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Writing All Six Cartesian Integration Orders

Derive the coordinate projections and bounds needed to express a triple integral over a solid in each of the six Cartesian orders.

Intermediate 12 min read

What You'll Learn

  • Read the written differentials to identify the outer, middle, and inner variables.
  • Derive a solid’s coordinate projections from its defining inequalities.
  • Write all six iterated-integral forms for a solid bounded by a plane and the coordinate planes.
  • Check whether bounds describe the original solid by tracing their projections and fibers.
  • Identify and correct a common error in matching integration bounds to differential order.

Why the Order Determines the Bounds

A triple integral can be written in six Cartesian orders because its three variables can be integrated in six permutations. The solid does not change when the order changes, but its description by coordinate projections and fibers does. Writing a correct iterated integral therefore requires more than rearranging differentials: each order needs bounds that describe the same solid.

In an expression such as \(\int\int\int f(x,y,z)\,dz\,dy\,dx\), the rightmost differential is the outermost variable. Here \(x\) is outermost, \(y\) is middle, and \(z\) is innermost. The innermost bounds describe a fiber in the \(z\)-direction; the bounds for \(y\) describe the projection needed to reach those fibers; and the outermost bounds describe the range of \(x\).

Definition: For a specified integration order, the outer variable is the variable paired with the rightmost differential. The middle variable is paired with the next differential, and the inner variable is paired with the leftmost differential. Bounds are written from outermost to innermost, while the differentials are written in the reverse order.

Projections and Fibers in Each Order

For an order with \(u\) outermost, \(v\) in the middle, and \(w\) innermost, first find the range of \(u\) covered by the solid. For each fixed \(u\), find the allowed range of \(v\). Then, for each fixed pair \((u,v)\), find the allowed range of \(w\). These ranges are the corresponding projections and fibers. The resulting integral is written with differentials \(dw\,dv\,du\).

For example, suppose a solid is described by \(0\leq x\leq 1\), \(0\leq y\leq 1\), and \(0\leq z\leq x\). For the order \(dz\,dy\,dx\), \(x\) ranges over its full projection, from 0 to 1. For each \(x\), \(y\) ranges from 0 to 1. For each \((x,y)\), the \(z\)-fiber ranges from 0 to \(x\). The bounds follow the nesting of the variables, not the order in which their names happen to appear in a sentence.

A General Derivation for a Plane-Bounded Solid

Consider a solid in the first octant bounded by a plane with positive coefficients. The first-octant condition means that all three coordinates are nonnegative. The following result gives the bounds for any choice of outer, middle, and inner variables.

Theorem: Let \(a,b,c,d\) be positive real numbers, and let \(x,y,z\geq 0\) satisfy \(ax+by+cz\leq d\). Choose any permutation of \(x,y,z\), naming the outer, middle, and inner variables \(u,v,w\), with respective coefficients \(a_u,a_v,a_w\). Then the bounds for the order \(dw\,dv\,du\) are
$$ 0\leq u\leq \frac{d}{a_u},\qquad 0\leq v\leq \frac{d-a_u u}{a_v},\qquad 0\leq w\leq \frac{d-a_u u-a_v v}{a_w}. $$

To justify the result, hold \(u\) and \(v\) fixed. The defining inequality becomes \(a_w w\leq d-a_u u-a_v v\), so the nonnegative \(w\)-fiber has the displayed endpoints whenever its upper endpoint is nonnegative. That condition is equivalent to \(0\leq v\leq (d-a_u u)/a_v\). Such \(v\)-values exist precisely when \(0\leq u\leq d/a_u\). Thus the bounds describe exactly the points satisfying the original inequality and nonnegativity conditions. If \(f\) is continuous on this bounded solid, the iterated integral in these bounds equals its triple integral by Fubini’s theorem.

Worked Example: Writing All Six Orders

Let \(E\) be the solid defined by \(x,y,z\geq 0\) and \(2x+3y+6z\leq 6\). The plane meets the coordinate axes at \(x=3\), \(y=2\), and \(z=1\), found by setting the other two coordinates to zero. In any chosen order, the outer coordinate runs from zero to its intercept. Fixing it leaves a two-variable inequality; fixing the middle coordinate then gives the inner fiber.

