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Splitting Regions with Piecewise Spherical Bounds

Partition a spherical-coordinate region wherever an angular threshold changes its radial bounds, then set up and evaluate the resulting triple integrals.

Intermediate 10 min read

What You'll Learn

  • Interpret piecewise inner and outer radial bounds in spherical coordinates
  • Find angular thresholds by solving where competing surfaces intersect
  • Split angular intervals so each integral has consistent bounds
  • Set up volume integrals using the spherical-coordinate volume element
  • Check that radial intervals are nonempty and adjacent pieces meet correctly
  • Evaluate piecewise spherical volume integrals and verify the calculations

Why a Spherical Region May Need to Be Split

A solid can have a boundary described by more than one surface. Along some directions, one surface determines the outer edge of the solid; along other directions, a different surface does. The same issue occurs at the inner edge: whichever of two constraints is more restrictive can change as the angle changes.

In spherical coordinates, this change is handled by splitting the angular interval at the direction where the competing radial bounds are equal. Each resulting integral then has bounds that describe the solid consistently throughout that part of the angle range.

Key idea: If a radial bound is defined by different formulas on different angular intervals, split the integral at every threshold where the formula changes. The pieces must cover the solid without gaps or positive-volume overlap.

Spherical Coordinates and Piecewise Bounds

We use spherical coordinates \((\rho,\theta,\phi)\), where \(\rho\) is the distance from the origin, \(\theta\) is the azimuthal angle in the \(xy\)-plane, and \(\phi\) is measured down from the positive \(z\)-axis. Thus \(0\leq\phi\leq\pi\), and a full turn in \(\theta\) can be represented by \(0\leq\theta\leq2\pi\). The volume element is \(\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\).

Definition: A spherical-coordinate description with piecewise radial bounds specifies angular intervals and, on each interval, a lower radial bound \(\rho_{\min}\) and an upper radial bound \(\rho_{\max}\). The radial interval is valid only where \(0\leq\rho_{\min}\leq\rho_{\max}\).

For a region with fixed angular ranges and radial bounds \(\rho_{\min}(\theta,\phi)\leq\rho\leq\rho_{\max}(\theta,\phi)\), its volume is found by integrating the volume element over those bounds. If the radial formulas change across an angular threshold, the volume is the sum of the integrals over the resulting pieces.

$$ V=\iiint_E \rho^2\sin\phi\,d\rho\,d\phi\,d\theta. $$

The threshold comes from comparing the candidate bounds. For an outer boundary, take the smaller of competing upper bounds; for an inner boundary, take the larger of competing lower bounds. Find where the formulas are equal, and use that angle to divide the region. Then check which formula controls on each side.

Why Splitting Gives the Correct Integral

Suppose an angular interval is divided at \(\phi=\alpha\), and the radial bound has one formula for \(a\leq\phi\leq\alpha\) and another for \(\alpha\leq\phi\leq b\). The two angular pieces meet at the same threshold. They do not overlap in their interiors, and together they cover the original interval. The boundary \(\phi=\alpha\) itself has zero volume, so assigning it to either piece does not change the integral.

By additivity of integration over adjacent regions, integrating over both pieces and adding the results gives the integral over the whole solid. The same reasoning applies when the threshold is in \(\theta\), or when the inner rather than the outer radial formula changes.

$$ \int_a^b f(\phi)\,d\phi = \int_a^\alpha f(\phi)\,d\phi + \int_\alpha^b f(\phi)\,d\phi. $$

For a three-dimensional region, \(f\) includes the radial integral and the spherical volume factor. The essential requirement is that the pieces describe the intended region and that their angular intervals partition it.

Worked Example: An Outer Bound Changes from a Plane to a Sphere

Worked Example: An Outer Bound Changes from a Plane to a Sphere

Find the volume inside the sphere \(\rho=3\) and below the plane \(z=1\). In spherical coordinates, \(z=\rho\cos\phi\). The plane condition is \(\rho\cos\phi\leq1\).

