Choosing an Integration Order from the Geometry

Why the Integration Order Matters

A triple integral over a solid can often be written in more than one Cartesian order. The choice is not just a matter of notation: an order determines which variable is integrated first, the bounds that describe its fibers, and the antiderivatives that must be found.

A good choice usually makes both the bounds and the inner antiderivative simple. These goals can compete. A geometrically convenient order may put a difficult function in the innermost integral, while an order that suits the function may require a complicated projection or piecewise bounds. The practical task is to compare the candidates before calculating.

For an order such as \(dz\,dy\,dx\), the innermost variable is \(z\). One first fixes \(x\) and \(y\), then describes the allowed \(z\)-values. The outer bounds describe the projection of the solid onto the \(xy\)-plane. The same principle applies to the other five orders: the innermost bounds describe fibers parallel to that variable’s axis, and the outer bounds describe the successive projections.

Order-selection principle: Prefer an order for which the solid has simple projections and fibers, the boundary equations give manageable bounds, and the innermost antiderivative is straightforward. If these considerations point to different orders, compare the actual candidate integrals rather than relying on geometry alone.

A Systematic Comparison

Begin by listing the solid’s defining inequalities and identifying which surfaces form its boundaries. For each plausible order, ask three questions:

Projection: Is the projection onto the plane of the two outer variables a simple region?
Fibers and boundaries: Can the remaining variable be bounded with one lower and one upper function throughout that projection?
Integrand: Is the inner antiderivative elementary and simpler than the alternatives?

A candidate order is especially attractive when it avoids solving boundary equations for a variable in a complicated way. Also check whether a proposed upper or lower bound changes across the projection. A minimum or maximum of two boundary functions can signal that a single set of bounds will not work throughout.

There is no rule that the innermost variable must be the one appearing most simply in the geometry, nor that the outermost variable must be the one appearing most simply in the integrand. The decision is about the whole iterated integral. The examples below show how the balance changes from one problem to another.

Worked Example: A Simple Top Surface

Worked Example: Let the Surface Suggest the Inner Variable

Let \(D\) be the solid defined by \(0\le x\le1\), \(0\le y\le1\), and \(0\le z\le x+y\). Choose a convenient order and evaluate \(\iiint_D z\,dV\).

The projection onto the \(xy\)-plane is the unit square, and each vertical fiber runs from \(z=0\) to the single surface \(z=x+y\). Thus \(dz\,dy\,dx\) uses simple bounds. It also integrates the factor \(z\) immediately:

\begin{aligned} \iiint_D z\,dV &=\int_0^1\int_0^1\int_0^{x+y} z\,dz\,dy\,dx\\ &=\int_0^1\int_0^1 \left[\frac{z^2}{2}\right]_{0}^{x+y}\,dy\,dx\\ &=\frac12\int_0^1\int_0^1 (x+y)^2\,dy\,dx\\ &=\frac12\int_0^1\int_0^1 (x^2+2xy+y^2)\,dy\,dx\\ &=\frac12\int_0^1\left(x^2+x+\frac13\right)\,dx\\ &=\frac12\left[\frac{x^3}{3}+\frac{x^2}{2}+\frac{x}{3}\right]_0^1\\ &=\frac12\left(\frac13+\frac12+\frac13\right) =\frac{7}{12}. \end{aligned}

The inner antiderivative is correct because \(\frac{d}{dz}(z^2/2)=z\); substituting the fiber endpoints gives \(((x+y)^2-0)/2\). The remaining integrals give \(1/3+1/2+1/3=7/6\), and the outer factor \(1/2\) gives \(7/12\). In this case the top boundary and the integrand both support the same choice: integrate \(z\) first.

The Integrand Can Change the Best Choice

Simple bounds do not automatically make an order efficient. The variable in the inner integral determines which antiderivative is needed. A function that is awkward to integrate over a moving interval may become easy if that variable is placed outside instead. In that case, the outer integrations need to account for the cross-sectional area or length left by the inner variables.

Worked Example: Put a Difficult Inner Antiderivative Outside

Let \(D\) be the tetrahedron \(x\ge0\), \(y\ge0\), \(z\ge0\), \(x+y+z\le1\), and let \(f(x,y,z)=1/(1+x^2)\). Choose an order that avoids making an inverse-tangent expression the integrand of another integral, and evaluate \(\iiint_D f\,dV\).

For fixed \(x\), the remaining cross-section satisfies \(0\le y\le1-x\) and \(0\le z\le1-x-y\). Taking \(x\) outermost gives the order \(dz\,dy\,dx\); the inner integrations involve only the constant \(1/(1+x^2)\):

\begin{aligned} \iiint_D \frac{1}{1+x^2}\,dV &=\int_0^1\int_0^{1-x}\int_0^{1-x-y} \frac{1}{1+x^2}\,dz\,dy\,dx\\ &=\int_0^1\int_0^{1-x} \frac{1-x-y}{1+x^2}\,dy\,dx\\ &=\int_0^1 \frac{(1-x)^2}{2(1+x^2)}\,dx\\ &=\frac12\int_0^1 \frac{1-2x+x^2}{1+x^2}\,dx\\ &=\frac12\int_0^1\left(1-\frac{2x}{1+x^2}\right)\,dx\\ &=\frac12\left[x-\ln(1+x^2)\right]_0^1\\ &=\frac{1-\ln 2}{2}. \end{aligned}

To check the last antiderivative, differentiate \(x-\ln(1+x^2)\): its derivative is \(1-2x/(1+x^2)\), the displayed integrand. The value at the upper endpoint is \(1-\ln2\); at the lower endpoint it is \(0\).