With \(x\) outermost, \(y\) middle, and \(z\) innermost, solve the plane inequality successively for \(z\), \(y\), and \(x\). This gives

$$ \int_0^3\int_0^{2-\frac{2x}{3}} \int_0^{1-\frac{x}{3}-\frac{y}{2}} f(x,y,z)\,dz\,dy\,dx. $$

With \(x\) outermost, \(z\) middle, and \(y\) innermost, solve successively for \(y\), \(z\), and \(x\):

$$ \int_0^3\int_0^{1-\frac{x}{3}} \int_0^{2-\frac{2x}{3}-2z} f(x,y,z)\,dy\,dz\,dx. $$

With \(y\) outermost, \(x\) middle, and \(z\) innermost, the intercept for \(y\) is 2, and fixing \(y\) leaves \(2x+6z\leq 6-3y\). Therefore

$$ \int_0^2\int_0^{3-\frac{3y}{2}} \int_0^{1-\frac{x}{3}-\frac{y}{2}} f(x,y,z)\,dz\,dx\,dy. $$

With \(y\) outermost, \(z\) middle, and \(x\) innermost, solving the remaining inequality for \(x\) gives

$$ \int_0^2\int_0^{1-\frac{y}{2}} \int_0^{3-\frac{3y}{2}-3z} f(x,y,z)\,dx\,dz\,dy. $$

With \(z\) outermost, \(x\) middle, and \(y\) innermost, the \(z\)-intercept is 1. After fixing \(z\), solve \(2x+3y\leq 6-6z\) for \(x\), then solve for \(y\):

$$ \int_0^1\int_0^{3-3z} \int_0^{2-\frac{2x}{3}-2z} f(x,y,z)\,dy\,dx\,dz. $$

Finally, with \(z\) outermost, \(y\) middle, and \(x\) innermost, solve the same remaining inequality for \(y\) and then \(x\):

$$ \int_0^1\int_0^{2-2z} \int_0^{3-\frac{3y}{2}-3z} f(x,y,z)\,dx\,dy\,dz. $$

Each outer interval ends at the corresponding axis intercept: 3 for \(x\), 2 for \(y\), and 1 for \(z\). Each inner upper bound is obtained by isolating the inner variable in the original plane inequality. As a numerical check that the bounds describe the same solid, set \(f=1\) and evaluate the first order. Integrating in \(z\) gives \(1-\frac{x}{3}-\frac{y}{2}\). Write \(Y=2-\frac{2x}{3}\). Then

$$ \begin{aligned} \int_0^3\int_0^Y\left(1-\frac{x}{3}-\frac{y}{2}\right)\,dy\,dx &=\int_0^3\left[\left(1-\frac{x}{3}\right)Y-\frac{Y^2}{4}\right]\,dx\\ &=\int_0^3\left[2\left(1-\frac{x}{3}\right)^2- \left(1-\frac{x}{3}\right)^2\right]\,dx\\ &=\int_0^3\left(1-\frac{x}{3}\right)^2\,dx\\ &=3\int_0^1(1-t)^2\,dt =3\left[ t-t^2+\frac{t^3}{3}\right]_0^1 =1. \end{aligned} $$

Here the substitution is \(t=x/3\), so \(dx=3\,dt\). The result is also consistent with the plane intercepts: the corresponding triangular cross-sections shrink to zero at the outer endpoint.

Checking an Order by Reconstructing the Solid

A useful check is to read an iterated integral from the inside out. The inner bounds must describe the permitted coordinate values for fixed values of the other two variables. The middle bounds must ensure that this fiber is nonempty, and the outer bounds must cover exactly the values for which such middle coordinates exist. At each step, substituting the proposed bounds into the original inequalities should preserve those inequalities.

Worked Example: Verifying an Alternate Order

For the same solid \(E\), take \(z\) outermost, \(y\) middle, and \(x\) innermost. The projection onto the \(z\)-axis is \(0\leq z\leq 1\), since \(6z\leq 6\). At a fixed \(z\), the remaining inequality is \(2x+3y\leq 6-6z\). Its projection onto the \(y\)-axis has \(0\leq y\leq 2-2z\), obtained by setting \(x=0\). For each such pair \((y,z)\), solve for \(x\):

$$ 0\leq x\leq \frac{6-3y-6z}{2} =3-\frac{3y}{2}-3z. $$

Thus the integral of 1 is

$$ \begin{aligned} \int_0^1\int_0^{2-2z} \int_0^{3-\frac{3y}{2}-3z}1\,dx\,dy\,dz &=\int_0^1\int_0^{2-2z} \left(3-\frac{3y}{2}-3z\right)\,dy\,dz\\ &=\int_0^1\left[3(1-z)y-\frac{3y^2}{4}\right]_0^{2-2z}\,dz\\ &=\int_0^1\left[6(1-z)^2-3(1-z)^2\right]\,dz\\ &=3\int_0^1(1-z)^2\,dz\\ &=3\left[z-z^2+\frac{z^3}{3}\right]_0^1 =1. \end{aligned} $$

The innermost upper bound is nonnegative throughout the stated \(y\)-range, reaching zero at \(y=2-2z\). The computed value agrees with the previous order.

