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Triple Integrals and Solid Regions · Tutorial 611 of 999

Reversing the Order of a Triple Integral

Translate a solid into a new Cartesian integration order by rebuilding its projections and bounds, then verify the result with exact calculations.

Intermediate 10 min read

What You'll Learn

  • Identify the solid represented by an iterated triple integral.
  • Determine the new projection and fibers required by a different integration order.
  • Derive bounds by solving the region’s defining inequalities for the new inner variable.
  • Check that every new bound describes exactly the original solid.
  • Compare integrals in two orders to verify a rewritten description.
  • Recognize when a proposed set of bounds includes points outside the solid.

Why Reversing an Order Requires Rebuilding the Bounds

A triple integral is an integral over a solid region, written in a particular order. Changing that order does not change the solid or the integral, provided the integrand is integrable and the new limits describe the very same region. But the bounds generally do change: they must be rebuilt to match the new projections and fibers.

The key is to reason about the region, not just rearrange the symbols in the integral. For each proposed order, identify the outer variables’ projection and then determine the interval available to the innermost variable at each point of that projection. A correct rewrite covers every point of the original solid exactly once and no point outside it.

Definition: An iterated integral describes a solid by successively specifying intervals for its variables. Reversing the order means expressing the same solid with a different sequence of variables and corresponding bounds.

A Systematic Method

Suppose an integral is written in the order \(dz\,dy\,dx\). Its bounds describe \(x\) first, then \(y\) for each chosen \(x\), and finally \(z\) for each chosen pair \((x,y)\). To change the order, rewrite the solid as a set of inequalities and ask how those inequalities look when the variables are chosen in the new sequence.

1
Record the solid.
Translate the original limits into inequalities involving all three variables.
2
Choose the new outer variables.
Project the solid onto the coordinate plane of those variables to find their joint range.
3
Find the new inner interval.
For each fixed choice of the outer variables, solve the inequalities for the innermost variable.
4
Verify both directions.
Check that every point satisfying the new limits satisfies the original inequalities, and that every point in the original solid satisfies the new limits.

This last check matters. Matching only the endpoints of a few boundary curves or surfaces does not guarantee that the entire region is right. The bounds must describe the same set of points, including their interior.

Worked Example: Changing the Innermost Variable in a Tetrahedron

Worked Example: Changing the Innermost Variable in a Tetrahedron

Rewrite the integral in the order \(dx\,dz\,dy\), and evaluate it, where

$$ \int_0^1\int_0^{1-x}\int_0^{1-x-y} x\,dz\,dy\,dx. $$

Step 1: Describe the solid. The limits say \(0\le x\le1\), \(0\le y\le1-x\), and \(0\le z\le1-x-y\). Equivalently,

$$ x\ge0,\qquad y\ge0,\qquad z\ge0,\qquad x+y+z\le1. $$

Step 2: Project onto the \(yz\)-plane. For a pair \((y,z)\) to occur in the solid, both must be nonnegative and there must be some \(x\ge0\) with \(x\le1-y-z\). This is possible exactly when \(y+z\le1\). Thus \(0\le y\le1\), and for each such \(y\), \(0\le z\le1-y\).

Step 3: Find the \(x\)-fiber. Once \(y,z\) are fixed in that projection, the inequality \(x+y+z\le1\), together with \(x\ge0\), gives \(0\le x\le1-y-z\). The rewritten integral is

$$ \int_0^1\int_0^{1-y}\int_0^{1-y-z} x\,dx\,dz\,dy. $$

Step 4: Evaluate in the new order.

$$ \begin{aligned} \int_0^1\int_0^{1-y}\int_0^{1-y-z}x\,dx\,dz\,dy &=\int_0^1\int_0^{1-y}\frac{(1-y-z)^2}{2}\,dz\,dy\\ &=\int_0^1\frac{(1-y)^3}{6}\,dy\\ &=\frac{1}{6}\cdot\frac{1}{4} =\frac{1}{24}. \end{aligned} $$

Check against the original order. Integrating first with respect to \(z\) gives

$$ \begin{aligned} \int_0^1\int_0^{1-x}\int_0^{1-x-y}x\,dz\,dy\,dx &=\int_0^1\int_0^{1-x}x(1-x-y)\,dy\,dx\\ &=\int_0^1\frac{x(1-x)^2}{2}\,dx\\ &=\frac12\left(\frac12-\frac23+\frac14\right) =\frac{1}{24}. \end{aligned} $$

The two evaluations agree, and the new limits express exactly the inequalities defining the tetrahedron.

