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Common Errors in Three-Dimensional Bounds

Use endpoint tests and carefully chosen points to decide whether iterated bounds describe the intended solid.

Intermediate 10 min read

What You'll Learn

  • Check that dependent lower and upper limits are ordered throughout their projection.
  • Test whether projected limits include every part of a solid and exclude points outside it.
  • Use representative points to detect reversed surfaces and inconsistent nested bounds.
  • Compare incorrect and corrected bounds by evaluating simple volume integrals.
  • Apply a systematic checklist before integrating over a solid.

Why Bounds Need to Be Tested

Writing an iterated integral is not merely a matter of placing limits around an integrand. The limits must describe exactly the intended solid: they must include every point in the solid and exclude every point outside it. A small error in a projection or in the order of two boundary surfaces can change the region, even when the bounds look plausible.

A useful habit is to test the proposed description before doing any substantial integration. Check that each interval is nonempty, evaluate dependent limits at the edges of their domains, and substitute representative points into the inequalities that define the solid. These checks catch different kinds of errors; passing one test does not guarantee that the bounds are correct.

Definition: For an iterated integral in the order \(dz\,dy\,dx\), the outer limits describe the \(x\)-interval, the next limits describe the allowed \(y\)-values for each \(x\), and the inner limits describe the allowed \(z\)-values for each \((x,y)\). The resulting bounds describe the set of points satisfying all of these inequalities simultaneously.

The same principle applies to other integration orders: read the limits from the outside inward, and ask what values remain allowed at each stage. In this tutorial, the focus is not on finding bounds from scratch, but on diagnosing common mistakes in proposed three-dimensional bounds.

A Reliable Set of Tests

Begin by writing the proposed bounds as inequalities. For example,

$$ a\le x\le b,\qquad g_1(x)\le y\le g_2(x),\qquad h_1(x,y)\le z\le h_2(x,y). $$

This notation makes the tests explicit. The bounds require \(a\le b\), \(g_1(x)\le g_2(x)\) throughout the stated \(x\)-interval, and \(h_1(x,y)\le h_2(x,y)\) for every \((x,y)\) allowed by the first two inequalities. If any of these conditions fails, part or all of the proposed description is empty or reversed.

1
Check each interval.
For dependent limits, verify that the lower limit is no greater than the upper limit throughout the full domain of the outer variables.
2
Check the projection.
Ask whether the outer bounds cover the full shadow of the solid on the corresponding coordinate plane, without extending into points that cannot lie in the solid.
3
Test representative points.
Substitute points known to be in the intended solid and points known to be outside it. The bounds should include the first kind and exclude the second.
4
Check the boundary surfaces.
At a fixed position in the projection, identify which surface is lower and which is upper. Do not infer their order from the way the surfaces are named.

Endpoint checks are especially efficient for limits that depend on another variable. If the lower and upper limits agree at an edge of the projection, the interval may shrink to zero there, which is often correct. But equality at an endpoint does not prove that the limits are ordered everywhere between endpoints. A representative interior value may also be needed.

Worked Example: Reversed Vertical Boundaries

Worked Example: Reversed Vertical Boundaries

Consider the solid whose projection onto the \(xy\)-plane is the triangle \(x\ge 0\), \(y\ge 0\), \(x+y\le 2\), and whose vertical coordinate lies between the surfaces \(z=x+y\) and \(z=2\). Check the bounds and find the volume.

The projection gives \(0\le x\le 2\) and \(0\le y\le 2-x\). At a fixed \((x,y)\) in this triangle, the lower surface is \(z=x+y\) and the upper surface is \(z=2\). Thus the proposed bounds are

$$ 0\le x\le 2,\qquad 0\le y\le 2-x,\qquad x+y\le z\le 2. $$

First check the vertical interval. Since \(x+y\le 2\) throughout the projected triangle, its lower endpoint does not exceed its upper endpoint. For an interior check, take \(x=\frac12\), \(y=\frac12\). Then \(x+y=1\), so the vertical interval is \(1\le z\le 2\), in the stated order.

