Changing the Order of Integration in Triple Integrals

Core Definition

Changing the order of integration in triple integrals is an analytical and geometric re-slicing method that rewrites an iterated triple integral across any of the $3! = 6$ possible differential volume arrangements:

dV = dz \, dy \, dx = dz \, dx \, dy = dy \, dz \, dx = dy \, dx \, dz = dx \, dz \, dy = dx \, dy \, dz

Because the inner boundaries depend functionally on the outer variables, the integral signs cannot simply be rearranged; one must extract the 3D boundary inequalities, re-project the solid onto a new coordinate plane, and formulate new functional and constant limits.

Concept Explanation

In multivariable calculus, changing the order of integration for a triple integral over a general 3D region is essential for two primary reasons:

Resolving Non-Elementary Antiderivatives: An integrand may be impossible to integrate with respect to the initial innermost variable using elementary functions (such as $\int e^{x^2} \, dx$ , $\int \sin(y^2) \, dy$ , or $\int \frac{\sin z}{z} \, dz$ ). Reorienting the solid so that a different variable is integrated first often introduces an algebraic multiplier (e.g., $x$ or $y$ ) from the boundary limits, immediately enabling standard $u$ -substitution.
Computational Simplification & Avoiding Region Splits: A 3D region may require splitting into multiple piecewise triple integrals in one orientation due to shifting ceiling or floor surfaces. Choosing an alternate projection plane often allows the solid to be expressed as a single, unified iterated integral.

The 5-Step 3D Re-slicing Protocol

To safely convert a triple integral from an initial order (e.g., $dz \, dy \, dx$ ) to a target order (e.g., $dx \, dz \, dy$ ):

Step 1: Extract the Original Boundary Inequalities
Read the given limits from outer constants to inner functional boundaries:
$a \le x \le b, \quad g_1(x) \le y \le g_2(x), \quad u_1(x, y) \le z \le u_2(x, y)$
Step 2: Identify the 3D Bounding Surfaces
Write out the equations of all bounding surfaces (planes, parabolic cylinders, spheres, cones) that enclose the solid $E$ .
Step 3: Identify the New Outer Projection Plane
Look at the outer two variables of your target order. For example, if the target order is $dx \, dz \, dy$ , the outer variables are $z$ and $y$ . Project the solid $E$ onto the $yz$ -coordinate plane to form a 2D shadow region $D_{yz}$ . Find the boundary of $D_{yz}$ by eliminating the inner variable ( $x$ ) across intersecting surface equations.
Step 4: Determine the New Inner Variable Limits
Solve the 3D boundary surface equations for the new inner variable (e.g., $x$ ). Determine the entrance surface $x = v_1(y, z)$ and exit surface $x = v_2(y, z)$ as viewed along the positive axis of the new inner variable.
Step 5: Determine Middle and Outer Limits on the 2D Projection
Treat the 2D shadow $D_{yz}$ exactly like a 2D double integral domain. Establish functional bounds for the middle variable and constant bounds for the outermost variable. If the entrance or exit surfaces change definitions over different portions of the shadow, partition the shadow into sub-regions and express the triple integral as a sum of two or more integrals.

Engineering Applications

Chemical & Environmental Engineering: Evaluating continuous reaction yields and pollutant degradation profiles in catalytic reaction columns where mass-transfer equations exhibit non-separable dependencies that are only analytically integrable after coordinate reordering.
Computational Fluid Dynamics & Computer Graphics: Optimizing volume-rendering algorithms and cell-sweeping passes in hexahedral finite-volume meshes, where reordering spatial sweep axes eliminates expensive boundary clipping operations.
Electromagnetics & Antenna Design: Integrating non-uniform 3D antenna beam patterns and charge density envelopes where reorienting integration axes aligns boundaries with physical ground planes.

Example Problems

Example 1: Overcoming a Non-Elementary Antiderivative

Consider the iterated triple integral:

I = \int_{1}^{4} \int_{0}^{1} \int_{2y}^{2} \frac{e^{x^2}}{z} \, dx \, dy \, dz

Rewrite the integral in the order

dy \, dx \, dz

and evaluate its exact value.

Solution

Step 1: Extract original bounds:

Outer ( $z$ ): $1 \le z \le 4$
Middle ( $y$ ): $0 \le y \le 1$
Inner ( $x$ ): $2y \le x \le 2$

Notice that $\int e^{x^2} \, dx$ has no elementary antiderivative; changing the inner order from $dx \, dy$ to $dy \, dx$ is required.

