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Ordering Spherical Integrals

Choose spherical integration limits to match a solid, and learn how to change the order without losing or adding part of the region.

Intermediate 10 min read

What You'll Learn

  • Read an iterated spherical integral from its bounds, starting with the innermost variable.
  • Identify when a spherical parameter region has independent bounds and when its bounds depend on another variable.
  • Reverse angular integration order by describing the same region with new limits.
  • Recognize when reversing an order requires a piecewise description or a region split.
  • Check spherical limits against the defining inequalities and compare equivalent orders.

Why the Order of Integration Matters

Spherical coordinates describe a point using \(\rho\), the distance from the origin; \(\phi\), the angle measured down from the positive \(z\)-axis; and \(\theta\), the azimuthal angle in the \(xy\)-plane. With this convention, the volume element is \(\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\). Choosing an order for integration means choosing which variable is integrated first and how the bounds for that variable depend on the others.

A solid may have simple bounds in one order and complicated bounds in another. Changing the order does not change the solid or the value of an integral, provided the new bounds describe exactly the same region. But the limits usually cannot simply be copied into a different order: they must be derived again from the region.

Key idea: An iterated spherical integral is an integral over a region in \((\rho,\phi,\theta)\)-parameter space, with the volume factor \(\rho^2\sin\phi\). To change the order, describe that same parameter region using bounds suited to the new order.

Reading Spherical Bounds

In an iterated integral, the rightmost differential indicates the outermost integration, and the leftmost differential indicates the innermost. For example, \(d\rho\,d\phi\,d\theta\) means integrate first in \(\rho\), then in \(\phi\), and finally in \(\theta\). The bounds for \(\rho\) may depend on \(\phi\) and \(\theta\); the bounds for \(\phi\) may depend on \(\theta\); and the outermost variable has constant limits.

There are six possible orders for three variables: \(\rho,\phi,\theta\); \(\rho,\theta,\phi\); \(\phi,\rho,\theta\); \(\phi,\theta,\rho\); \(\theta,\rho,\phi\); and \(\theta,\phi,\rho\). A particular order is useful only if its bounds can be stated accurately and conveniently. The geometric description of the solid determines which choices are straightforward.

For instance, if a region is described by \(0\leq \rho\leq 2\), \(0\leq\phi\leq\pi\), and \(0\leq\theta\leq2\pi\), all three parameter ranges are independent. Any order is then possible with the same constant limits. If instead the region includes an inequality such as \(\phi\geq\theta\), the allowable values of one angle depend on the other, and the limits must reflect that relationship.

Definition: A parameter region has independent bounds when each variable ranges over a fixed interval regardless of the values of the other variables. When one variable's allowable range depends on another, the bounds are dependent and must be written in an order that captures that dependence.

Worked Example: A Full Ball with Independent Bounds

Worked Example: A Full Ball with Independent Bounds

Set up and evaluate the volume of the ball \(\rho\leq2\), using the order \(\rho\), then \(\phi\), then \(\theta\). In spherical coordinates the full ball has \(0\leq\rho\leq2\), \(0\leq\phi\leq\pi\), and \(0\leq\theta\leq2\pi\). These bounds do not depend on one another.

The volume integral includes the spherical volume element:

$$ V=\int_0^{2\pi}\int_0^\pi\int_0^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta. $$

First integrate with respect to \(\rho\):

$$ \int_0^2 \rho^2\sin\phi\,d\rho =\sin\phi\left[\frac{\rho^3}{3}\right]_0^2 =\frac{8}{3}\sin\phi. $$

Then integrate with respect to \(\phi\):

$$ \int_0^\pi \frac{8}{3}\sin\phi\,d\phi =\frac{8}{3}\left[-\cos\phi\right]_0^\pi =\frac{8}{3}(1-(-1)) =\frac{16}{3}. $$

Finally integrate with respect to \(\theta\):

$$ V=\int_0^{2\pi}\frac{16}{3}\,d\theta =\frac{16}{3}[\,\theta\,]_0^{2\pi} =\frac{32\pi}{3}. $$

This is the ball's volume. Since the three bounds are independent, the same constant intervals can be used in any of the six integration orders. The factors in the volume element still belong in the integrand; changing the order does not remove \(\rho^2\sin\phi\).

Changing Order by Rewriting the Region

When bounds depend on one another, changing order is a geometric task before it is a calculation. A reliable method is to identify the inequalities defining the parameter region, choose the variable to integrate outermost, and determine the full range of that variable. Then, for each fixed outer value, find the allowable range of the next variable, and continue inward.

