single integrals.

Example 1: a polynomial

∬ (x + 2y) dA over [0, 1] × [0, 2]. Inner: ∫₀² (x + 2y) dy = 2x + 4. Outer: ∫₀¹ (2x + 4) dx = 1 + 4 = 5.

Example 2: a factorable integrand

∬ x e^y dA over [0, 2] × [0, 1] = (∫₀² x dx)(∫₀¹ e^y dy) = 2(e − 1) ≈ 3.437.

How do you integrate over a general region?

A type I region is bounded by two graphs over an interval: a ≤ x ≤ b, g₁(x) ≤ y ≤ g₂(x). Then ∬_D f dA = ∫_a^b ∫_g1(x)^g₂(x) f dy dx. A type II region is c ≤ y ≤ d, h₁(y) ≤ x ≤ h₂(y), and the integral is ∫_c^d ∫_h1(y)^h₂(y) f dx dy. The outer limits are always constants; the inner limits may depend on the outer variable.

A good method: sketch D, draw a line through it parallel to the inner variable’s axis, and read where it enters and leaves.

Example 3: a triangle

D is the triangle with vertices (0, 0), (1, 0), (1, 1), that is 0 ≤ x ≤ 1, 0 ≤ y ≤ x. ∬ x y dA = ∫₀¹ x (x²/2) dx = ∫₀¹ x³/2 dx = 1/8.

Example 4: between a parabola and a line

D is bounded by y = x2 and y = x, with 0 ≤ x ≤ 1 and x2 ≤ y ≤ x. ∬ (x + y) dA = ∫01 [xy + y2/2]x2x dx = ∫01 (x2 + x2/2 − x3 − x4/2) dx = ∫01 (1.5x2 − x3 − 0.5x4) dx = 0.5 − 0.25 − 0.1 = 0.15.

Example 5: the area of a disc

For x² + y² ≤ R²: ∬ 1 dA = ∫_−R^R 2√(R² − x²) dx = πR². Polar coordinates make this easier.

How do you change the order of integration?

Some inner integrals have no elementary antiderivative. Reversing the order may rescue the computation. Describe the same region in the other way, then integrate.

Example 6: an integral that needs a swap

I = ∫₀¹ ∫_x¹ e^y2 dy dx. The function e^y2 has no elementary antiderivative in y. The region is 0 ≤ x ≤ y ≤ 1, which in the other order is 0 ≤ y ≤ 1, 0 ≤ x ≤ y. Then I = ∫₀¹ e^y2 y dy = (1/2)(e − 1) ≈ 0.859.

Example 7: swapping with a parabola

∫₀⁴ ∫_√y² x dx dy. The region is 0 ≤ y ≤ 4, √y ≤ x ≤ 2, equal to 0 ≤ x ≤ 2, 0 ≤ y ≤ x². Then ∫₀² x · x² dx = 4. Direct check: inner = (4 − y)/2, then ∫₀⁴ (4 − y)/2 dy = (16 − 8)/2 = 4 ✓.

What do double integrals compute?

The volume under z = f(x, y) is ∬ f dA. The mass of a lamina with surface density σ(x, y) is m = ∬ σ dA. The center of mass is (x̄, ȳ) = (1/m)(∬ xσ dA, ∬ yσ dA). The average value of f over D is (1/A(D)) ∬ f dA. The moment of inertia about the origin is ∬ (x² + y²) σ dA.

Example 8: a volume

The volume under z = 4 − x − y over the square [0, 1] × [0, 1]: ∫₀¹ (4 − x − 1/2) dx = 3.5 − 0.5 = 3.

Example 9: mass of a lamina

The triangle 0 ≤ y ≤ x ≤ 2 with density σ = x. m = ∫₀² x · x dx = 8/3. x̄ = (3/8) ∫₀² x³ dx = (3/8)(4) = 3/2.

How do triple integrals work?

For a solid E described by (x, y) in D and z₁(x, y) ≤ z ≤ z₂(x, y), ∭_E f dV = ∬_D (∫_z1^z₂ f dz) dA. For f = 1 the result is the volume of E. A box [a, b] × [c, d] × [e, g] gives three nested integrals with constant limits.

Example 10: a box

∭ xyz dV over [0, 1] × [0, 2] × [0, 3] = (1/2)(2)(9/2) = 4.5.

Example 11: the volume of a tetrahedron

E is bounded by the planes x = 0, y = 0, z = 0 and x + y + z = 1. Then 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x, 0 ≤ z ≤ 1 − x − y. V = ∫₀¹ ∫₀^1−x (1 − x − y) dy dx = ∫₀¹ (1 − x)²/2 dx = 1/6.

Example 12: a solid under a paraboloid

E lies between z = 0 and z = x² + y² over the square [0, 1] × [0, 1]. V = ∫₀¹ ∫₀¹ (x² + y²) dx dy = 1/3 + 1/3 = 2/3.

Example 13: the volume of a sphere (the idea)

For x² + y² + z² ≤ R², the cross-section at height z is a disc of area π(R² − z²), so V = ∫_−R^R π(R² − z²) dz = 4πR³/3. The same volume results from a triple integral, which is simplest in spherical coordinates.

Table of standard setups

Where does the theory come from?

Iterated integration appears in the work of Euler and Lagrange in the eighteenth century, who computed areas and volumes by repeated integration. The theorem on exchanging the order of integration for integrable functions is named after Guido Fubini, who proved a general version for the Lebesgue integral in 1907; a related result is due to Leonida Tonelli (1909). Earlier forms for continuous functions were known well before.

Which mistakes are common?

Variable limits on the outer integral

The outer limits must be numbers. Only the inner limits may depend on the other variable.

Swapping order without redrawing the region

Simply exchanging the limits is wrong unless the region is a rectangle. Sketch D and describe it again.

Forgetting the Jacobian

Mixing up area and volume

With f = 1 a double integral gives area and a triple integral gives volume; with f = height, a double integral gives volume.