Double Integrals

Fix x and the solid under z = f(x, y) has a cross-section whose area A(x) is an ordinary integral in y. Integrate A(x) along x and you have the volume. Fix y instead and the region has to be read the other way round, with x between two curves; the two orders agree in every case on this page but one.

Slices at fixed x	$\iint_{D} f d A = \int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} f (x, y) d y d x$
The other order (Fubini), when ∬\|f\| dA is finite	$\int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} f d y d x = \int_{c}^{d} \int_{h_{1} (y)}^{h_{2} (y)} f d x d y \iint_{D} ∣ f ∣ d A < \infty$

Worked example — Double Integrals

Area of the slice at this x	0.2188
Volume, slices at fixed x	0.15
Volume in the other order, slices at fixed y	0.15

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Lesson

The theory — Double Integrals

The double integral ∬_D f(x, y) dA is the volume between a region D of the xy-plane and the surface z = f(x, y), with any part below the plane counted negative. It is computed as an iterated integral: integrate in one variable with the other held fixed, then integrate what comes out.

What each symbol means

A(x): the area of the slice at a fixed x, itself an integral in y
g₁(x), g₂(x): the lower and upper edge of D when D is read along x
h₁(y), h₂(y): the left and right edge of D when D is read along y

Where the formula comes from

Hold x fixed. The plane at that x cuts the solid in a slice whose top edge is the curve y ↦ f(x, y), from the region's lower edge to its upper edge. Its area is an integral in y with x treated as a constant, the same move as taking a partial derivative.
$A (x) = \int_{g_{1} (x)}^{g_{2} (x)} f (x, y) d y$
Add up the slices. The volume is the integral of A(x) along x: volume by slicing, with a slice that needed an integral of its own.
$V = \int_{a}^{b} A (x) d x = \int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} f (x, y) d y d x$
Or slice the other way. Read D along y, with x between a left and a right edge, and integrate in x first. Fubini's theorem says the two orders give the same number whenever ∬|f| dA is finite, which holds for every continuous f on a closed, bounded region.
$\int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} f d y d x = \int_{c}^{d} \int_{h_{1} (y)}^{h_{2} (y)} f d x d y$

Assumes: f is continuous on D, or at least ∬|f| dA is finite. Where an edge of D changes formula, as the triangle's top edge does at x = 1, the integral is split there.

Problems solved in full

Over a rectangle, both orders 3 steps
Evaluate ∬ xy² dA over the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1, in both orders.

Open this problem in the tool
1. At fixed x, integrate in y with x held constant:
  
  $A (x) = \int_{0}^{1} x y^{2} d y = \frac{x}{3}$
2. Then along x:
  
  $\int_{0}^{2} \frac{x}{3} d x = \frac{4}{6} = \frac{2}{3}$
3. The other order, slices at fixed y:
  
  $\int_{0}^{1} \int_{0}^{2} x y^{2} d x d y = \int_{0}^{1} 2 y^{2} d y = \frac{2}{3}$
Answer

2/3 ≈ 0.6667 either way. Over a rectangle the limits are numbers in both orders, and a product splits: (∫₀² x dx)(∫₀¹ y² dy) = 2 · 1/3.

$\iint_{R} x y^{2} d A = \frac{2}{3} \approx 0.6667$
A region between two curves 2 steps
Evaluate ∬ (x + y) dA over the region between y = x² and y = x.

Open this problem in the tool
Show solution
1. The curves meet at x = 0 and x = 1, and between them x² ≤ x, so at fixed x the slice runs from y = x² up to y = x:
  
  $A (x) = \int_{x^{2}}^{x} (x + y) d y = [x y + \frac{y ^{2}}{2}]_{x^{2}}^{x} = \frac{3 x ^{2}}{2} - x^{3} - \frac{x ^{4}}{2}$
2. Integrate from 0 to 1:
  
  $\int_{0}^{1} (\frac{3 x ^{2}}{2} - x^{3} - \frac{x ^{4}}{2}) d x = \frac{1}{2} - \frac{1}{4} - \frac{1}{10} = \frac{3}{20}$
Answer

3/20 = 0.15.

$\iint_{D} (x + y) d A = \frac{3}{20} = 0.15$
Reversing the order 3 steps
Write the integral of problem 2 with the order of integration reversed, and evaluate it.