If \(x\) were integrated first instead, the inner integral would be \(\int_0^{1-y-z}1/(1+x^2)\,dx=\arctan(1-y-z)\). The subsequent integrals of that moving inverse-tangent expression are less convenient. The chosen order lets the \(y\) and \(z\) integrations reduce to the cross-sectional area factor \((1-x)^2/2\), leaving one elementary integral.

When a Projection Creates a Minimum

Some solids are simple when viewed along one axis but not another. A frequent warning sign is that a fiber has to satisfy two upper bounds. Its actual upper endpoint is the smaller of the two, so the endpoint may change over the projection. An order that keeps both constraints visible can avoid this change.

Worked Example: Avoid a Piecewise Fiber Bound

Let \(D\) be defined by \(x,y,z\ge0\), \(x+y\le1\), and \(x+z\le1\). Choose an order that avoids a changing upper bound for the innermost variable, then find the volume of \(D\).

With \(x\) outermost, fixing \(x\) leaves the independent ranges \(0\le y\le1-x\) and \(0\le z\le1-x\). Hence the \(yz\)-cross-section is a square and \(dz\,dy\,dx\) has simple bounds:

\begin{aligned} \operatorname{Vol}(D) &=\int_0^1\int_0^{1-x}\int_0^{1-x}1\,dz\,dy\,dx\\ &=\int_0^1\int_0^{1-x}(1-x)\,dy\,dx\\ &=\int_0^1(1-x)^2\,dx\\ &=\left[x-x^2+\frac{x^3}{3}\right]_0^1 =\frac13. \end{aligned}

The antiderivative differentiates to \(1-2x+x^2=(1-x)^2\), and its endpoint difference is \(1-1+1/3=1/3\). By contrast, if \(x\) is innermost, then for fixed \(y,z\), its upper bound is \(\min(1-y,1-z)\). This endpoint changes depending on whether \(y\le z\) or \(z\le y\). The outer projection is a square, but one uniform formula for the fiber bound does not describe the whole square.

Why Valid Orders Agree

Choosing an order changes the description and the calculation, not the value of the integral, provided every set of bounds describes the same solid and the integrand is integrable. The relevant principle is Fubini’s theorem, applied to the solid rather than just to a box.

Theorem: If \(D\) is a bounded solid with a piecewise-smooth boundary and \(f\) is continuous on \(D\), then the iterated integrals obtained from valid descriptions of \(D\) have the same value as \(\iiint_D f\,dV\). Each description must include the full projection and the correct fiber bounds.

Here is the reason the theorem applies. Choose a rectangular box \(B\) containing \(D\), and define \(g\) on \(B\) to equal \(f\) on \(D\) and \(0\) outside \(D\). Under the stated boundary and continuity conditions, \(g\) is integrable. Fubini’s theorem allows its triple integral over \(B\) to be evaluated in any variable order. In each order, the points where \(g\) is nonzero are precisely those in \(D\); describing those points by their projections and fibers gives the corresponding iterated integral over \(D\). Thus a correctly bounded integral in any order represents the same quantity.

This argument also explains why missing a constraint changes the answer: the resulting bounds integrate over a different set of points. A convenient order is useful only if its bounds still represent the entire solid exactly.

A Common Bounds Error

For the solid in the previous example, suppose one uses \(0\le y,z\le1\) and \(0\le x\le1-y\). This enforces \(x+y\le1\) but ignores \(x+z\le1\). It therefore gives the wrong volume:

\int_0^1\int_0^1\int_0^{1-y}1\,dx\,dz\,dy =\int_0^1(1-y)\,dy =\left[y-\frac{y^2}{2}\right]_0^1 =\frac12.

The error is not in the integration: the bounds describe a larger region than \(D\). For fixed \(y,z\), the correct upper bound for \(x\) is \(\min(1-y,1-z)\), which requires accounting for where the two expressions switch. Alternatively, use \(x\) as the outer variable, as above, and keep both constraints in the square cross-section. The correct volume is \(1/3\), not \(1/2\).

Make the Choice Before Integrating

A short comparison can prevent substantial unnecessary work. First sketch or otherwise inspect the solid and its coordinate-plane projections. Next, for each candidate order, solve the boundary equations only as far as needed to identify the fiber endpoints. Finally, inspect the inner antiderivative of the actual integrand, not just the shape of the bounds.

A single uncomplicated set of bounds is preferable when available. If a candidate order produces a minimum, maximum, or a boundary that changes across its projection, it may need a piecewise description. Sometimes that order is still worthwhile, but the extra regions must be included. Comparing the alternatives can reveal an order that avoids both piecewise bounds and difficult antiderivatives.

Read the constraints.
Identify the surfaces and inequalities that limit the solid.

Compare projections.
Look for an order with a simple projected region and uncomplicated fibers.

Test the inner integral.
Check whether the integrand has a manageable antiderivative over the proposed fiber.

Verify the region.
Confirm that every constraint is represented and the bounds cover the entire solid exactly once.

Check Your Understanding

For each question, explain how geometry and the integrand affect the choice of order.

What do the inner bounds and outer bounds describe in an iterated triple integral?
Why might a simple projection still lead to an inconvenient integration order?
In the tetrahedron example, what made it useful to integrate with respect to \(x\) last?
What does a minimum of two boundary functions indicate about a fiber bound?
Why did the incorrect bounds in the common-error example give a volume that was too large?