[CALCULUS] Double integrals and Riemann sums

Define double integrals over rectangles using tagged partitions and upper and lower sums.

This page defines the double integral over a rectangle as a limit of Riemann sums. The construction copies the one-variable integral: small intervals become small rectangles and their lengths become areas.

The definition says what the integral means; later pages compute it. Fubini's theorem replaces the sums by iterated integrals, regions bounded by curves extend the construction beyond rectangles and a change of variables reshapes the iterated integrals.

From one variable to two

For a one-variable partition

a=x_0<x_1<\cdots<x_m=b,

the $i$ th subinterval has length $\Delta x_i=x_i-x_{i-1}$ . Choosing a tag $x_i^*$ in each interval produces

\sum_{i=1}^m f(x_i^*)\,\Delta x_i.

The integral is the limit of these sums as the partition becomes fine, provided the limit does not depend on the tags. Over a rectangle

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

we partition both coordinates:

a=x_0<\cdots<x_m=b, \qquad c=y_0<\cdots<y_n=d.

The resulting small rectangles are

R_{ij}=[x_{i-1},x_i]\times[y_{j-1},y_j], \qquad \Delta A_{ij}=\Delta x_i\,\Delta y_j.

The mesh of this product partition is the largest side length among the small rectangles:

\lVert P\rVert=\max\left\{\max_i\Delta x_i,\ \max_j\Delta y_j\right\}.

“The partition becomes fine” means that $\lVert P\rVert\to0$ .

[DEFINITION]

Double Riemann sum

For a partition of $R$ into rectangles $R_{ij}$ and tag points $(x_{ij}^*,y_{ij}^*)\in R_{ij}$ , the corresponding double Riemann sum is

\sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*,y_{ij}^*)\,\Delta A_{ij}.

When $f\ge 0$ , each term approximates the volume of a box above $R_{ij}$ . When $f$ takes both signs, the sum approximates a signed volume: contributions below the $xy$ -plane are subtracted.

[DEFINITION]

Double integral over a rectangle

Let $f$ be bounded on $R=[a,b]\times[c,d]$ . If there is a number $I$ such that, for every $\varepsilon>0$ , some $\delta>0$ makes every tagged partition with $\lVert P\rVert<\delta$ satisfy

\left| \sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*,y_{ij}^*)\,\Delta A_{ij}-I \right|<\varepsilon,

then $f$ is Riemann integrable over $R$ and

\iint_R f(x,y)\,dA = \lim \sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*,y_{ij}^*)\,\Delta A_{ij}.

The notation

dA=dx\,dy=dy\,dx

records that area is being accumulated. It does not say which variable is integrated first.

Upper and lower sums

Upper and lower sums remove the arbitrary choice of tags. For each small rectangle $R_{ij}$ , set

m_{ij}=\inf_{(x,y)\in R_{ij}}f(x,y), \qquad M_{ij}=\sup_{(x,y)\in R_{ij}}f(x,y).

Then

m_{ij}\le f(x,y)\le M_{ij} \qquad ((x,y)\in R_{ij}),

so multiplying by $\Delta A_{ij}$ and adding gives lower and upper bounds.

[DEFINITION]

Upper and lower double Riemann sums

For a partition $P$ of $R$ into rectangles $R_{ij}$ , the lower and upper double Riemann sums are

L(f,P)=\sum_{i,j}m_{ij}\,\Delta A_{ij}, \qquad U(f,P)=\sum_{i,j}M_{ij}\,\Delta A_{ij},

where $m_{ij}$ and $M_{ij}$ are the infimum and supremum of $f$ on $R_{ij}$ .

For a nonnegative $f$ , $L(f,P)$ is the volume of the tallest boxes on the partition that fit below the graph and $U(f,P)$ is the volume of the shortest boxes that reach above it. For a signed $f$ , the box heights are signed. Whenever the integral exists,

L(f,P)\le \iint_R f\,dA\le U(f,P).

[THEOREM]

Criterion for Riemann integrability

A bounded function $f$ is Riemann integrable over $R$ if and only if, for every $\varepsilon>0$ , there is a partition $P_\varepsilon$ such that

0\le U(f,P_\varepsilon)-L(f,P_\varepsilon)<\varepsilon.

In that case,

\iint_R f\,dA =\sup_P L(f,P) =\inf_P U(f,P).

This is equivalent to requiring that the tagged sums have a common limit. It is usually the easier criterion for proving integrability or non-integrability.

[THEOREM]

Continuous functions are integrable

Every continuous function on a rectangle is Riemann integrable over that rectangle.

On sufficiently small rectangles, continuity prevents large variation in $f$ , so the upper and lower sums can be made arbitrarily close.

Continuity is sufficient, not necessary: a discontinuous function may still be integrable.

[1]:Double integral over a rectangle

[EXAMPLE]

A midpoint approximation

Estimate the signed volume under

f(x,y)=x+y^2

over

R=[-2,2]\times[0,4]

by dividing $R$ into $4\times2$ rectangles and using their midpoints.

Because $f$ is negative on part of this rectangle, the result estimates a signed volume, not the geometric volume between the graph and the $xy$ -plane. For instance, $f(-\tfrac32,1)=-\tfrac12$ , so that box contributes negatively.

The subinterval widths are $\Delta x=1$ and $\Delta y=2$ , so $\Delta A=2$ . The midpoint $x$ -values are

-\frac32,\ -\frac12,\ \frac12,\ \frac32,

and the midpoint $y$ -values are $1$ and $3$ . For $y=1$ , the four function values sum to $4$ ; for $y=3$ , they sum to $36$ . Therefore

\iint_R f\,dA \approx 2(4+36)=80.

[routine]

[1]:Double Riemann sum

[EXAMPLE]

Upper and lower sums on eight squares

Partition

R=[0,1]\times[0,2]

into eight squares of side length $\tfrac12$ and let

f(x,y)=x^2y.

The function is increasing in both variables on $R$ , so on each small square the minimum is at the lower-left corner and the maximum at the upper-right corner. Every square has area $\tfrac14$ , so