For \(0\leq\phi<\pi/2\), the plane condition gives \(\rho\leq\sec\phi\), while the sphere gives \(\rho\leq3\). The controlling outer bound changes where \(\sec\phi=3\), so define \(\alpha=\arccos(1/3)\). For \(0\leq\phi\leq\alpha\), the plane is the nearer boundary. For \(\alpha\leq\phi\leq\pi\), the sphere is the nearer boundary or the plane condition is automatically satisfied. The solid includes every azimuthal angle and begins at \(\rho=0\).

$$ V= \int_0^{2\pi}\int_0^\alpha\int_0^{\sec\phi} \rho^2\sin\phi\,d\rho\,d\phi\,d\theta + \int_0^{2\pi}\int_\alpha^\pi\int_0^3 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta. $$

Integrating first with respect to \(\rho\) gives \(\int_0^R \rho^2\,d\rho=R^3/3\). Therefore, after integrating over \(\theta\),

$$ V=\frac{2\pi}{3} \left( \int_0^\alpha \sec^3\phi\sin\phi\,d\phi + 27\int_\alpha^\pi\sin\phi\,d\phi \right). $$

Since \(\cos\alpha=1/3\), \(\tan^2\alpha=1/\cos^2\alpha-1=8\). An antiderivative of \(\sec^3\phi\sin\phi\) is \(\tfrac12\tan^2\phi\), because its derivative is \(\tan\phi\sec^2\phi=\sin\phi/\cos^3\phi\). Thus the first angular integral is \(\tfrac12(8-0)=4\). The second is \(27[-\cos\phi]_\alpha^\pi=27(1+\tfrac13)=36\). Hence \(V=\frac{2\pi}{3}(4+36)=\frac{80\pi}{3}\).

As a check, the two outer bounds agree at the split: \(\sec\alpha=3\). Also, differentiating the stated antiderivative gives the original first integrand, and the second angular antiderivative \(-\cos\phi\) differentiates to \(\sin\phi\). The radial bounds therefore meet at the threshold and the evaluated integrals use the correct bounds.

Worked Example: An Inner Bound Changes from a Sphere to a Plane

Worked Example: An Inner Bound Changes from a Sphere to a Plane

Find the volume of the region satisfying \(2\leq\rho\leq3\) and \(z\geq1\). Because \(z=\rho\cos\phi\), the condition \(z\geq1\) requires \(\cos\phi>0\) and \(\rho\geq\sec\phi\). The solid can exist only for \(0\leq\phi\leq\beta\), where \(\beta=\arccos(1/3)\): beyond that angle, \(\sec\phi>3\), so no point can satisfy both \(\rho\geq\sec\phi\) and \(\rho\leq3\).

The inner bound is the larger of \(2\) and \(\sec\phi\). They are equal at \(\alpha=\arccos(1/2)=\pi/3\). On \(0\leq\phi\leq\alpha\), the inner bound is \(2\); on \(\alpha\leq\phi\leq\beta\), it is \(\sec\phi\). The outer bound is \(3\) throughout.

$$ V= \int_0^{2\pi}\int_0^\alpha\int_2^3 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta + \int_0^{2\pi}\int_\alpha^\beta\int_{\sec\phi}^3 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta. $$

Integrating in \(\rho\) and then \(\theta\) gives

$$ V=\frac{2\pi}{3} \left( 19\int_0^\alpha\sin\phi\,d\phi + \int_\alpha^\beta(27-\sec^3\phi)\sin\phi\,d\phi \right), $$

because \(3^3-2^3=27-8=19\). For the first angular integral, \(\int_0^\alpha\sin\phi\,d\phi=1-\cos\alpha=1/2\), so its contribution inside the parentheses is \(19/2\). For the second, \(\cos\alpha=1/2\), \(\cos\beta=1/3\), \(\tan^2\alpha=3\), and \(\tan^2\beta=8\). Therefore,

$$ \int_\alpha^\beta(27-\sec^3\phi)\sin\phi\,d\phi = 27\left(\frac12-\frac13\right) -\frac12(8-3) = \frac92-\frac52 = 2. $$

It follows that \(V=\frac{2\pi}{3}(\frac{19}{2}+2)=\frac{23\pi}{3}\). As a check, the radial interval at the split is valid: both inner formulas give \(2\), which is below the outer bound \(3\). At the final angle, \(\sec\beta=3\), so the radial interval closes exactly. The antiderivative \(\tfrac12\tan^2\phi\) differentiates to \(\sec^3\phi\sin\phi\), confirming the term used in the evaluation.