The reason this works is that the integral covers the same points in parameter space regardless of the order of integration. For a continuous integrand on a bounded region with an appropriately described boundary, the order of integration can be changed while preserving the integral, as long as the new bounds cover each point of that region exactly once, apart from boundary points. This is the relevant iterated-integral form of the Fubini theorem.

1
Write the region as inequalities.
Keep the restrictions on \(\rho\), \(\phi\), and \(\theta\) visible instead of relying only on a diagram.
2
Choose the outer variable.
Find the complete range of that variable across the whole region.
3
Find the remaining bounds.
Hold the outer variables fixed and determine the allowable range of each inner variable.
4
Check coverage.
Confirm that every point in the original region is included and that no point outside it is included. Split the region if one set of bounds cannot do this.

Worked Example: Reversing Dependent Angular Bounds

Worked Example: Reversing Dependent Angular Bounds

Consider the solid described by \(0\leq\rho\leq2\), \(0\leq\theta\leq\pi/2\), and \(\theta\leq\phi\leq\pi/2\). Find its volume in two angular orders. The angular restriction says that for each \(\theta\), \(\phi\) begins at \(\theta\) and ends at \(\pi/2\). Equivalently, for each \(\phi\), \(\theta\) ranges from \(0\) to \(\phi\).

First use \(\rho\) inside, then \(\phi\), then \(\theta\):

$$ V=\int_0^{\pi/2}\int_\theta^{\pi/2}\int_0^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta. $$

The radial integral is \(\int_0^2\rho^2\,d\rho=8/3\). Thus

$$ \begin{aligned} V &=\frac{8}{3}\int_0^{\pi/2}\int_\theta^{\pi/2}\sin\phi\,d\phi\,d\theta\\ &=\frac{8}{3}\int_0^{\pi/2} [-\cos\phi]_\theta^{\pi/2}\,d\theta\\ &=\frac{8}{3}\int_0^{\pi/2}\cos\theta\,d\theta\\ &=\frac{8}{3}[\sin\theta]_0^{\pi/2} =\frac{8}{3}. \end{aligned} $$

Now reverse the angular order. The same angular region can be written \(0\leq\phi\leq\pi/2\), \(0\leq\theta\leq\phi\). Therefore

$$ \begin{aligned} V &=\int_0^{\pi/2}\int_0^\phi\int_0^2 \rho^2\sin\phi\,d\rho\,d\theta\,d\phi\\ &=\frac{8}{3}\int_0^{\pi/2}\int_0^\phi\sin\phi\,d\theta\,d\phi\\ &=\frac{8}{3}\int_0^{\pi/2}\phi\sin\phi\,d\phi\\ &=\frac{8}{3}[-\phi\cos\phi+\sin\phi]_0^{\pi/2} =\frac{8}{3}. \end{aligned} $$

The two orders agree. In the first order, the inner angular integral is \(\int_\theta^{\pi/2}\sin\phi\,d\phi=\cos\theta\), not \(1-\cos\theta\). The latter expression would come from different limits, \(0\) to \(\theta\), and would describe a different angular slice.

When Reversing Order Requires a Split

Some regions have bounds that change form partway across their range. In that case, reversing the order may require splitting the original description into pieces—or may allow a single, simpler description in the new order. The key is to locate where the boundary expressions meet and determine which one controls the region on each side.

Worked Example: Changing Order Across a Piecewise Angular Region

Let \(0\leq\rho\leq1\), \(0\leq\theta\leq\pi/2\), and impose the angular bounds \(0\leq\phi\leq\theta\) when \(0\leq\theta\leq\pi/4\), and \(0\leq\phi\leq\pi/2-\theta\) when \(\pi/4\leq\theta\leq\pi/2\). The two upper boundaries meet at \(\theta=\pi/4\). Find the volume, first with \(\theta\) outermost among the angles, then with \(\phi\) outermost.