Open this problem in the tool
Show solution
1. Read the region along y: it runs from y = 0 to y = 1, and at height y its left edge is the line, x = y, and its right edge is the parabola, x = √y:
  
  $\int_{0}^{1} \int_{x^{2}}^{x} (x + y) d y d x = \int_{0}^{1} \int_{y}^{y} (x + y) d x d y$
2. The inner integral, y held fixed:
  
  $\int_{y}^{y} (x + y) d x = [\frac{x ^{2}}{2} + x y]_{y}^{y} = \frac{y}{2} + y^{3/2} - \frac{3 y ^{2}}{2}$
3. Then along y:
  
  $\int_{0}^{1} (\frac{y}{2} + y^{3/2} - \frac{3 y ^{2}}{2}) d y = \frac{1}{4} + \frac{2}{5} - \frac{1}{2} = \frac{3}{20}$
Answer

3/20 again. The parabola that was the lower edge read along x is the right edge read along y, written x = √y.

$\int_{0}^{1} \int_{y}^{y} (x + y) d x d y = \frac{3}{20}$
Choosing the order that needs one integral 3 steps
Evaluate ∬ xy dA over the triangle with corners (0, 0), (2, 0) and (1, 1).

Open this problem in the tool
Show solution
1. Read along x, the top edge is y = x up to x = 1 and y = 2 − x after it, which means two integrals. Read along y, the triangle runs from the left edge x = y to the right edge x = 2 − y for every y, so slice at fixed y:
  
  $\iint_{T} x y d A = \int_{0}^{1} \int_{y}^{2 - y} x y d x d y$
2. The inner integral:
  
  $\int_{y}^{2 - y} x y d x = \frac{y}{2} ((2 - y)^{2} - y^{2}) = 2 y (1 - y)$
3. Then along y:
  
  $\int_{0}^{1} 2 y (1 - y) d y = 1 - \frac{2}{3} = \frac{1}{3}$
Answer

1/3 ≈ 0.3333. The other order gives 1/8 + 5/24 from its two pieces, the same 1/3.

$\iint_{T} x y d A = \frac{1}{3} \approx 0.3333$
Reversing the order to get a formula 3 steps
Evaluate ∫₀¹ ∫ₓ¹ e^(y²) dy dx.

Open this problem in the tool
Show solution
1. The inner integral has no elementary antiderivative, so reverse the order. The region is 0 ≤ x ≤ y ≤ 1; read along y it is 0 ≤ y ≤ 1 with 0 ≤ x ≤ y:
  
  $\int_{0}^{1} \int_{x}^{1} e^{y^{2}} d y d x = \int_{0}^{1} \int_{0}^{y} e^{y^{2}} d x d y$
2. The integrand does not depend on x:
  
  $\int_{0}^{y} e^{y^{2}} d x = y e^{y^{2}}$
3. Substitute u = y²:
  
  $\int_{0}^{1} y e^{y^{2}} d y = [\frac{e ^{y^{2}}}{2}]_{0}^{1} = \frac{e - 1}{2}$
Answer

(e − 1)/2 ≈ 0.8591.

$\frac{e - 1}{2} \approx 0.8591$
A solid described by its faces 3 steps
Find the volume of the tetrahedron bounded by the three coordinate planes and the plane x + y + z = 1.
Show solution
1. The solid lies under z = 1 − x − y and over the part of the xy-plane where that height is not negative: x ≥ 0, y ≥ 0, x + y ≤ 1. Read that triangle along x:
  
  $V = \int_{0}^{1} \int_{0}^{1 - x} (1 - x - y) d y d x$
2. The inner integral, x held fixed:
  
  $\int_{0}^{1 - x} (1 - x - y) d y = \frac{( 1 - x ) ^{2}}{2}$
3. Then along x:
  
  $\int_{0}^{1} \frac{( 1 - x ) ^{2}}{2} d x = \frac{1}{6}$
Answer

1/6, which agrees with a third of the base times the height: (1/3)(1/2)(1).

$V = \frac{1}{6}$
The average height over a region 3 steps
Find the average value of f(x, y) = x + y over the region of problem 2.
Show solution
1. The average is the volume divided by the area |D| of the region, the height of a flat slab over D with the same volume:
  
  $\overset{ˉ}{f} = \frac{1}{∣ D ∣} \iint_{D} f d A$
2. The area is the double integral of 1:
  
  $∣ D ∣ = \int_{0}^{1} \int_{x^{2}}^{x} d y d x = \int_{0}^{1} (x - x^{2}) d x = \frac{1}{6}$
3. With the volume 3/20 from problem 2:
  
  $\overset{ˉ}{f} = \frac{3/20}{1/6} = \frac{9}{10}$
Answer

9/10 = 0.9.

$\overset{ˉ}{f} = 0.9$
When the two orders disagree 4 steps
Show that the two iterated integrals of f(x, y) = (x² − y²)/(x² + y²)² over the unit square are different, and say which hypothesis of Fubini's theorem fails.

