9.1 Definition and properties of double integrals

Double integrals extend the concept of integration to functions of two variables. They calculate the volume under a surface over a rectangular region in the $xy$-plane, giving you a powerful way to analyze three-dimensional quantities.

This section covers the formal definition of double integrals via Riemann sums, the key properties that make them manageable to work with (linearity, additivity), and the comparison theorem for bounding their values.

Definition and Properties

Double Integral and Rectangular Region

A double integral computes the signed volume between a surface $z = f(x,y)$ and a rectangular region $R$ in the $xy$-plane. "Signed" means that where $f(x,y) < 0$, the integral counts volume as negative, just like single-variable integrals count area below the $x$-axis as negative.

The rectangular region is defined as the Cartesian product of two closed intervals:

$R = [a,b] \times [c,d]$

This is just the set of all points $(x,y)$ satisfying $a \leq x \leq b$ and $c \leq y \leq d$. Geometrically, it's a rectangle with sides parallel to the coordinate axes.

The double integral of $f(x,y)$ over $R$ is written as:

$\iint_R f(x,y)\, dA$

where $dA$ represents a small element of area in the $xy$-plane. When $f(x,y) \geq 0$ everywhere on $R$, the integral equals the volume of the solid sitting above $R$ and below the surface $z = f(x,y)$.

Integrable Function and Properties

A function $f(x,y)$ is integrable over $R$ if the double integral $\iint_R f(x,y)\, dA$ exists as a finite number. A sufficient condition for integrability is that $f$ is bounded on $R$ and continuous there except possibly at a finite number of points (or along a finite number of smooth curves).

Linearity property. If $f$ and $g$ are both integrable over $R$ and $c$ is a constant, then:

• $\iint_R c\, f(x,y)\, dA = c \iint_R f(x,y)\, dA$
• $\iint_R [f(x,y) + g(x,y)]\, dA = \iint_R f(x,y)\, dA + \iint_R g(x,y)\, dA$

These work exactly like the linearity rules you already know from single-variable integration. You can factor out constants and split sums.

Additivity property. If $R_1$ and $R_2$ are rectangular regions that don't overlap (except possibly along a shared boundary), and $f$ is integrable over each, then:

$\iint_{R_1 \cup R_2} f(x,y)\, dA = \iint_{R_1} f(x,y)\, dA + \iint_{R_2} f(x,y)\, dA$

This lets you break a region into smaller pieces, integrate over each one separately, and add the results. It's especially handy when $f$ has different formulas on different parts of the domain.

Riemann Sums

Approximating Double Integrals

The double integral is defined as the limit of Riemann sums, which approximate the integral by chopping the region into small rectangles and adding up volumes of thin rectangular prisms. Here's the setup:

1. Partition $[a,b]$ into $m$ subintervals of width $\Delta x = \frac{b-a}{m}$ and $[c,d]$ into $n$ subintervals of width $\Delta y = \frac{d-c}{n}$. This creates $mn$ subrectangles $R_{ij}$, each with area $\Delta A = \Delta x \, \Delta y$.
2. In each subrectangle $R_{ij}$, pick a sample point $(x_{ij}^*, y_{ij}^*)$. Common choices are the midpoint, the lower-left corner, or the upper-right corner.
3. Evaluate $f$ at each sample point and form the sum:

$S_{m,n} = \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*)\, \Delta x\, \Delta y$

Each term $f(x_{ij}^*, y_{ij}^*)\, \Delta x\, \Delta y$ is the volume of a thin rectangular prism (or "column") with base $\Delta x \times \Delta y$ and height $f(x_{ij}^*, y_{ij}^*)$. The full sum approximates the total volume under the surface.

Limit of Riemann Sum

As you refine the partition ($m, n \to \infty$), the subrectangles shrink and the approximation improves. The double integral is defined as this limit:

$\iint_R f(x,y)\, dA = \lim_{m,n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*)\, \Delta x\, \Delta y$

Two things to note:

• This limit exists if and only if $f$ is integrable over $R$.
• The choice of sample points doesn't matter. Whether you pick midpoints, corners, or any other points inside each $R_{ij}$, the limit converges to the same value. This is why the definition is robust: you can't get a "wrong" answer by choosing different sample points, as long as $f$ is integrable.

In practice, you'll rarely compute double integrals directly from Riemann sums. The definition matters because it justifies the evaluation techniques (like iterated integrals) that you'll use going forward.

Theorems

Comparison Theorem for Double Integrals

The comparison theorem gives you a way to bound a double integral even when you can't evaluate it exactly.

If $f(x,y) \leq g(x,y)$ for all $(x,y)$ in $R$, then:

$\iint_R f(x,y)\, dA \leq \iint_R g(x,y)\, dA$

This extends naturally to a squeeze (sandwich) version: if $f(x,y) \leq g(x,y) \leq h(x,y)$ on $R$, then:

$\iint_R f(x,y)\, dA \leq \iint_R g(x,y)\, dA \leq \iint_R h(x,y)\, dA$

Quick example. Suppose you need to bound $\iint_R \sin(xy)\, dA$ over $R = [0,1] \times [0,1]$. Since $0 \leq \sin(xy) \leq 1$ on this region, and the area of $R$ is 1, the comparison theorem gives:

$0 \leq \iint_R \sin(xy)\, dA \leq 1$

More generally, if $m \leq f(x,y) \leq M$ on $R$ and the area of $R$ is $A$, then $mA \leq \iint_R f(x,y)\, dA \leq MA$. This is the double-integral version of the estimation property you used in Calculus I.