For the first order, the region must be written in two pieces. Integrating in \(\rho\) gives \(\int_0^1\rho^2\,d\rho=1/3\), so

$$ \begin{aligned} V &=\frac13\left[ \int_0^{\pi/4}\int_0^\theta\sin\phi\,d\phi\,d\theta +\int_{\pi/4}^{\pi/2}\int_0^{\pi/2-\theta}\sin\phi\,d\phi\,d\theta \right]\\ &=\frac13\left[ \int_0^{\pi/4}(1-\cos\theta)\,d\theta +\int_{\pi/4}^{\pi/2}(1-\sin\theta)\,d\theta \right]\\ &=\frac13\left[ \left(\frac{\pi}{4}-\frac{\sqrt2}{2}\right) +\left(\frac{\pi}{4}-\frac{\sqrt2}{2}\right) \right]\\ &=\frac13\left(\frac{\pi}{2}-\sqrt2\right). \end{aligned} $$

To reverse the order, note that \(\phi\) cannot exceed either \(\theta\) or \(\pi/2-\theta\). The largest possible \(\phi\) occurs where those two upper restrictions meet, at \(\phi=\pi/4\). For each fixed \(0\leq\phi\leq\pi/4\), the inequalities \(\phi\leq\theta\) and \(\phi\leq\pi/2-\theta\) give \(\phi\leq\theta\leq\pi/2-\phi\). Hence one integral suffices:

$$ \begin{aligned} V &=\frac13\int_0^{\pi/4}\int_\phi^{\pi/2-\phi}\sin\phi\,d\theta\,d\phi\\ &=\frac13\int_0^{\pi/4}\left(\frac{\pi}{2}-2\phi\right)\sin\phi\,d\phi\\ &=\frac13\left[ \frac{\pi}{2}(1-\cos(\pi/4)) -2\bigl(-\tfrac{\pi}{4}\cos(\pi/4)+\sin(\pi/4)\bigr) \right]\\ &=\frac13\left[ \frac{\pi}{2}-\frac{\pi\sqrt2}{4} +\frac{\pi\sqrt2}{4}-\sqrt2 \right]\\ &=\frac13\left(\frac{\pi}{2}-\sqrt2\right). \end{aligned} $$

The two calculations agree. The piecewise description in the first order is necessary because the upper bound switches at \(\theta=\pi/4\). When \(\phi\) is outermost, the same region has a single set of inner bounds.

A Common Mistake: Reversing Differentials but Not Bounds

Suppose the angular region in the previous example is described in the \((\theta,\phi)\)-plane. A tempting but incorrect reversal is to keep \(0\leq\theta\leq\pi/2\) while allowing \(0\leq\phi\leq\pi/4\) for every \(\theta\). That rectangular set contains points that violate the original restrictions. For example, \(\theta=0\), \(\phi=\pi/4\) is included in that rectangle, but it fails \(\phi\leq\theta\).

The correct reversed limits are \(0\leq\phi\leq\pi/4\) and \(\phi\leq\theta\leq\pi/2-\phi\). These enforce both inequalities defining the region. A quick check is to test boundary values: at \(\phi=0\), the full range \(0\leq\theta\leq\pi/2\) is allowed; at \(\phi=\pi/4\), the bounds force \(\theta=\pi/4\), exactly where the two boundaries meet.

Practical check: After changing order, substitute a few boundary values into the proposed limits and compare them with the original inequalities. If the new bounds allow a point outside the original region or exclude a point inside it, the reversal is incorrect.

Choosing an Order Efficiently

The best integration order is usually the one that makes the bounds easiest to state and the calculation simplest. For a spherical solid, radial bounds often depend on the angular variables, while planes, cones, and wedges can impose relationships between angles. The region's geometry—not a preference for one differential order—determines which bounds are valid.

  • Use independent ranges when the parameter region is a product of intervals.
  • For dependent angular ranges, express the defining inequalities before changing order.
  • Look for intersections of boundary curves; they often indicate where a bound changes or a split is needed.
  • Keep the factor \(\rho^2\sin\phi\) in every volume integral, regardless of the order.
  • Verify that the new limits describe the same parameter region, then check that equivalent orders give the same value.

These checks matter because an iterated integral is only as accurate as its bounds. A correct antiderivative cannot repair a region that was set up incorrectly. Conversely, once the limits precisely describe the solid, changing order is a controlled way to choose a more convenient calculation.

Check Your Understanding

Use the bounds and region descriptions in each question to justify your answers.

  1. What does the order \(d\rho\,d\phi\,d\theta\) tell you about which variable is integrated first?
  2. For \(0\leq\theta\leq\pi/2\) and \(\theta\leq\phi\leq\pi/2\), what are the angular bounds when \(\phi\) is outermost?
  3. Why can independent spherical parameter bounds be used in any integration order?
  4. In the piecewise region above, why does the first angular description require two integrals?
  5. What two checks can reveal that reversed bounds describe the wrong region?

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