Nonconstant Radial Bounds from Surface Intersections

Why Surface Intersections Determine Radial Bounds

A spherical-coordinate description of a solid is built one ray at a time. Fix angles \(\phi\) and \(\theta\), then let \(\rho\) vary along that ray. Each boundary surface may meet the ray at a radial value that depends on the angles. The solid’s inequalities determine which values of \(\rho\) belong to it.

This dependence matters: a surface that looks simple in Cartesian coordinates may give a nonconstant radial bound. It also means that solving an equation for \(\rho\) is only part of the work. You must check the original inequalities, keep only admissible roots, and find the angles for which the radial bounds are in the right order.

Definition: For fixed \((\phi,\theta)\), the radial slice of a solid is the set of radii \(\rho\ge 0\) for which the point with those angles belongs to the solid. A description by one pair of radial bounds applies when each radial slice is an interval or is empty. If a slice consists of multiple disjoint intervals, describe and integrate over each interval separately.

We use the spherical-coordinate convention in which \(\rho\ge 0\), \(0\le\phi\le\pi\) is measured from the positive \(z\)-axis, and \(0\le\theta<2\pi\) is the azimuthal angle. In particular, \(z=\rho\cos\phi\) and \(x^2+y^2=\rho^2\sin^2\phi\). The volume element is \(dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\).

A Reliable Method for Finding the Bounds

Write each boundary in spherical coordinates, then solve for the radial values where the ray meets it. A resulting equation can have more than one root, and the roots need not all describe the relevant part of the surface. Use the inequalities defining the solid to choose the correct radial side of each boundary.

Convert the boundary equations.
Substitute the spherical-coordinate expressions into each surface equation. Keep the angle variables visible.

Solve along a ray.
For fixed angles, solve the resulting equation or inequality for \(\rho\), retaining only nonnegative values that satisfy the original condition.

Find the angular domain.
Require that the lower radial bound does not exceed the upper bound. Intersections of boundary surfaces often determine where equality occurs and where a formula changes.

Set up the integral.
Use the radial intervals actually occupied by the solid, with the factor \(\rho^2\sin\phi\) in the volume integral.

The reason this method works is that, at fixed angles, all the solid’s boundary conditions become conditions on one variable, \(\rho\). Their intersection gives exactly the radii on that ray that lie in the solid. The angular domain records which rays have a nonempty radial slice.

Worked Example: Outside a Cylinder and Inside a Sphere

Worked Example: A Nonzero Inner Radius

Find the volume inside the sphere \(x^2+y^2+z^2=4\) and outside the cylinder \(x^2+y^2=1\).

The sphere becomes \(\rho^2=4\), so its interior gives \(0\le\rho\le2\). The cylinder becomes \(\rho^2\sin^2\phi=1\). Since \(\rho\ge0\) and \(\sin\phi\ge0\) on \(0\le\phi\le\pi\), being outside the cylinder gives \(\rho\sin\phi\ge1\), or \(\rho\ge\csc\phi\). Thus the radial interval is \(\csc\phi\le\rho\le2\), provided \(\csc\phi\le2\).

The last condition is \(\sin\phi\ge1/2\), so \(\pi/6\le\phi\le5\pi/6\). There is no restriction on \(\theta\), which runs from \(0\) to \(2\pi\). Therefore

V=\int_0^{2\pi}\int_{\pi/6}^{5\pi/6}\int_{\csc\phi}^{2} \rho^2\sin\phi\,d\rho\,d\phi\,d\theta.

Integrating in \(\rho\) gives \(\int_{\csc\phi}^{2}\rho^2\,d\rho=(8-\csc^3\phi)/3\). Hence

V=\frac{2\pi}{3}\int_{\pi/6}^{5\pi/6} \left(8\sin\phi-\csc^2\phi\right)\,d\phi.

Since \(\int_{\pi/6}^{5\pi/6}\sin\phi\,d\phi=\sqrt3\) and \(\int_{\pi/6}^{5\pi/6}\csc^2\phi\,d\phi=[-\cot\phi]_{\pi/6}^{5\pi/6}=2\sqrt3\), the volume is

V=\frac{2\pi}{3}(8\sqrt3-2\sqrt3)=4\pi\sqrt3.