Worked Example: Reversing the Order of Two Inner Variables

Worked Example: Reversing the Order of Two Inner Variables

Change the order of integration from \(dz\,dy\,dx\) to \(dy\,dz\,dx\) and evaluate:

$$ \int_0^1\int_0^{1-x}\int_0^x z\,dz\,dy\,dx. $$

Step 1: Read the inequalities. The solid satisfies \(0\le x\le1\), \(0\le y\le1-x\), and \(0\le z\le x\). For each fixed \(x\), the allowable \(y\)-range and \(z\)-range are independent of one another: \(0\le y\le1-x\) and \(0\le z\le x\).

Step 2: Write the new order. Keep \(x\) outermost. For each fixed \(x\), choose \(z\) between \(0\) and \(x\), then \(y\) between \(0\) and \(1-x\). Hence the integral becomes

$$ \int_0^1\int_0^x\int_0^{1-x} z\,dy\,dz\,dx. $$

Step 3: Evaluate the new integral.

$$ \begin{aligned} \int_0^1\int_0^x\int_0^{1-x}z\,dy\,dz\,dx &=\int_0^1\int_0^x z(1-x)\,dz\,dx\\ &=\int_0^1\frac{x^2(1-x)}{2}\,dx\\ &=\frac12\left(\frac13-\frac14\right) =\frac{1}{24}. \end{aligned} $$

Check in the original order.

$$ \begin{aligned} \int_0^1\int_0^{1-x}\int_0^x z\,dz\,dy\,dx &=\int_0^1\int_0^{1-x}\frac{x^2}{2}\,dy\,dx\\ &=\int_0^1\frac{x^2(1-x)}{2}\,dx =\frac{1}{24}. \end{aligned} $$

Here the new order is especially direct because, at fixed \(x\), the \(y\)- and \(z\)-ranges are independent. That independence is a feature of this solid, not a rule that applies to every change of order.

Worked Example: Reversing Bounds Involving a Curved Boundary

Worked Example: Reversing Bounds Involving a Curved Boundary

Rewrite the integral in the order \(dx\,dz\,dy\), then evaluate it:

$$ \int_0^1\int_{x^2}^1\int_0^y 1\,dz\,dy\,dx. $$

Step 1: Translate the bounds. The region is described by \(0\le x\le1\), \(x^2\le y\le1\), and \(0\le z\le y\).

Step 2: Project onto the \(yz\)-plane. For a fixed \(y\), the original conditions allow \(x\) precisely when \(x^2\le y\) and \(x\ge0\), so \(0\le x\le\sqrt y\). Such an \(x\) exists exactly for \(0\le y\le1\). The condition on \(z\) is \(0\le z\le y\). Therefore the projection is \(0\le y\le1\), \(0\le z\le y\), and the \(x\)-fiber is \(0\le x\le\sqrt y\).

Step 3: Write and evaluate the new integral.

$$ \begin{aligned} \int_0^1\int_0^y\int_0^{\sqrt y}1\,dx\,dz\,dy &=\int_0^1\int_0^y\sqrt y\,dz\,dy\\ &=\int_0^1 y\sqrt y\,dy\\ &=\int_0^1 y^{3/2}\,dy =\frac25. \end{aligned} $$

Check in the original order.

$$ \begin{aligned} \int_0^1\int_{x^2}^1\int_0^y1\,dz\,dy\,dx &=\int_0^1\int_{x^2}^1 y\,dy\,dx\\ &=\int_0^1\frac{1-x^4}{2}\,dx\\ &=\frac12\left(1-\frac15\right) =\frac25. \end{aligned} $$

The square root in the new limit comes from solving \(x^2\le y\) for the nonnegative variable \(x\). Since the original region also requires \(x\ge0\), the correct solution is \(x\le\sqrt y\), not \(x\le\pm\sqrt y\).