Worked Example: A Bound Changes at an Azimuthal Threshold

Worked Example: A Bound Changes at an Azimuthal Threshold

Consider the region \(0\leq\phi\leq\pi/2\), \(0\leq\theta\leq\pi\), and \(0\leq\rho\leq R(\theta)\), where \(R(\theta)=2\) for \(0\leq\theta\leq\pi/2\) and \(R(\theta)=1\) for \(\pi/2\leq\theta\leq\pi\). The outer radial boundary changes at the azimuthal threshold \(\theta=\pi/2\), so split the integral there.

$$ V= \int_0^{\pi/2}\int_0^{\pi/2}\int_0^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta + \int_{\pi/2}^{\pi}\int_0^{\pi/2}\int_0^1 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta. $$

For the first piece, the radial integral is \(\int_0^2\rho^2\,d\rho=8/3\), the polar integral is \(\int_0^{\pi/2}\sin\phi\,d\phi=1\), and the azimuthal interval has length \(\pi/2\). Its volume is therefore \((8/3)(1)(\pi/2)=4\pi/3\). For the second piece, the corresponding values are \(\int_0^1\rho^2\,d\rho=1/3\), \(1\), and \(\pi/2\), giving volume \(\pi/6\). The total is \(V=4\pi/3+\pi/6=3\pi/2\).

The angular intervals meet at \(\theta=\pi/2\), and the radius changes there from \(2\) to \(1\) as specified. The radial antiderivative \(\rho^3/3\) differentiates to \(\rho^2\), and the polar antiderivative \(-\cos\phi\) differentiates to \(\sin\phi\); these confirm the factors used in both pieces.

Common Pitfall: Choosing the Wrong Bound on One Side

A frequent error is to find where two candidate bounds meet, but then use the same candidate on both sides of the threshold. Equality at the threshold alone does not tell you which surface controls on either side; you must compare the values.

For example, in the first worked example the outer bound is \(\min(3,\sec\phi)\). At \(\phi=0\), \(\sec 0=1<3\), so the plane bound controls near \(0\). At \(\phi=\pi/2\) approached from below, \(\sec\phi\) grows beyond \(3\), so the sphere controls after the crossing. Using \(3\) throughout would incorrectly include points above \(z=1\) near the positive \(z\)-axis. Using \(\sec\phi\) throughout would incorrectly extend the radial bound past the sphere where \(\sec\phi>3\).

Practical check: After finding a threshold, test one angle on each side. For an outer bound, choose the smaller candidate; for an inner bound, choose the larger. Also verify that the inner bound never exceeds the outer bound on the interval of integration.

A Reliable Setup Process

When the boundary formula changes with an angle, use a deliberate sequence rather than writing the integral immediately.

1
List the candidate radial bounds.
Translate each surface or constraint into a lower or upper bound for \(\rho\).
2
Find the thresholds.
Set competing formulas equal and solve for the angle or angles where their ordering can change.
3
Test each interval.
Compare candidate values on either side of each threshold and determine which bound controls.
4
Check that the region exists.
Keep only angular ranges where the inner radial bound is no greater than the outer radial bound.
5
Write and add the pieces.
Use the spherical volume element in every integral, and make the angular intervals adjacent without gaps.

This method matters because a correct integrand cannot repair an incorrect description of the region. A missed threshold can omit part of the solid or include points outside it. A clearly partitioned setup makes the geometry visible in the limits and gives each integral a consistent meaning.

Check Your Understanding

For each question, identify the boundary comparison or interval check that determines the correct setup.

  1. If an outer radial bound is the smaller of \(4\) and \(\sec\phi\), what equation determines a possible switching angle?
  2. For an inner radial bound that is the larger of \(2\) and \(\sec\phi\), which candidate controls at \(\phi=0\), and why?
  3. Why can two integrals over angular intervals that share an endpoint be added without double-counting positive volume?
  4. What must be checked about the inner and outer radial bounds before integrating over an angular interval?
  5. For a bound that changes at \(\theta=\pi/2\), which part of the triple integral must be split at that threshold?

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