At \(\phi=\pi/6\) and \(5\pi/6\), the inner and outer radii both equal \(2\), so the radial interval closes at the angular endpoints as required. Differentiating \(-\cos\phi\) gives \(\sin\phi\), and differentiating \(-\cot\phi\) gives \(\csc^2\phi\); these antiderivatives confirm the evaluations.

A Changing Upper Bound

Two surfaces can exchange which one supplies the outer radial boundary. The angular range must then be divided at their intersection. The following example shows why both the inequality direction and the radial power in the volume integral matter.

Worked Example: Inside a Sphere and Above a Paraboloid

Find the volume inside the unit sphere \(x^2+y^2+z^2=1\) and above the paraboloid \(z=x^2+y^2\).

The sphere gives \(0\le\rho\le1\). The paraboloid condition is \(\rho\cos\phi\ge\rho^2\sin^2\phi\). For \(\rho>0\), this is equivalent to \(\rho\le\cos\phi/\sin^2\phi\) when \(\sin\phi\ne0\). The region is above the \(xy\)-plane, so \(0\le\phi\le\pi/2\). The radial upper bound is therefore the smaller of \(1\) and \(\cos\phi/\sin^2\phi\).

To find where these bounds switch, set \(\cos\phi/\sin^2\phi=1\). Writing \(c=\cos\phi\), this becomes \(c=1-c^2\), so \(c=(\sqrt5-1)/2\). Define \(\phi_0=\arccos((\sqrt5-1)/2)\). For \(0\le\phi\le\phi_0\), the paraboloid bound is at least \(1\), so the sphere supplies the upper bound. For \(\phi_0\le\phi\le\pi/2\), the paraboloid supplies it.

V=\frac{2\pi}{3}\left[ \int_0^{\phi_0}\sin\phi\,d\phi+ \int_{\phi_0}^{\pi/2} \sin\phi\left(\frac{\cos\phi}{\sin^2\phi}\right)^3d\phi \right].

The cube in the second integral comes from integrating \(\rho^2\) with respect to \(\rho\). Set \(u=\cos\phi\) in that integral. Its limits become \(u_0=\cos\phi_0\) and \(0\), and reversing them gives

\int_0^{u_0}\frac{u^3}{(1-u^2)^3}\,du.

An antiderivative is \(F(u)=1/[4(1-u^2)^2]-1/[2(1-u^2)]\), since differentiation gives \(F'(u)=u^3/(1-u^2)^3\). Thus the integral is \(F(u_0)-F(0)\). Here \(u_0=(\sqrt5-1)/2\) satisfies \(1-u_0^2=u_0\), so its value simplifies to \(1/(4u_0^2)-1/(2u_0)+1/4\). The first integral is \(1-u_0\). Therefore

V=\frac{2\pi}{3}\left( 1-u_0+\frac{1}{4u_0^2}-\frac{1}{2u_0}+\frac14 \right), \qquad u_0=\frac{\sqrt5-1}{2}.

The radial bounds agree at \(\phi_0\), because \(\cos\phi_0/\sin^2\phi_0=1\). At \(\phi=\pi/2\), the paraboloid bound is \(0\), consistent with the radial slice shrinking to the axis. The antiderivative check above verifies the nonconstant part of the volume calculation.

Worked Example: Two Nonconstant Radial Bounds

Worked Example: A Cylinder, a Paraboloid, and a Cone

Consider the region with \(\pi/6\le\phi\le\pi/4\), \(0\le\theta<2\pi\), outside the cylinder \(x^2+y^2=1\), and above the paraboloid \(z=x^2+y^2\). Find its volume.