Step 2: Re-slice in the $xy$ -plane:
The outer variable $z$ remains unchanged ( $1 \le z \le 4$ ).
In the $xy$ -plane, the region is bounded by $x = 2y \implies y = \frac{x}{2}$ , $x = 2$ , and $y = 0$ .
When $y = 0$ , $x = 0$ . When $y = 1$ , $x = 2$ .
For the new order where $y$ is the inner variable:

$y$ sweeps vertically from the line $y = 0$ up to $y = \frac{x}{2}$ .
$x$ sweeps across the constant interval $0 \le x \le 2$ .

Step 3: Formulate and evaluate the rewritten integral:

I = \int_{1}^{4} \int_{0}^{2} \int_{0}^{x/2} \frac{e^{x^2}}{z} \, dy \, dx \, dz

Inner $y$ -integral:

\int_{0}^{x/2} \frac{e^{x^2}}{z} \, dy = \frac{e^{x^2}}{z} [y]_{0}^{x/2} = \frac{x e^{x^2}}{2z}

Middle $x$ -integral:
Use $u$ -substitution: let $u = x^2 \implies du = 2x \, dx \implies x \, dx = \frac{du}{2}$ .
When $x = 0, u = 0$ ; when $x = 2, u = 4$ .

\int_{0}^{2} \frac{x e^{x^2}}{2z} \, dx = \frac{1}{2z} \int_{0}^{4} \frac{1}{2} e^u \, du = \frac{1}{4z} [e^u]_{0}^{4} = \frac{e^4 - 1}{4z}

Outer $z$ -integral:

\int_{1}^{4} \frac{e^4 - 1}{4z} \, dz = \frac{e^4 - 1}{4} [\ln |z|]_{1}^{4} = \frac{e^4 - 1}{4} (\ln 4 - \ln 1) = \frac{(e^4 - 1)\ln 4}{4}

Example 2: Reordering across Intersecting Boundary Surfaces (Region Split)

Rewrite the triple integral $\int_{0}^{1} \int_{0}^{1-x^2} \int_{0}^{1-x} f(x, y, z) \, dz \, dy \, dx$ into the order $dx \, dy \, dz$ .

Solution

Step 1: Extract original bounds:

0 \le x \le 1, \quad 0 \le y \le 1 - x^2, \quad 0 \le z \le 1 - x

Step 2: Identify 3D surfaces:
The solid is bounded by $x = 0, y = 0, z = 0$ , the parabolic cylinder $y = 1 - x^2 \implies x = \sqrt{1 - y}$ , and the plane $z = 1 - x \implies x = 1 - z$ .

Step 3: Analyze new inner limits for $x$ :
Looking along the $x$ -axis, $x$ enters at the back plane $x = 0$ and exits through whichever surface is encountered first:

0 \le x \le \min(1 - z, \sqrt{1 - y})

Step 4: Find the boundary trace in the $yz$ -plane:
Equating the two upper boundaries:

1 - z = \sqrt{1 - y} \implies (1 - z)^2 = 1 - y \implies y = 1 - (1 - z)^2 = 2z - z^2

In the

yz

-plane where

0 \le z \le 1

and

0 \le y \le 1

, this parabolic curve

y = 2z - z^2

divides the shadow into two distinct regions:

Region 1 (where $0 \le y \le 2z - z^2$ ): The plane is the active upper bound: $x = 1 - z$ .
Region 2 (where $2z - z^2 \le y \le 1$ ): The cylinder is the active upper bound: $x = \sqrt{1 - y}$ .

Step 5: Write the sum of two iterated integrals:

\int_{0}^{1} \int_{0}^{2z-z^2} \int_{0}^{1-z} f(x, y, z) \, dx \, dy \, dz + \int_{0}^{1} \int_{2z-z^2}^{1} \int_{0}^{\sqrt{1-y}} f(x, y, z) \, dx \, dy \, dz

Connections & References

Parent Concepts: Triple Integrals over a General Region, Switching the Order of Integration
Sub-concepts & Extensions:
- Change of Variables in Multiple Integrals
Course Links:
- Calc-3 Session 16 - Triple Integrals

Tony's Intertwingled Memex • Reference Manual