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Triple Riemann Sums and Signed Volume

Build triple integrals from rectangular sample-point sums and interpret their limits as signed volume or accumulated quantity.

Intermediate 10 min read

What You'll Learn

  • Define partitions, subboxes, sample points, mesh, and triple Riemann sums.
  • Explain how subbox volumes weight the function values in each sum.
  • Understand why continuous functions have a well-defined Riemann-sum limit.
  • Connect triple integrals with signed volume and accumulated quantity.
  • Identify cancellation when a function takes both positive and negative values.
  • Avoid common errors involving subbox volume and signed contributions.

From Small Boxes to a Triple Integral

A triple integral begins with a simple approximation: divide a three-dimensional box into smaller boxes, evaluate a function once in each small box, and multiply each value by that box’s volume. Adding these products estimates the total accumulation across the original box.

The approximation becomes more reliable as the subboxes become smaller. The limiting value is the triple integral. When the function is positive, the sum has a direct volume interpretation; when it changes sign, positive and negative contributions offset one another.

The central idea: A triple Riemann sum adds the function’s value at a sample point in each subbox, weighted by that subbox’s volume. The triple integral is the limit of these sums as the partitions become arbitrarily fine.

Rectangular Partitions and Sample Points

Let \(B=[a,b]\times[c,d]\times[e,g]\) be a rectangular box in three-dimensional space. Partition each of its three coordinate intervals into finitely many subintervals. The resulting small rectangular boxes are the subboxes of a partition of \(B\).

Write one subbox as \(B_{ijk}\), where \(i,j,k\) identify its intervals in the \(x\)-, \(y\)-, and \(z\)-directions. If its side lengths are \(\Delta x_i\), \(\Delta y_j\), and \(\Delta z_k\), then its volume is

$$ \Delta V_{ijk}=\Delta x_i\,\Delta y_j\,\Delta z_k. $$

Choose a sample point \(\mathbf{x}_{ijk}^{*}=(x_{ijk}^{*},y_{ijk}^{*},z_{ijk}^{*})\) in each subbox. For a function \(f:B\to\mathbb{R}\), its triple Riemann sum for this partition and these sample points is

$$ \sum_{i,j,k} f(\mathbf{x}_{ijk}^{*})\,\Delta V_{ijk}. $$

The indices run over all subboxes. A sample point may be a corner, a midpoint, or any other point in its subbox. The mesh of a partition is the largest diameter of any of its subboxes. Saying the mesh tends to zero means that the largest subbox becomes arbitrarily small in every direction.

Definition: The function \(f\) is Riemann integrable on \(B\) if the triple Riemann sums approach one common finite limit as the mesh tends to zero, regardless of the choices of partitions and sample points. That limit is the triple integral of \(f\) over \(B\), written \(\iiint_B f\,dV\).

Here \(dV\) indicates volume in the three input coordinates. Each term uses the function’s value as a weight and the subbox volume as the amount of domain represented by that value. Omitting \(\Delta V_{ijk}\) would not produce an approximation to the integral.

Why the Limit Exists for Continuous Functions

Continuity ensures that the function cannot vary greatly inside a sufficiently small subbox. On a closed, bounded box, a continuous function is uniformly continuous: for every \(\varepsilon>0\), there is a \(\delta>0\) such that points less than \(\delta\) apart have function values differing by less than \(\varepsilon\).

To make the reasoning precise, on each subbox let the oscillation be the supremum of \(f\) there minus its infimum there. The upper sum uses each subbox’s supremum, and the lower sum uses its infimum; both multiply by that subbox’s volume. Their difference is the sum of oscillation times volume. If the mesh is \(\delta\), uniform continuity makes every subbox’s oscillation small. Since the subbox volumes add to the volume of \(B\), the difference between the upper and lower sums can be made arbitrarily small. The Riemann criterion says that this is sufficient for Riemann integrability.

$$ U-L =\sum_{i,j,k} \bigl(\sup_{B_{ijk}}f-\inf_{B_{ijk}}f\bigr)\Delta V_{ijk} \leq \varepsilon\sum_{i,j,k}\Delta V_{ijk} =\varepsilon\,\operatorname{Vol}(B). $$

This also explains why the sample points do not affect the limiting value. In any subbox, changing the sample point changes its function value by no more than the oscillation on that subbox. For a fine partition, the total possible change in the sum is bounded by the same upper-minus-lower difference, which tends to zero. Thus different sample-point choices may give different finite approximations, but their limits agree.

Theorem: Every continuous real-valued function on a closed, bounded rectangular box is Riemann integrable on that box. Uniform continuity makes the upper and lower sums arbitrarily close as the mesh tends to zero.

Worked Example: A Constant Function

Worked Example: A Constant Function

Find the triple integral of \(f(x,y,z)=3\) over \(B=[0,2]\times[0,1]\times[0,3]\) from its Riemann sums.

The box has side lengths \(2\), \(1\), and \(3\), so its volume is \(2\cdot1\cdot3=6\). For any partition, the function value at every sample point is \(3\). Therefore the sum is

$$ \sum_{i,j,k}3\,\Delta V_{ijk} =3\sum_{i,j,k}\Delta V_{ijk} =3\cdot6 =18. $$

This sum is already \(18\) for every partition, so its limit is \(18\). As a check, substituting the constant value \(3\) and the box volume \(6\) gives \(3\cdot6=18\), matching the sum.

Worked Example: A Function Varying in All Three Coordinates

Worked Example: A Function Varying in All Three Coordinates

Use equal subdivisions and midpoint sample points to find the limiting sum for \(f(x,y,z)=x+y+z\) on the unit box \(B=[0,1]^3\).