All Six Orders for a Solid with a Sloping Boundary

The plane example has coordinate-plane faces, so all fibers start at zero. Other solids may have an inner fiber whose lower endpoint is not zero. The same projection-and-fiber method applies: solve both the lower and upper boundary conditions for the variable being integrated.

Worked Example: A Solid Below \(z=x\)

Let \(S\) be described by \(0\leq x\leq 1\), \(0\leq y\leq 1\), and \(0\leq z\leq x\). The \(y\)-coordinate is independent of the other constraints. For \(x\) outermost, \(y\) middle, and \(z\) innermost, the bounds are \(0\leq x\leq 1\), \(0\leq y\leq 1\), \(0\leq z\leq x\). For \(x\) outermost, \(z\) middle, and \(y\) innermost, the \(z\)-projection is \(0\leq z\leq x\), while \(y\) still ranges from 0 to 1. For \(y\) outermost, \(x\) middle, and \(z\) innermost, the original \(z\)-fiber remains \(0\leq z\leq x\). These give three orders:

$$ \begin{aligned} &\int_0^1\int_0^1\int_0^x f(x,y,z)\,dz\,dy\,dx,\\ &\int_0^1\int_0^x\int_0^1 f(x,y,z)\,dy\,dz\,dx,\\ &\int_0^1\int_0^1\int_0^x f(x,y,z)\,dz\,dx\,dy. \end{aligned} $$

For the other three orders, \(z\) is outermost or middle. The inequality \(z\leq x\leq 1\) implies \(0\leq z\leq 1\). At fixed \(z\), solving \(z\leq x\leq 1\) gives the \(x\)-range \(z\leq x\leq 1\). Since \(y\) remains independent, this yields

$$ \begin{aligned} &\int_0^1\int_z^1\int_0^1 f(x,y,z)\,dy\,dx\,dz,\\ &\int_0^1\int_0^1\int_z^1 f(x,y,z)\,dx\,dy\,dz,\\ &\int_0^1\int_0^1\int_z^1 f(x,y,z)\,dx\,dz\,dy. \end{aligned} $$

To check the bounds, evaluate the integral of 1 using the first and fourth forms. The first gives

$$ \int_0^1\int_0^1\int_0^x 1\,dz\,dy\,dx =\int_0^1\int_0^1 x\,dy\,dx =\int_0^1 x\,dx =\left[\frac{x^2}{2}\right]_0^1 =\frac12. $$

The fourth gives

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

The two orders agree, and the bounds in the second expression correctly account for the nonzero lower endpoint \(x=z\).

A Common Mistake: Confusing Bounds with Differential Order

Suppose a learner writes

$$ \int_0^1\int_0^{2-2z}\int_0^{3-\frac{3y}{2}-3z} f(x,y,z)\,dx\,dy\,dz $$

and calls it the order with \(x\) outermost because the \(x\)-bound appears first in their written description. That is incorrect. The differentials are \(dx\,dy\,dz\), so \(z\) is outermost, \(y\) is middle, and \(x\) is innermost. In this particular expression the bounds are valid for \(E\), but they describe the \(z\)-outermost order.

To make \(x\) outermost, use the bounds derived from the projection onto the \(x\)-axis and write the differentials in reverse order:

$$ \int_0^3\int_0^{2-\frac{2x}{3}} \int_0^{1-\frac{x}{3}-\frac{y}{2}} f(x,y,z)\,dz\,dy\,dx. $$

Before checking the algebra of the bounds, identify the variables from the differentials, starting at the right. Then verify that each upper limit describes the appropriate projection or fiber.

Why Writing All Six Orders Matters

Writing all six forms makes explicit how coordinate projections depend on the direction of slicing. A variable that has a constant range in one order may have a range depending on another coordinate in a different order. The same solid can therefore have simple bounds in some orders and complicated or piecewise bounds in others. Correctly deriving the bounds is what preserves the region—and hence the value of the integral—when the order changes.

Key takeaway: For each integration order, determine the outer projection first, then the middle projection, then the inner fiber. Write the differentials from innermost to outermost, and check the resulting bounds against the defining inequalities.

Check Your Understanding

For each question, pay attention to both the projections and the order of the differentials.

  1. In an iterated integral written with differentials \(dy\,dz\,dx\), which variable is outermost, and which is innermost?
  2. For \(x,y,z\geq 0\) and \(x+2y+z\leq 4\), what are the bounds when \(y\) is outermost, \(z\) is middle, and \(x\) is innermost?
  3. For the solid \(0\leq x\leq 1\), \(0\leq y\leq 1\), \(0\leq z\leq x\), what is the lower bound for \(x\) when \(z\) is fixed and \(z\) is outermost?
  4. Why must the middle-variable bounds ensure that the inner fiber is nonempty?
  5. What does the differential string \(dx\,dy\,dz\) tell you about the outermost variable?

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