Tutorials › Calculus III › The Double Integral as a Limit of Riemann Sums

Home  /  Tutorials  /  Double Integrals on Rectangles

Double Integrals on Rectangles · Tutorial 504 of 999

The Double Integral as a Limit of Riemann Sums

Learn to interpret the double integral as the common limit of rectangular Riemann sums, independent of the partitions and sample points used.

Beginner 10 min read

What You'll Learn

  • Define the mesh of a rectangular partition using the largest subrectangle diameter.
  • State precisely when tagged Riemann sums converge to a double integral.
  • Interpret the limit notation for the double integral over a rectangle.
  • Verify convergence with error estimates for functions on rectangles.
  • Recognize why a single Riemann sum does not determine the limit.

From Rectangular Sums to a Limit

A Riemann sum for a function of two variables combines function values sampled from small rectangles with the areas of those rectangles. A single sum is an approximation: it depends on the chosen partition and on the sample point selected in each piece.

The double integral is defined by asking whether these sums approach one common number as the pieces become arbitrarily small. Both the shrinking size of the pieces and the independence from particular sample choices are essential to the definition.

Partitions, Tags, and Mesh

Let \(R=[a,b]\times[c,d]\) be a rectangle, where \(a<b\) and \(c<d\). Divide \([a,b]\) into intervals using \(a=x_0<x_1<\cdots<x_m=b\), and divide \([c,d]\) using \(c=y_0<y_1<\cdots<y_n=d\). These divisions produce smaller rectangles \(R_{ij}=[x_{i-1},x_i]\times[y_{j-1},y_j]\).

The side lengths of \(R_{ij}\) are \(\Delta x_i=x_i-x_{i-1}\) and \(\Delta y_j=y_j-y_{j-1}\). Its area is \(\Delta A_{ij}=\Delta x_i\Delta y_j\), and its diameter, the greatest distance between two points in it, is \(\sqrt{(\Delta x_i)^2+(\Delta y_j)^2}\). The mesh of the partition is the largest of these diameters. A tag \((x_{ij}^*,y_{ij}^*)\) is any point in \(R_{ij}\).

Definition: For a bounded function \(f:R\to\mathbb{R}\), a tagged Riemann sum for a partition \(P\) is $$ S(P,f)=\sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*,y_{ij}^*)\,\Delta A_{ij}. $$ The function is Riemann integrable on \(R\) if there is a number \(I\) such that, for every \(\varepsilon>0\), there is a \(\delta>0\) for which every partition with mesh less than \(\delta\), and every choice of tags in its subrectangles, satisfies \(|S(P,f)-I|<\varepsilon\). That number \(I\) is the double integral of \(f\) over \(R\).

In this definition, “every choice of tags” matters. It is not enough for sums based on one convenient selection of sample points to approach a number. The same number must be approached even when the tags vary from one partition to another.

Reading the Limit Notation

When \(f\) is Riemann integrable on \(R\), the definition is written as

$$ \iint_R f(x,y)\,dA = \lim_{\lVert P\rVert\to 0} \sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*,y_{ij}^*)\,\Delta A_{ij}. $$

Here \(\lVert P\rVert\) denotes the mesh of the partition. The notation \(\lVert P\rVert\to0\) means that the largest subrectangle diameter tends to zero. The numbers of intervals \(m\) and \(n\) may change as the partition changes; the rectangles need not all have the same size or shape.

The limit is not necessarily a limit as one integer \(n\) increases. It says that tagged sums approach \(I\) for all sufficiently fine partitions, with any tags. Equivalently, along any sequence of tagged partitions whose meshes tend to zero, the sums approach the same \(I\). A sequence of sums from just one chosen family of partitions can illustrate convergence, but by itself does not establish the full definition.

Worked Example: A Constant Function

Worked Example: A Constant Function

Find the limit of the tagged Riemann sums for \(f(x,y)=3\) on \(R=[0,2]\times[1,4]\).

