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Bounding Triple Integrals by Comparison

Use bounds on an integrand and the volume of its solid to estimate a triple integral and check whether a result is plausible.

Intermediate 10 min read

What You'll Learn

  • State the comparison rule for triple integrals and its assumptions
  • Bound an integral using pointwise bounds and the volume of a solid
  • Use sign information to determine whether an integral can be positive or negative
  • Check an exact integral against comparison bounds
  • Recognize when a claimed integrand bound is not valid

Why Bound a Triple Integral?

A triple integral can be difficult to evaluate exactly, especially when the solid or integrand is complicated. Even without finding an exact value, you can often establish a range that the integral must lie in. This gives a useful estimate and a way to check a calculation.

The key is to compare the integrand with constant lower and upper bounds throughout the solid. Integrating those constants multiplies each bound by the solid's volume. The result is a simple inequality that also carries sign information.

Central idea: If the integrand stays between \(m\) and \(M\) everywhere on a solid \(E\), then its triple integral stays between \(m\) times the volume of \(E\) and \(M\) times that volume.

The Comparison Rule

Let \(E\) be a bounded solid, and suppose \(f\) is continuous on \(E\). Write \(V(E)=\iiint_E 1\,dV\) for the volume of \(E\). If constants \(m\) and \(M\) satisfy \(m\leq f(x,y,z)\leq M\) at every point of \(E\), then:

$$ mV(E)\leq \iiint_E f(x,y,z)\,dV\leq MV(E). $$

Continuity is a standard sufficient condition for the integrals here to exist. The essential comparison requirement is that the inequalities hold throughout the solid, not merely at a selection of points. The constants \(m\) and \(M\) need not be the best possible bounds; any valid lower and upper bounds give a valid estimate.

When \(E\) is closed and bounded and \(f\) is continuous, the extreme value theorem guarantees that \(f\) attains a minimum and maximum on \(E\). Those values can be used for \(m\) and \(M\). Sometimes simpler bounds are easier to find and are already sufficient.

Why the Rule Works

The argument follows from the order property of integration: if one integrable function is no larger than another throughout a region, its integral is no larger either. Here is the complete comparison argument.

1
Subtract the lower bound.
Since \(f\geq m\) everywhere on \(E\), the function \(f-m\) is nonnegative throughout \(E\). The integral of a nonnegative function is nonnegative, so \(\iiint_E(f-m)\,dV\geq 0\).
2
Rearrange.
Linearity of the integral gives \(\iiint_E f\,dV-m\iiint_E1\,dV\geq0\). Since \(\iiint_E1\,dV=V(E)\), this is \(mV(E)\leq\iiint_Ef\,dV\).
3
Subtract from the upper bound.
Since \(M-f\geq0\) throughout \(E\), \(\iiint_E(M-f)\,dV\geq0\). Linearity then gives \(MV(E)-\iiint_Ef\,dV\geq0\), or \(\iiint_Ef\,dV\leq MV(E)\).

Combining the two inequalities proves the rule. The proof does not depend on whether \(m\) or \(M\) is positive. If the integrand is negative in part or all of the solid, the same comparison still applies.

Worked Example: A Bound on a Rectangular Box

Worked Example: A Bound on a Rectangular Box

Let \(E=[0,1]\times[0,2]\times[0,3]\), and let \(f(x,y,z)=2+x-y\). First estimate \(\iiint_E f\,dV\) by comparison, then evaluate it to check the estimate.

Since \(0\leq x\leq1\) and \(0\leq y\leq2\), the smallest possible value of \(2+x-y\) is \(2+0-2=0\), and the largest is \(2+1-0=3\). Thus \(0\leq f\leq3\) throughout \(E\). The volume is \(1\cdot2\cdot3=6\), so comparison gives:

$$ 0\cdot6\leq\iiint_E(2+x-y)\,dV\leq3\cdot6, \qquad 0\leq\iiint_E(2+x-y)\,dV\leq18. $$

To check, evaluate in the order \(dx\,dy\,dz\):

$$ \begin{aligned} \iiint_E(2+x-y)\,dV &=\int_0^3\int_0^2\int_0^1(2+x-y)\,dx\,dy\,dz\\ &=\int_0^3\int_0^2\left[(2-y)x+\frac{x^2}{2}\right]_0^1\,dy\,dz\\ &=\int_0^3\int_0^2\left(\frac52-y\right)\,dy\,dz\\ &=\int_0^3\left[\frac52y-\frac{y^2}{2}\right]_0^2\,dz\\ &=\int_0^3 3\,dz\\ &=9. \end{aligned} $$

The computed value passes the check: \(0\leq9\leq18\). The estimate is valid but not especially tight; comparison provides a guaranteed range, not necessarily a close approximation.

Sign and Size Checks

The sign of the integrand gives an immediate test. If \(f\geq0\) throughout \(E\), then \(\iiint_E f\,dV\geq0\). If \(f\leq0\) throughout \(E\), then \(\iiint_E f\,dV\leq0\). If the integrand changes sign, the integral may be positive, negative, or zero, depending on the contributions from the positive and negative parts.

A useful consequence of comparison is a bound on the absolute value. If \(|f|\leq C\) everywhere on \(E\), then \(-C\leq f\leq C\), so:

$$ -CV(E)\leq\iiint_E f\,dV\leq CV(E), \qquad \left|\iiint_E f\,dV\right|\leq CV(E). $$

This can help assess whether a result is plausible in scale. A value outside this interval cannot be correct if the stated bound on \(|f|\) and the volume are correct.

Worked Example: A Bound on a Tetrahedron

Worked Example: A Bound on a Tetrahedron

Consider the tetrahedron \(E=\{(x,y,z):x\geq0,\ y\geq0,\ z\geq0,\ x+y+z\leq1\}\) and the integrand \(f(x,y,z)=1+x+y+z\). Find a comparison range and evaluate the integral to verify it.