Divide each coordinate interval into \(n\) subintervals of length \(1/n\). There are \(n^3\) subboxes, each with volume \(1/n^3\). The midpoint coordinates are \((i-\tfrac12)/n\), \((j-\tfrac12)/n\), and \((k-\tfrac12)/n\), for \(1\leq i,j,k\leq n\). The sum is

$$ S_n=\frac{1}{n^3}\sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{k=1}^{n} \left(\frac{i-\tfrac12}{n}+\frac{j-\tfrac12}{n}+\frac{k-\tfrac12}{n}\right). $$

For the terms involving \(x\), each \(i\)-value appears with \(n^2\) choices of \(j,k\). Also, \(\sum_{i=1}^{n}(i-\tfrac12)=n^2/2\). Thus the total \(x\)-contribution is

$$ \frac{n^2}{n^3}\sum_{i=1}^{n}\frac{i-\tfrac12}{n} =\frac{1}{n}\cdot\frac{1}{n}\cdot\frac{n^2}{2} =\frac12. $$

The \(y\)- and \(z\)-contributions are also \(1/2\), by the same finite sum. Hence \(S_n=1/2+1/2+1/2=3/2\) for every positive integer \(n\). Its limit is \(3/2\). The check is direct: the three coordinate contributions add to \(3/2\), so the calculated sums already equal the claimed limit.

Signed Volume and Accumulation

The phrase signed volume describes how the sign of \(f\) affects its contribution. If \(f\geq0\) throughout \(B\), all terms in the Riemann sums are nonnegative. If \(f\leq0\), they are nonpositive. If \(f\) takes both signs, positive and negative contributions can cancel.

For a function of three variables, \(f(x,y,z)\) is not generally a height above a two-dimensional region in ordinary three-dimensional space. One useful geometric picture adds a fourth coordinate \(w\), with \(w=f(x,y,z)\). Then \(f\) represents a signed height over the three-dimensional box: values above \(w=0\) contribute positively, and values below \(w=0\) contribute negatively. The integral is the signed four-dimensional volume between that graph and \(w=0\). It can also represent the total of any signed quantity distributed across the box.

In the special case \(f=1\), every subbox contributes its volume. The triple integral is then the ordinary three-dimensional volume of \(B\). For a general \(f\), the integral instead records the function-weighted accumulation. In particular, it need not equal the volume obtained by counting all positive and negative contributions as positive.

Worked Example: Positive and Negative Contributions

Worked Example: Positive and Negative Contributions

Use equal subdivisions and midpoint sample points to find the triple Riemann sum for \(f(x,y,z)=x-1\) on \(B=[0,2]\times[0,1]\times[0,1]\). Determine what the limit says about signed volume.

Divide each interval into \(n\) equal pieces. The midpoint in the \(x\)-direction for interval \(i\) is \((2i-1)/n\); the \(y\)- and \(z\)-midpoints do not affect \(f\). Each subbox has volume \(2/n^3\), and each \(x\)-midpoint occurs for \(n^2\) pairs of \(y,z\) intervals. Thus

$$ \begin{aligned} S_n &=\frac{2}{n}\sum_{i=1}^{n}\left(\frac{2i-1}{n}-1\right)\\ &=\frac{2}{n}\left(\frac{1}{n}\sum_{i=1}^{n}(2i-1)-n\right)\\ &=\frac{2}{n}\left(\frac{n^2}{n}-n\right)\\ &=0. \end{aligned} $$

The identity \(\sum_{i=1}^{n}(2i-1)=n^2\) follows by adding the first \(n\) odd integers. Since \(S_n=0\) for every \(n\), the limiting signed volume is \(0\). As a check, the function is negative for \(0\leq x<1\) and positive for \(1<x\leq2\); the negative and positive contributions have equal magnitude. Each side has magnitude \(1/2\), since the \(x\)-cross-section is a triangle of base \(1\) and height \(1\), and the \(y,z\) lengths are both \(1\). They cancel: \(1/2-1/2=0\). The unsigned total would instead be \(1/2+1/2=1\).

A Common Mistake: Forgetting the Subbox Volume

A frequent error is to add the sampled function values without multiplying each one by its subbox volume. For the constant-function example, taking one sample value \(3\) and reporting \(3\) as the integral forgets that the whole box has volume \(6\). The correct one-subbox sum is \(3\cdot6=18\). With a finer partition, there are more terms, but each is weighted by its own subbox volume; those volumes still add to \(6\).

Another important distinction is between signed and unsigned totals. A zero integral does not imply that \(f\) is zero throughout the box. In the last example, positive and negative contributions cancel exactly. If the goal is the total magnitude of the contributions, the relevant function is \(|f|\), not \(f\).

When This Viewpoint Matters

Riemann sums make the meaning of a triple integral concrete: a continuous function is approximated by its values on small pieces of a three-dimensional domain, with each value weighted by the volume it represents. This viewpoint applies whether \(f\) describes a signed quantity, a density relative to volume, or a field whose local contributions accumulate across the box.

The definition is also the foundation for later methods of computing triple integrals. Those methods provide efficient ways to find the same limit; they do not change what the limit means. Keeping the partition, sample values, and volume factors in view helps distinguish the integral itself from a particular computational procedure.

Check Your Understanding

Answer these questions using the definitions and examples above.

  1. What three factors determine the volume of a rectangular subbox?
  2. How is the mesh of a partition defined, and what happens to it in the limit?
  3. Why does uniform continuity help ensure that changing sample points has little effect on a fine-partition sum?
  4. What does the sign of \(f\) tell you about the contributions to its triple integral?
  5. Why can an integral equal zero even when the function is not identically zero?

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