For every subrectangle, regardless of the tag, \(f(x_{ij}^*,y_{ij}^*)=3\). Thus any tagged sum is

$$ S(P,f)=\sum_{i=1}^{m}\sum_{j=1}^{n}3\,\Delta x_i\Delta y_j =3\left(\sum_{i=1}^{m}\Delta x_i\right) \left(\sum_{j=1}^{n}\Delta y_j\right). $$

The interval lengths add to the side lengths of \(R\): \(\sum_i\Delta x_i=2-0=2\) and \(\sum_j\Delta y_j=4-1=3\). Therefore

$$ S(P,f)=3(2)(3)=18. $$

This value is the same for every partition and every tag choice, so in particular it is the limit as the mesh tends to zero. As a check, the rectangle has area \(2\cdot3=6\), and the constant value \(3\) times that area is \(18\), agreeing with the Riemann-sum calculation.

Why the Limit Is Independent of the Tags

A useful way to prove convergence is to compare each tagged sum with a simpler sum whose value can be calculated exactly. If the difference is bounded by a quantity that tends to zero with the mesh, then every tagged sum has the same limit. The following example demonstrates the argument in full.

Worked Example: The Function \(f(x,y)=x\)

Consider \(f(x,y)=x\) on \(R=[0,1]\times[0,2]\). Take any rectangular partition and any tags \((\xi_{ij},\eta_{ij})\). In each interval \([x_{i-1},x_i]\), let \(q_i=(x_{i-1}+x_i)/2\) be its midpoint. Compare the tagged sum with the sum that uses \(q_i\) as the \(x\)-coordinate in every subrectangle of the \(i\)-th vertical strip.

First calculate the midpoint sum. Since \(\Delta y_j=y_j-y_{j-1}\),

$$ \begin{aligned} \sum_{i=1}^{m}\sum_{j=1}^{n}q_i\Delta x_i\Delta y_j &=\left(\sum_{i=1}^{m}q_i\Delta x_i\right) \left(\sum_{j=1}^{n}\Delta y_j\right)\\ &=\left(\sum_{i=1}^{m} \frac{x_{i-1}+x_i}{2}(x_i-x_{i-1})\right)(2)\\ &=\left(\sum_{i=1}^{m}\frac{x_i^2-x_{i-1}^2}{2}\right)(2)\\ &=\frac{1^2-0^2}{2}(2)=1. \end{aligned} $$

Now let \(\delta=\lVert P\rVert\). Since both \(\xi_{ij}\) and \(q_i\) lie in the same interval, \(|\xi_{ij}-q_i|\leq\Delta x_i/2\). The difference between the tagged sum and the midpoint sum is therefore bounded by

$$ \begin{aligned} \left|S(P,f)-1\right| &\leq\sum_{i=1}^{m}\sum_{j=1}^{n} |\xi_{ij}-q_i|\Delta x_i\Delta y_j\\ &\leq\frac12\sum_{i=1}^{m}\sum_{j=1}^{n} (\Delta x_i)^2\Delta y_j\\ &=\frac12\left(\sum_{i=1}^{m}(\Delta x_i)^2\right)(2)\\ &=\sum_{i=1}^{m}(\Delta x_i)^2 \leq\delta\sum_{i=1}^{m}\Delta x_i =\delta. \end{aligned} $$

The last inequality uses \(\Delta x_i\leq\delta\), which follows because each interval width is no larger than its subrectangle diameter. Given any \(\varepsilon>0\), choosing \(\delta\) smaller than \(\varepsilon\) makes \(|S(P,f)-1|<\varepsilon\) whenever the mesh is sufficiently small. Thus every tagged sum approaches \(1\). As a check, the exact midpoint sum is \(1\) for every partition, and the error estimate shows that arbitrary tags approach that same value.