On \(E\), the sum \(x+y+z\) lies between \(0\) and \(1\). Therefore \(1\leq f\leq2\). The volume, found by integrating \(1\), is:

$$ \begin{aligned} V(E) &=\int_0^1\int_0^{1-x}\int_0^{1-x-y}1\,dz\,dy\,dx\\ &=\int_0^1\int_0^{1-x}(1-x-y)\,dy\,dx\\ &=\int_0^1\frac{(1-x)^2}{2}\,dx\\ &=\frac12\left[-\frac{(1-x)^3}{3}\right]_0^1 =\frac16. \end{aligned} $$

Consequently, the comparison rule gives:

$$ \frac16\leq\iiint_E(1+x+y+z)\,dV\leq\frac13. $$

For the exact value, put \(a=1-x-y\), the upper \(z\)-bound. The inner integral is:

$$ \begin{aligned} \int_0^a(1+x+y+z)\,dz &=\int_0^a(2-a+z)\,dz\\ &=(2-a)a+\frac{a^2}{2} =2a-\frac{a^2}{2}. \end{aligned} $$

Integrating next in \(y\), where \(a=1-x-y\), gives:

$$ \begin{aligned} \int_0^{1-x}\left(2(1-x-y)-\frac{(1-x-y)^2}{2}\right)dy &=\int_0^{1-x}\left(2a-\frac{a^2}{2}\right)da\\ &=\left[a^2-\frac{a^3}{6}\right]_0^{1-x}\\ &=(1-x)^2-\frac{(1-x)^3}{6}. \end{aligned} $$

Finally:

$$ \begin{aligned} \iiint_E(1+x+y+z)\,dV &=\int_0^1\left((1-x)^2-\frac{(1-x)^3}{6}\right)dx\\ &=\frac13-\frac1{24} =\frac7{24}. \end{aligned} $$

Writing the bounds with denominator \(24\) gives \(4/24\leq7/24\leq8/24\), so the exact value lies in the required interval.

When the Integrand Changes Sign

Bounds remain useful even when they do not determine the sign of the integral. For a sign-changing integrand, a lower bound below zero and an upper bound above zero simply leave open the possibility of cancellation. The integral is a signed total, not the total amount of positive values alone.

Worked Example: A Sign-Changing Integrand

Let \(E=[0,1]\times[0,2]\times[0,1]\) and \(f(x,y,z)=2x-y+z\). Find comparison bounds and evaluate the integral.

The smallest value on the box is \(2(0)-2+0=-2\), and the largest is \(2(1)-0+1=3\). The volume is \(1\cdot2\cdot1=2\). Thus:

$$ -2(2)\leq\iiint_E(2x-y+z)\,dV\leq3(2), \qquad -4\leq\iiint_E(2x-y+z)\,dV\leq6. $$

Evaluate in the order \(dx\,dy\,dz\):

$$ \begin{aligned} \iiint_E(2x-y+z)\,dV &=\int_0^1\int_0^2\int_0^1(2x-y+z)\,dx\,dy\,dz\\ &=\int_0^1\int_0^2(1-y+z)\,dy\,dz\\ &=\int_0^1(2-2+2z)\,dz\\ &=\left[z^2\right]_0^1\\ &=1. \end{aligned} $$

The value \(1\) lies between \(-4\) and \(6\). It is positive even though the integrand takes negative values on part of the box; the positive and negative contributions combine to give the signed total.

A Common Pitfall: Checking Too Few Points

A proposed lower or upper bound is only useful if it holds everywhere on the solid. A common error is to inspect a few convenient points and mistake those values for the global minimum or maximum. For a general continuous function, extrema can occur away from the corners, so the whole solid must be considered or a valid argument for the bounds must be given.

In the first example, suppose someone checks \(x=0,y=0\) and \(x=1,y=0\), sees values \(2\) and \(3\), and incorrectly claims \(2\leq2+x-y\leq3\). Multiplying by the volume \(6\) would then give \(12\leq\iiint_Ef\,dV\leq18\). But at \(x=0,y=2\), the integrand is \(0\), so the claimed lower bound fails. The corrected bound is \(0\leq f\leq3\), giving \(0\leq\iiint_Ef\,dV\leq18\), consistent with the exact value \(9\).

For an affine function on a rectangular box, checking the endpoint choices for each coordinate does establish its extrema. That shortcut depends on the function and region; it should not be assumed for an arbitrary integrand or solid.

Using Comparison as a Practical Check

Before accepting a computed triple integral, ask whether its sign and size agree with basic information about the integrand and solid. Establish a pointwise range, compute the volume, and compare the answer with the resulting interval. These checks can reveal errors in an evaluation or in the stated bounds, although passing them does not by itself prove that a computation is correct.

Takeaway: A valid integrand bound yields a guaranteed integral bound after multiplication by the solid's volume. Check the sign of the bounds, verify that they hold everywhere, and use the resulting interval to assess—but not replace—an exact calculation.

Check Your Understanding

Use comparison and volume to reason about each question.

  1. If \(2\leq f\leq5\) throughout a solid of volume \(3\), what bounds follow for \(\iiint_E f\,dV\)?
  2. If \(f\leq0\) everywhere on a solid, what can you conclude about the sign of its triple integral?
  3. On the box \([0,1]\times[0,1]\times[0,2]\), find a lower and upper bound for \(x+2y-z\), then use the volume to bound its integral.
  4. Why is it not enough to check a few points when proposing bounds for a general continuous integrand?
  5. If \(|f|\leq4\) on a solid of volume \(2\), what bound does comparison give for \(\left|\iiint_E f\,dV\right|\)?

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