Multiple Integrals — Mathematics Knowledge System

Overview

A double integral ∬_D f(x,y) dA extends the single-variable integral to compute volume under a surface z=f(x,y) over a 2D region D, by slicing the solid into infinitesimal boxes instead of rectangles. Triple integrals ∭_E f(x,y,z) dV extend this further to sum a quantity over a 3D solid region. Multiple integrals are evaluated in practice using Fubini's theorem, which reduces them to iterated single-variable integrals, and their difficulty often hinges on choosing the right coordinate system (Cartesian, polar, cylindrical, spherical).

Intuition

For a single integral, you slice a region into thin vertical strips and sum rectangle areas. For a double integral over a region in the plane, you slice into thin 2D boxes (dx by dy) and sum up f(x,y) times each box's tiny area, giving a volume under the surface z=f(x,y). Triple integrals push this one dimension further: slice 3D space into tiny boxes (dx·dy·dz) and sum f(x,y,z) over each box — used for mass, charge, or probability distributed through a volume, not just area.

Formal Definition

Definition

The double integral over a region D is defined as a limit of Riemann sums over a partition of D into small rectangles:

\iint_D f(x,y)\, dA = \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*, y_i^*) \Delta A_i

Δ Aᵢ is the area of the i-th small rectangle in a fine partition of D

\iint_D f(x,y)\, dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx

Valid for continuous f on a region between two curves y=g₁(x) and y=g₂(x)

Fubini's theorem (iterated integral)

\iiint_E f(x,y,z)\, dV = \int\!\!\int\!\!\int f(x,y,z)\,dz\,dy\,dx

Triple integral as an iterated integral

\iint_D f(x,y)\,dA = \iint_{D'} f(r\cos\theta, r\sin\theta)\, r\, dr\, d\theta

The extra factor r is the polar Jacobian |∂(x,y)/∂(r,θ)|

Change of variables to polar coordinates

Notation

Notation	Meaning
$\iint_D f(x,y)\,dA$	Double integral of f over region D (dA = dx dy)
$\iiint_E f(x,y,z)\,dV$	Triple integral of f over solid region E (dV = dx dy dz)
$\left\|\frac{\partial(x,y)}{\partial(u,v)}\right\|$	The determinant scaling factor for changing variables in a multiple integral

Derivation

The polar-coordinate Jacobian factor r is derived by computing the determinant of the transformation's partial-derivative matrix, with x=r cosθ, y=r sinθ:

\frac{\partial(x,y)}{\partial(r,\theta)} = \det\begin{pmatrix} \partial x/\partial r & \partial x/\partial \theta \\ \partial y/\partial r & \partial y/\partial \theta \end{pmatrix} = \det\begin{pmatrix}\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta\end{pmatrix}

Set up the Jacobian matrix of partial derivatives

= r\cos^2\theta + r\sin^2\theta = r

Expand the 2×2 determinant and simplify using sin²θ+cos²θ=1

dA = dx\,dy = |r|\,dr\,d\theta = r\,dr\,d\theta \ (r\geq 0)

The area element picks up the absolute value of the Jacobian as a scaling factor

Properties

Linearity

\iint_D (af+bg)\, dA = a\iint_D f\,dA + b\iint_D g\,dA

Additivity over regions

\iint_{D_1\cup D_2} f\, dA = \iint_{D_1} f\,dA + \iint_{D_2} f\,dA

Condition: D₁, D₂ overlap only on a boundary of measure zero

Volume interpretation

\iint_D 1\, dA = \text{Area}(D), \qquad \iiint_E 1\, dV = \text{Volume}(E)

Order of integration is interchangeable

\text{(Fubini)}\ \int_a^b\int_c^d f\,dy\,dx = \int_c^d\int_a^b f\,dx\,dy

Condition: for continuous f, or more generally when ∬|f|dA is finite

Theorems

Theorem 1: Fubini's Theorem

\text{If } f \text{ is continuous on the rectangle } [a,b]\times[c,d], \text{ then } \iint_D f\,dA = \int_a^b\int_c^d f(x,y)\,dy\,dx = \int_c^d\int_a^b f(x,y)\,dx\,dy

Theorem 2: Change of Variables Theorem

\iint_D f(x,y)\,dA = \iint_{D'} f(x(u,v),y(u,v)) \left|\frac{\partial(x,y)}{\partial(u,v)}\right| du\,dv

Applications

Mass, center of mass, and moment of inertia of a 2D lamina or 3D solid with variable density ρ(x,y,z) are computed as ∬ρ dA or ∭ρ dV.