Why the Rewritten Integral Has the Same Value

The geometric argument behind changing order is that both iterated integrals integrate over the same set of points. For a continuous function \(f\) on a bounded solid \(E\), Fubini’s theorem allows the order of integration to change when the resulting iterated descriptions cover \(E\) and the required integrals exist. The theorem does not say that one may simply permute the differential symbols while keeping unrelated bounds.

For example, consider the tetrahedron from the first worked example. Its inequalities are \(x\ge0\), \(y\ge0\), \(z\ge0\), and \(x+y+z\le1\). Fixing \(x,y\) gives the \(z\)-interval \(0\le z\le1-x-y\). Alternatively, fixing \(y,z\) gives \(0\le x\le1-y-z\). Both descriptions select the same points because the final inequality can be solved for either \(z\) or \(x\). Integrating the same function over these two descriptions therefore gives the same value.

Theorem: If a bounded solid has valid iterated descriptions in two Cartesian orders, and the integrand is continuous on the solid, then the two iterated integrals have the same value. The bounds in each order must describe the same solid; changing the order does not authorize changing the region.

A Common Mistake: Keeping an Old Bound After Changing the Order

For the tetrahedron \(x,y,z\ge0\), \(x+y+z\le1\), the correct order \(dx\,dz\,dy\) is

$$ \int_0^1\int_0^{1-y}\int_0^{1-y-z}1\,dx\,dz\,dy. $$

A tempting but incorrect rewrite is to keep \(x\le1-y\), as though the original bound on \(y\) could be reused unchanged:

$$ \int_0^1\int_0^{1-y}\int_0^{1-y}1\,dx\,dz\,dy. $$

This incorrect expression allows \(x+ y\le1\) and \(z\le1-y\) independently. It can include points for which \(x+y+z>1\). Its value confirms the overcount:

$$ \int_0^1(1-y)^2\,dy=\frac13, $$

whereas the tetrahedron’s correct volume integral is \(1/6\). The correction is to derive the innermost bound from the full inequality \(x+y+z\le1\), giving \(x\le1-y-z\). A good check is always to substitute the proposed limits back into the original inequalities.

What to Check Before Accepting New Limits

Before evaluating a rewritten integral, test the geometry of its bounds. Confirm that the projection includes exactly the outer-variable values that occur in the solid. Then, for each fixed point in that projection, confirm that the innermost interval gives exactly the permitted fiber. In particular, verify that each upper and lower limit comes from the inequalities that constrain that variable at the chosen stage.

When the region has a single interval as each fiber in the proposed order, one set of iterated bounds describes it directly. If a fiber has multiple separated pieces, one set of continuous limits may not describe it as a single interval. Recognizing that issue is important; the limits must still cover only the original solid. The careful approach is to establish what the fibers actually are before writing the integral.

Check Your Understanding

For each question, begin by expressing the region as inequalities before proposing bounds.

  1. For \(x,y,z\ge0\) and \(x+y+z\le2\), what are the bounds in the order \(dx\,dy\,dz\)?
  2. In the tetrahedron \(x,y,z\ge0\), \(x+y+z\le1\), why must the upper \(x\)-bound in the order \(dx\,dz\,dy\) depend on both \(y\) and \(z\)?
  3. For the region \(0\le x\le1\), \(x^2\le y\le1\), \(0\le z\le y\), what is the projection onto the \(yz\)-plane?
  4. Why is interchanging two inner variables straightforward in the second worked example but not automatically valid for every solid?
  5. What does it mean to verify that new limits cover the original solid exactly?

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