The volume is the integral of \(1\) over these bounds. Integrating first with respect to \(z\) gives the vertical height \(2-x-y\):

$$ \begin{aligned} V &=\int_0^2\int_0^{2-x}\int_{x+y}^{2}1\,dz\,dy\,dx\\ &=\int_0^2\int_0^{2-x}(2-x-y)\,dy\,dx\\ &=\int_0^2\left[(2-x)y-\frac{y^2}{2}\right]_{0}^{2-x}dx\\ &=\int_0^2\frac{(2-x)^2}{2}\,dx\\ &=\frac12\left[-\frac{(2-x)^3}{3}\right]_{0}^{2}\\ &=\frac12\left(0-\left(-\frac83\right)\right) =\frac43. \end{aligned} $$

Now suppose the vertical bounds are mistakenly written as \(2\le z\le x+y\), with the same \(x\)- and \(y\)-limits. At the test point \(x=y=\frac12\), this says \(2\le z\le 1\), which is impossible. The endpoints are reversed. The resulting iterated integral confirms the error:

$$ \begin{aligned} \int_0^2\int_0^{2-x}\int_2^{x+y}1\,dz\,dy\,dx &=\int_0^2\int_0^{2-x}(x+y-2)\,dy\,dx\\ &=\int_0^2\left[(x-2)y+\frac{y^2}{2}\right]_{0}^{2-x}dx\\ &=\int_0^2-\frac{(2-x)^2}{2}\,dx\\ &=-\frac43. \end{aligned} $$

A negative result for the integral of \(1\) is a warning: the integral is signed because the inner limits run from the larger value to the smaller one. Reversing the surfaces does not describe the intended solid with positive volume.

Projection Errors: Omitting Part of a Solid

The outer limits control the projection in the outermost coordinate. If those limits are too narrow, an entire part of the solid can be left out, even if every inner bound is otherwise correct. A simple point test can expose this kind of omission: choose a point in the intended solid whose outer coordinate falls outside the proposed interval.

Worked Example: An Incomplete Projection

Let the intended solid be the rectangular box

$$ -1\le x\le 1,\qquad 0\le y\le 2,\qquad 0\le z\le 1. $$

Suppose someone proposes \(0\le x\le 1\), \(0\le y\le 2\), \(0\le z\le 1\). The limits are internally consistent, but the \(x\)-projection omits the half of the box with negative \(x\). For example, the point \(\left(-\frac12,1,\frac12\right)\) satisfies all the inequalities of the intended box, but fails the proposed inequality \(0\le x\). So the proposed bounds do not include every point of the solid.

Compare the two volume integrals. The correct bounds give

$$ \begin{aligned} V &=\int_{-1}^{1}\int_0^2\int_0^1 1\,dz\,dy\,dx\\ &=\int_{-1}^{1}\int_0^2 1\,dy\,dx\\ &=\int_{-1}^{1}2\,dx\\ &=2(1-(-1))=4. \end{aligned} $$

The incomplete projection gives

$$ \begin{aligned} V_{\text{proposed}} &=\int_0^1\int_0^2\int_0^1 1\,dz\,dy\,dx\\ &=\int_0^1\int_0^2 1\,dy\,dx\\ &=\int_0^1 2\,dx=2. \end{aligned} $$

The calculation is correct for the proposed bounds, but those bounds describe only half the intended box. The point test identifies the geometric omission, and the volume comparison confirms its effect.

Dependent Bounds Must Match the Projection

A projection is not always a rectangle. If one coordinate is restricted by another, replacing a dependent limit with a constant can add points outside the projection. It is not enough for the new limits to be individually sensible; they must encode the same set of allowed coordinate pairs.

Worked Example: A Dependent Limit Made Constant

Consider the solid defined by \(0\le x\le 1\), \(0\le y\le x\), and \(0\le z\le y\). Its bounds in the order \(dz\,dy\,dx\) are

$$ 0\le x\le 1,\qquad 0\le y\le x,\qquad 0\le z\le y. $$

At \(x=0\), the \(y\)-interval reduces to the single value \(y=0\); at \(x=1\), it runs from \(0\) to \(1\). These endpoint checks agree with the triangular projection \(0\le y\le x\). Now test the point \(\left(\frac14,\frac34,\frac12\right)\). It satisfies \(0\le x\le 1\) and \(0\le z\le y\), but it fails \(y\le x\), since \(\frac34>\frac14\). It is not in the solid.