Worked Example: The Function \(f(x,y)=xy\)

Worked Example: The Function \(f(x,y)=xy\)

Consider \(f(x,y)=xy\) on the unit square \(R=[0,1]\times[0,1]\). In each subrectangle, write \(q_i\) and \(r_j\) for the midpoints of its \(x\)- and \(y\)-intervals. The sum using the center \((q_i,r_j)\) of each subrectangle can be factored:

$$ \begin{aligned} \sum_{i=1}^{m}\sum_{j=1}^{n}q_ir_j\Delta x_i\Delta y_j &=\left(\sum_{i=1}^{m}q_i\Delta x_i\right) \left(\sum_{j=1}^{n}r_j\Delta y_j\right)\\ &=\left(\sum_{i=1}^{m} \frac{x_{i-1}+x_i}{2}(x_i-x_{i-1})\right) \left(\sum_{j=1}^{n} \frac{y_{j-1}+y_j}{2}(y_j-y_{j-1})\right)\\ &=\left(\frac{1^2-0^2}{2}\right) \left(\frac{1^2-0^2}{2}\right) =\frac14. \end{aligned} $$

To control any tagged sum, let its tag in the subrectangle be \((\xi_{ij},\eta_{ij})\), and set \(\delta=\lVert P\rVert\). Since all coordinates lie between zero and one,

$$ \begin{aligned} |\xi_{ij}\eta_{ij}-q_ir_j| &\leq|\xi_{ij}-q_i|\,|\eta_{ij}| +|q_i|\,|\eta_{ij}-r_j|\\ &\leq\frac{\Delta x_i}{2}+\frac{\Delta y_j}{2} \leq\delta. \end{aligned} $$

Here each distance from a tag coordinate to its interval midpoint is at most half the interval width, and both widths are at most \(\delta\). The total area of the unit square is \(1\), so

$$ \left|S(P,f)-\frac14\right| \leq\sum_{i=1}^{m}\sum_{j=1}^{n} \delta\,\Delta x_i\Delta y_j =\delta. $$

For any \(\varepsilon>0\), taking the mesh smaller than \(\varepsilon\) makes every tagged sum differ from \(1/4\) by less than \(\varepsilon\). Hence the limit is \(1/4\). The exact center sum and the error bound together verify that this is the common limit, not just a value obtained from one tag selection.

A Common Mistake: Treating One Sum as the Integral

A Riemann sum from one partition is not, in general, the double integral. For example, use the single subrectangle \([0,1]\times[0,1]\) for \(f(x,y)=x\), and choose the tag \((1,\tfrac12)\). The resulting sum is

$$ f(1,\tfrac12)(1)(1)=1. $$

It would be incorrect to conclude from this one calculation that the double integral is \(1\). The mesh of this partition is \(\sqrt{2}\), so this calculation says nothing about what happens as the mesh tends to zero. The preceding error estimate shows the correction: every tagged sum on \([0,1]\times[0,2]\) differs from \(1\) by at most the mesh, so the limit there is \(1\). On the unit square, the same midpoint comparison gives a limit of \(\tfrac12\). The rectangle is part of the data, and a single coarse sum cannot replace the limiting process.

What the Definition Does—and Does Not—Say

The limit notation is a precise convergence requirement, not a shortcut for selecting convenient sample points. Its key features are:

  • The subrectangles must become uniformly small in the sense that the largest diameter tends to zero.
  • Tags may be chosen anywhere within their respective subrectangles.
  • All sufficiently fine tagged sums must be close to one common value.
  • The definition applies when the function is Riemann integrable; it does not claim that every function has such a limit.

This definition gives a way to build the double integral from finite sums without first relying on a procedure for evaluating it. In later work, other methods can be used to calculate integrals, but the limit definition explains what number those methods are intended to produce.

Check Your Understanding

Use the definition and the estimates in this tutorial to answer the following questions.

  1. What does the mesh of a rectangular partition measure, and how is it related to the subrectangle diameters?
  2. In the definition of integrability, why must the estimate hold for every choice of tags?
  3. For \(f(x,y)=4\) on \([0,1]\times[0,3]\), what is every tagged Riemann sum, regardless of the partition?
  4. Why does a single tagged sum from a coarse partition not determine the double integral?
  5. In the \(f(x,y)=xy\) example, what error bound shows that arbitrary tagged sums approach the center-sum value?

Previous

Next

Back to Tutorials