Open this problem in the tool
Show solution
1. The derivative of y/(x² + y²) with respect to y is f, so at fixed x > 0:
  
  $\int_{0}^{1} \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} d y = [\frac{y}{x ^{2} + y ^{2}}]_{0}^{1} = \frac{1}{1 + x ^{2}}$
2. In this order, then:
  
  $\int_{0}^{1} \int_{0}^{1} f d y d x = \int_{0}^{1} \frac{d x}{1 + x ^{2}} = arctan 1 = \frac{π}{4}$
3. Swapping x and y changes the sign of f, so the other order is the same calculation with the sign reversed:
  
  $\int_{0}^{1} \int_{0}^{1} f d x d y = - \frac{π}{4}$
4. In polar coordinates f = cos 2θ/r². Over the quarter disc of radius 1, which lies inside the square,
  
  $\int_{0}^{π /2} \int_{0}^{1} \frac{∣ cos 2 θ ∣}{r ^{2}} r d r d θ = \int_{0}^{π /2} ∣ cos 2 θ ∣ d θ \int_{0}^{1} \frac{d r}{r} = \infty$
Answer

π/4 ≈ 0.7854 in one order and −π/4 in the other. ∬|f| dA is infinite, and finiteness is exactly what Fubini's theorem assumes.

$\int_{0}^{1} \int_{0}^{1} f d y d x \int_{0}^{1} \int_{0}^{1} f d x d y = \frac{π}{4} \approx 0.7854 = - \frac{π}{4}$

Swap the order and an integral with no formula gets one 🖖

Press A slice with no formula. At fixed x the slice is the integral of e^(y²) from y = x to 1, and e^(y²) has no antiderivative built from elementary functions, the same wall e^(−t²) runs into in the integration chapter. The page can only compute each slice numerically. Now press The same, the other way. At fixed y the slice runs from x = 0 to x = y, and the height e^(y²) does not depend on x, so the slice is a rectangle of area y·e^(y²). That is the derivative of e^(y²)/2, and the volume is (e − 1)/2 exactly. Nothing about the solid changed. The order in which you integrate decided whether the slices have a formula at all.

One function where the order changes the answer 🖖

Press Order changes the answer. The surface (x² − y²)/(x² + y²)² is positive below the diagonal of the unit square and negative above it. Slice at fixed x and every slice nets out positive, 1/(1 + x²), so the iterated integral is π/4 ≈ 0.7854. Switch to dx dy and every slice nets out negative, −1/(1 + y²), and it comes to −0.7854. Neither calculation has a mistake in it. Near the origin the surface behaves like cos 2θ/r², so the volume above the square and the volume below it are each infinite, and the two orders cancel one against the other differently. There is no double integral for them to agree on. Fubini's theorem, published in 1907, promises that the orders agree when ∬|f| dA is finite. Here it is infinite, and the theorem promises nothing.

The solid under an infinite spike has volume π²/6 🖖

Press A volume of π²/6. The surface 1/(1 − xy) rises without bound at the corner (1, 1), and the solid under it over the unit square still has volume 1.6449. Expand the integrand as a geometric series, 1 + xy + x²y² + …, and integrate term by term: xⁿyⁿ over the square gives 1/(n + 1)², so the volume is 1 + 1/4 + 1/9 + …, the sum Euler showed in 1735 to be π²/6. Read backwards, the double integral is a way of summing the series. Frits Beukers used integrals over the same square with 1 − xy in the denominator for his short proof, in 1979, that 1 + 1/8 + 1/27 + … is irrational.

Learning path

Applications of integration

References (4)

The method and the two descriptions of a region the page uses: Type I (“between two vertical lines and the graphs of two continuous functions”) and Type II (between two horizontal lines), and Theorem 5.4, Fubini's theorem (strong form), for a function continuous on such a region: G. Strang and E. Herman, Calculus Volume 3, OpenStax (2016), section 5.2, “Double Integrals over General Regions”.
Insight block 2: Fubini's theorem on successive integration, for f in L¹, that is, with ∬|f| finite; published 1907: G. Fubini, “Sugli integrali multipli”, Atti della Reale Accademia dei Lincei, Rendiconti, Classe di Scienze Fisiche, Matematiche e Naturali (5) 16 (1907), no. 1, 608–614.
Insight block 3: Euler showed in 1735 that 1 + 1/4 + 1/9 + … = π²/6: J. J. O'Connor and E. F. Robertson, “Leonhard Euler”, MacTutor History of Mathematics Archive, University of St Andrews (last updated September 1998).
Insight block 3: integrals over the unit square with 1 − xy in the denominator, used to prove ζ(2) and ζ(3) irrational: F. Beukers, “A Note on the Irrationality of ζ(2) and ζ(3)”, Bulletin of the London Mathematical Society 11 (1979), no. 3, 268–272.

By the 42Math team · Reviewed against published sources. Corrections welcome.