Double Integral — Definition, Formula & Examples

A double integral computes the accumulated value of a function

f(x, y)

over a two-dimensional region. It generalizes the single-variable definite integral to functions of two variables, and geometrically it gives the volume under a surface above the

xy

-plane.

Given a function $f(x, y)$ continuous on a bounded, closed region $R \subset \mathbb{R}^2$ , the double integral $\iint_R f(x,y)\,dA$ is defined as the limit of Riemann sums $\lim_{\|P\|\to 0} \sum_{k=1}^{n} f(x_k^*, y_k^*)\,\Delta A_k$ , where $P$ is a partition of $R$ into $n$ sub-rectangles of area $\Delta A_k$ and $(x_k^*, y_k^*)$ is a sample point in each sub-rectangle.

Key Formula

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

Where:

$R$ = The region of integration in the $xy$-plane
$f(x,y)$ = The integrand, a function of two variables
$a, b$ = Limits of integration for the outer variable $x$
$g_1(x), g_2(x)$ = Lower and upper bounds for $y$ as functions of $x$
$dA$ = Area element, equal to $dy\,dx$ (or $dx\,dy$)

How It Works

You evaluate a double integral by converting it into two nested single integrals, called an iterated integral. First, choose an order of integration — either

dx\,dy

dy\,dx

— and determine the limits for each variable from the region

R

. Then integrate the inner integral while treating the other variable as a constant, and finally integrate the result with respect to the outer variable. For non-rectangular regions, the inner limits are typically functions of the outer variable. Fubini's theorem guarantees that both orders give the same answer when

f

is continuous on

R

Worked Example

Problem: Evaluate

\iint_R (2x + 3y)\,dA

where

R

is the rectangle

0 \le x \le 2

0 \le y \le 1

Set up iterated integral: Write the double integral as an iterated integral with the given constant limits.

\int_0^2 \int_0^1 (2x + 3y)\,dy\,dx

Evaluate the inner integral: Integrate with respect to

y

, treating

x

as a constant.

\int_0^1 (2x + 3y)\,dy = \left[2xy + \tfrac{3}{2}y^2\right]_0^1 = 2x + \tfrac{3}{2}

Evaluate the outer integral: Integrate the result with respect to

x

\int_0^2 \left(2x + \tfrac{3}{2}\right)dx = \left[x^2 + \tfrac{3}{2}x\right]_0^2 = 4 + 3 = 7

Answer: The double integral equals

7

Why It Matters

Double integrals appear throughout multivariable calculus, physics, and engineering. They are used to compute areas of curved surfaces, centers of mass of laminae, and probabilities over two-dimensional distributions. Mastering them is also essential for understanding Green's theorem and surface integrals.

Common Mistakes

Mistake: Using constant limits for both variables when the region is not a rectangle.

Correction: For non-rectangular regions, the inner limits must be expressed as functions of the outer variable. Sketch the region first to identify the correct bounds.

Related Terms

Partial Derivative — Differentiation counterpart used inside integrands
Jacobian — Scale factor when changing variables in double integrals
Green's Theorem — Converts a double integral to a line integral
Gradient — Vector of partial derivatives of the integrand
Divergence Theorem — Higher-dimensional generalization linking flux and volume integrals
Laplacian — Second-order operator often appearing in double-integral problems