If the \(y\)-limits are mistakenly replaced by \(0\le y\le 1\), that point is included. The altered bounds describe a larger region. The correct volume is

$$ \begin{aligned} V &=\int_0^1\int_0^x\int_0^y 1\,dz\,dy\,dx\\ &=\int_0^1\int_0^x y\,dy\,dx\\ &=\int_0^1\frac{x^2}{2}\,dx\\ &=\left[\frac{x^3}{6}\right]_0^1=\frac16. \end{aligned} $$

For the altered bounds, the calculation becomes

$$ \begin{aligned} V_{\text{altered}} &=\int_0^1\int_0^1\int_0^y 1\,dz\,dy\,dx\\ &=\int_0^1\int_0^1 y\,dy\,dx\\ &=\int_0^1\frac12\,dx=\frac12. \end{aligned} $$

The altered integral includes all of the intended solid, but also adds points with \(y>x\). The difference in the calculated volumes is \(\frac12-\frac16=\frac13\). This agrees with the volume of the added portion: integrating over \(0\le x\le 1\), \(x\le y\le 1\), \(0\le z\le y\) gives \(\int_0^1\int_x^1 y\,dy\,dx=\frac13\).

Why the Tests Work

The bounds are a compact way to state a collection of inequalities. For example, the three conditions

$$ a\le x\le b,\qquad g_1(x)\le y\le g_2(x),\qquad h_1(x,y)\le z\le h_2(x,y) $$

mean that a point \((x,y,z)\) belongs to the described set exactly when it satisfies every inequality. This gives a direct criterion for checking proposed bounds: they are correct precisely when their inequalities accept all points of the intended solid and reject all points outside it.

The endpoint-order test checks whether the inequalities can be satisfied at all for each permitted choice of outer variables. It catches empty or reversed intervals. The projection check asks whether the outer variables range over the correct set. Representative points test actual membership in the full three-dimensional set. Each test addresses a different part of the description, which is why a complete check uses all of them.

There is also an important limitation: testing a few points cannot, by itself, prove that bounds are correct everywhere. Points are diagnostic samples, not a replacement for reasoning about the inequalities over their full domains. When the limits involve functions, verify the required inequalities throughout the projection, using algebra, a graph, or a suitable argument—not just at one convenient point.

Common Pitfalls and a Final Checklist

Several shortcuts repeatedly lead to errors. First, assuming that a named “top” surface is always above a named “bottom” surface can reverse the vertical limits; compare their values on the projection. Second, checking only one interior point can miss an error near the edge of the projection; inspect endpoints and dependent limits there as well. Third, checking only that every proposed point lies in the intended solid can miss omissions; also test whether representative points of the intended solid are included. Finally, a correct evaluation of an integral does not validate its bounds: arithmetic can be flawless while the described region is wrong.

Takeaway: Before integrating, translate every limit into an inequality. Check that dependent intervals are ordered across their full domains, verify that the projection is neither too small nor too large, and test points for both inclusion and exclusion.

A practical final review is to ask: Are all outer endpoints in the right order? Do dependent limits meet the intended edges of the projection? At a representative interior location, are the lower and upper surfaces correctly ordered? Does a known point of the solid satisfy every proposed inequality? Does a known outside point fail at least one? These questions make many bound errors visible before they affect a longer calculation.

Check Your Understanding

For each question, focus on what the proposed inequalities describe, not only on whether the integral can be evaluated.

  1. For bounds \(0\le x\le 1\), \(0\le y\le x\), and \(x+y\le z\le 1\), what inequality must be checked to ensure the vertical interval is nonempty throughout the projected region?
  2. A solid has \(-2\le x\le 2\), but proposed bounds use \(0\le x\le 2\). What kind of error should you suspect, and what point would help test it?
  3. Why does finding one point that satisfies proposed bounds but lies outside the intended solid prove that the bounds are incorrect?
  4. Why is checking the endpoints of a dependent interval useful, and why might endpoint checks alone be insufficient?
  5. If an integral of \(1\) evaluates to a negative number, what should you check first in the bounds?

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