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Multiple Integral — Definition, Formula & Examples

A multiple integral is an integral evaluated over a region in two or more dimensions, such as a double integral over an area or a triple integral over a volume. It computes quantities like area, volume, mass, and probability by summing infinitesimal contributions across a multi-dimensional domain.

Given a function f:Rn→Rf: \mathbb{R}^n \to \mathbb{R} and a measurable region D⊆RnD \subseteq \mathbb{R}^n, the multiple integral \idotsintDf dV\idotsint_D f\,dV is defined as the limit of Riemann sums over partitions of DD. For n=2n = 2 this yields a double integral, and for n=3n = 3 a triple integral. By Fubini's theorem, if ff is continuous on DD, the multiple integral can be computed as iterated single integrals.

Key Formula

∬Df(x,y) dA=∫ab∫g1(x)g2(x)f(x,y) dy dx\iint_D f(x, y)\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y)\,dy\,dx
Where:
  • DD = Region of integration in the $xy$-plane
  • f(x,y)f(x,y) = Integrand, a function of two variables
  • a,ba, b = Limits for the outer integral (over $x$)
  • g1(x),g2(x)g_1(x), g_2(x) = Lower and upper bounds for $y$ as functions of $x$
  • dAdA = Area element, equal to $dy\,dx$ in Cartesian coordinates

How It Works

You evaluate a multiple integral by converting it into iterated integrals — integrating one variable at a time while treating the others as constants. The key step is determining the correct limits of integration for each variable based on the region DD. For double integrals, you integrate over a 2D region; for triple integrals, over a 3D region. When the region has curved boundaries, switching to polar, cylindrical, or spherical coordinates often simplifies the computation — but you must include the appropriate Jacobian factor.

Worked Example

Problem: Evaluate the double integral ∬D(x+y) dA\iint_D (x + y)\,dA where DD is the rectangle 0≤x≤20 \le x \le 2, 0≤y≤30 \le y \le 3.
Set up iterated integral: Write the double integral as iterated single integrals with constant limits since the region is a rectangle.
∫02∫03(x+y) dy dx\int_0^2 \int_0^3 (x + y)\,dy\,dx
Integrate with respect to y: Treat xx as a constant and integrate the inner integral.
∫03(x+y) dy=[xy+y22]03=3x+92\int_0^3 (x + y)\,dy = \left[xy + \frac{y^2}{2}\right]_0^3 = 3x + \frac{9}{2}
Integrate with respect to x: Now integrate the result over xx from 0 to 2.
∫02(3x+92)dx=[3x22+9x2]02=6+9=15\int_0^2 \left(3x + \frac{9}{2}\right)dx = \left[\frac{3x^2}{2} + \frac{9x}{2}\right]_0^2 = 6 + 9 = 15
Answer: The value of the double integral is 1515.

Why It Matters

Multiple integrals are essential in physics and engineering for computing mass, center of mass, moments of inertia, and electric flux over solid regions. They also underpin probability theory, where joint density functions are integrated over multi-dimensional domains to find probabilities.

Common Mistakes

Mistake: Forgetting the Jacobian when changing coordinates (e.g., omitting the rr factor in polar coordinates).
Correction: When switching to polar coordinates, dA=r dr dθdA = r\,dr\,d\theta, not dr dθdr\,d\theta. Similarly, spherical coordinates require ρ2sin⁡ϕ\rho^2 \sin\phi. Always include the Jacobian determinant of the coordinate transformation.