3D Visualization

Animation

Animates a solid region being sliced into thinner and thinner 3D boxes (Riemann sum boxes) that converge to the exact volume as the partition is refined — visually connecting the discrete Riemann sum to the double/triple integral.

Worked Examples

Set up as an iterated integral over the rectangle.
$\int_0^1\int_0^2 (x+y)\, dy\, dx$
Integrate with respect to y first, treating x as constant.
$\int_0^2 (x+y)\,dy = \left[xy+\frac{y^2}{2}\right]_0^2 = 2x+2$
Integrate the result with respect to x.
$\int_0^1 (2x+2)\,dx = \left[x^2+2x\right]_0^1 = 3$

Answer: 3

Practice Problems

Difficulty 6/10

Evaluate ∬_D xy dA where D = [0,2]×[0,3].

Difficulty 7/10

Set up (do not evaluate) the triple integral for the volume of the region bounded by z=0, z=4−x²−y² (paraboloid), using cylindrical coordinates.

Difficulty 7/10

A lamina occupies the unit square [0,1]×[0,1] with density ρ(x,y)=x+y. Find its total mass.

Common Mistakes

Common Mistake

Forgetting the extra factor of r when converting a double integral to polar coordinates.

dA = dx dy becomes r dr dθ in polar coordinates, NOT just dr dθ — the r comes from the Jacobian of the transformation and is easy to drop by accident.

Common Mistake

Setting up the bounds of an iterated integral in the wrong order, treating inner/outer limits as always constants.

For a region between curves, the INNER integral's bounds are typically functions of the outer variable (e.g. y from g₁(x) to g₂(x)), not constants — sketching the region first is essential to get the bounds right.

Common Mistake

Assuming Fubini's theorem lets you swap integration order for any function, without checking integrability.

Fubini's theorem requires f to be continuous (or ∬|f|dA<∞); for badly-behaved functions, swapping the order can give different (wrong) answers — a classic counterexample exists for f(x,y)=(x²−y²)/(x²+y²)² on [0,1]×[0,1].

Quiz

In converting ∬_D f(x,y) dA to polar coordinates, dA becomes:

dr dθr dr dθr² dr dθdr² dθ

What does ∭_E 1 dV compute?

The surface area of EThe volume of the solid region EThe mass of E assuming unit densityBoth B and C are correct

Flashcards

1 / 4

Historical Background

Multiple integration developed alongside the calculus of surfaces and volumes in the 18th century. Euler and Lagrange computed double and triple integrals for physical problems (center of mass, moment of inertia) without a fully general theory. Guido Fubini proved the modern theorem bearing his name in 1907, rigorously establishing the conditions under which a multiple integral can be evaluated by iterated single integration in any order — a result that had been used informally for over a century before being proven.

1770s
Euler and Lagrange compute double/triple integrals for centers of mass and gravitational attraction
Leonhard Euler, Joseph-Louis Lagrange
1907
Guido Fubini proves the general theorem on iterated integration order
Guido Fubini

Summary

Double integrals ∬_D f dA compute volume under a surface (or mass, if f is a density); triple integrals ∭_E f dV extend this to 3D solids.
Fubini's theorem reduces multiple integrals to iterated single-variable integrals, evaluated one variable at a time.
Changing to polar/cylindrical/spherical coordinates requires multiplying by the appropriate Jacobian factor (r for polar/cylindrical, r²sinφ for spherical).
Correctly sketching the region of integration is essential to set up correct bounds, since inner limits are often functions of the outer variable.
Applications include mass, center of mass, moment of inertia, and joint probability computations.

References

BookStewart, J. Calculus: Early Transcendentals, 8th ed. Ch. 15.
PaperFubini, G. (1907). Sugli integrali multipli.
WebsiteWikipedia — Multiple integral