Outside the cylinder gives \(\rho\ge\csc\phi\). Above the paraboloid means \(\rho\cos\phi\ge\rho^2\sin^2\phi\). For \(\rho>0\), this gives \(\rho\le\cos\phi/\sin^2\phi\), an upper bound. (Points below the paraboloid satisfy \(\rho\ge\cos\phi/\sin^2\phi\) instead; together with \(\rho\ge\csc\phi\) that would leave every ray unbounded outward, so the region would have infinite volume. This is why the problem asks for the region above.)

Thus the radial interval is \(\csc\phi\le\rho\le\cos\phi/\sin^2\phi\). It is nonempty on the stated angular range because the ordering is equivalent to \(\sin\phi\le\cos\phi\), true for \(\phi\le\pi/4\). The volume is

V=\frac{2\pi}{3}\int_{\pi/6}^{\pi/4} \left(\frac{\cos^3\phi}{\sin^5\phi}-\csc^2\phi\right)d\phi.

For the first term, use \(u=\sin\phi\). Then \(\cos^3\phi\,d\phi/\sin^5\phi=(u^{-5}-u^{-3})\,du\), whose antiderivative is \(-1/(4u^4)+1/(2u^2)\). Its value from \(\pi/6\) to \(\pi/4\) is \(0-(-2)=2\). The integral of \(\csc^2\phi\) over these limits is \([-\cot\phi]_{\pi/6}^{\pi/4}=\sqrt3-1\). Consequently,

V=\frac{2\pi}{3}\bigl(2-(\sqrt3-1)\bigr) =\frac{2\pi}{3}(3-\sqrt3).

The two radial boundaries meet at \(\phi=\pi/4\): both equal \(\sqrt2\). The antiderivative \(-1/(4\sin^4\phi)+1/(2\sin^2\phi)\) differentiates to \(\cos^3\phi/\sin^5\phi\), confirming the first integral.

A Common Mistake: Reversing the Inequality

When a surface equation is rearranged, its equation alone does not say which side of the surface belongs to the solid. For the paraboloid \(z=x^2+y^2\), the condition “above” is \(z\ge x^2+y^2\), not \(z\le x^2+y^2\). Substitution gives

\rho\cos\phi\ge\rho^2\sin^2\phi \quad\Longrightarrow\quad \rho\le\frac{\cos\phi}{\sin^2\phi} \qquad(\rho>0,\ 0\le\phi<\pi/2).

A reversed calculation would claim that points above the paraboloid require \(\rho\ge\cos\phi/\sin^2\phi\). At \(\phi=\pi/4\), that threshold is \(\sqrt2\). But \(\rho=1/2\) gives \(z=\sqrt2/4\) and \(x^2+y^2=1/8\), so this point is above the paraboloid even though its radius is less than \(\sqrt2\). The inequality direction is therefore essential.

A second common error is to use the square of a radial endpoint after integrating \(\rho^2\). The correct rule is \(\int_a^b\rho^2\,d\rho=(b^3-a^3)/3\). A quick check is to differentiate \(\rho^3/3\), which returns \(\rho^2\).

When to Split the Radial Description

Compare the candidate lower and upper bounds before writing a single integral. If their order changes with the angles, split the angular domain at the equality points. If a ray meets the solid in separated radial pieces, keep each interval: using only the smallest and largest radii would incorrectly include the gap between them.

Takeaway: Radial bounds come from the solid’s inequalities along each ray, not from surface equations alone. Solve for \(\rho\), check the inequality direction and admissible roots, determine where the bounds are ordered, and split the description wherever the radial slice changes.

Check Your Understanding

For each question, identify the radial inequalities before deciding on an integral.

For the region inside \(\rho=3\) and outside the cylinder \(\rho\sin\phi=2\), what are the radial bounds and the allowed range of \(\phi\)?
Why must the radial upper bound for the unit sphere intersected with the region above \(z=x^2+y^2\) be the smaller of \(1\) and \(\cos\phi/\sin^2\phi\)?
What equation determines the angle at which two candidate radial bounds exchange roles?
When does one pair of radial bounds fail to describe a solid’s slice along a ray?
After integrating \(\rho^2\) from \(a\) to \(b\), what expression must appear in the remaining integral?