Spherical Shells and Radial Bounds

Radial Bounds Describe a Hollow Region

A ball centered at the origin contains every point whose distance from the origin is at most a fixed radius. A spherical shell keeps only the points whose distances lie between two radii. The smaller radius leaves a hollow interior; the larger radius sets the outer boundary.

This is a natural use of spherical coordinates: the distance from the origin is the radial coordinate, so the two spherical boundaries become constant radial limits. When the entire shell is included, the angular coordinates range over all directions.

Definition: Let \(0\leq a<b\). The spherical shell centered at the origin with inner radius \(a\) and outer radius \(b\) is the region \[ \{(x,y,z)\in\mathbb R^3:a\leq \rho\leq b\}, \] where \(\rho=\sqrt{x^2+y^2+z^2}\) is the distance from the origin. If \(a>0\), the region is a hollow ball. If \(a=0\), it is the solid ball of radius \(b\).

The boundaries \(\rho=a\) and \(\rho=b\) are spheres. The radii \(a\) and \(b\) must be nonnegative, and the inner radius must be strictly smaller than the outer radius. A shell centered somewhere other than the origin is not described by these radial bounds unless the coordinate origin is moved to its center.

Writing the Bounds in Spherical Coordinates

Use \(\rho\) for distance from the origin, \(\phi\) for the angle measured down from the positive \(z\)-axis, and \(\theta\) for the azimuthal angle in the \(xy\)-plane. The spherical volume element is \(dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\). Since a full shell includes every direction, its bounds are

a\leq \rho\leq b,\qquad 0\leq\phi\leq\pi,\qquad 0\leq\theta\leq 2\pi.

The important feature is that the radial limits are constants: every direction from the origin meets the same inner and outer spheres at radii \(a\) and \(b\). This simple description applies to concentric spheres. If a boundary depends on direction, the radial limits need not be constants, which is a different kind of region.

Worked Example: Volume of a Shell with Radii 2 and 5

Find the volume of the region between the spheres \(\rho=2\) and \(\rho=5\). The full spherical bounds are \(2\leq\rho\leq5\), \(0\leq\phi\leq\pi\), and \(0\leq\theta\leq2\pi\). Integrate \(1\) over the region:

\begin{aligned} V &=\int_0^{2\pi}\int_0^\pi\int_2^5 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=\left(\int_2^5\rho^2\,d\rho\right) \left(\int_0^\pi\sin\phi\,d\phi\right) \left(\int_0^{2\pi}d\theta\right)\\ &=\left[\frac{\rho^3}{3}\right]_2^5 \left[-\cos\phi\right]_0^\pi \left[\theta\right]_0^{2\pi}\\ &=\frac{125-8}{3}\cdot 2\cdot 2\pi =\frac{468\pi}{3} =156\pi. \end{aligned}

As a check, the outer ball has volume \(\frac{4\pi}{3}(5^3)=\frac{500\pi}{3}\), and the inner ball has volume \(\frac{4\pi}{3}(2^3)=\frac{32\pi}{3}\). Their difference is \(\frac{468\pi}{3}=156\pi\), matching the integral.

Deriving the Shell-Volume Formula

The volume of a shell is the outer ball's volume minus the inner ball's volume. Equivalently, integrate the spherical volume element over the shell. The angular factors are fixed for a full sphere:

\int_0^{2\pi}d\theta=2\pi, \qquad \int_0^\pi\sin\phi\,d\phi=2.

Thus the volume integral is the radial integral multiplied by \(4\pi\). Evaluating the radial integral gives the formula below.

Theorem: The volume of the spherical shell with inner radius \(a\) and outer radius \(b\), where \(0\leq a<b\), is \[ V=\frac{4\pi}{3}(b^3-a^3). \]

For completeness, the spherical-coordinate calculation justifying the result is

\begin{aligned} V &=\int_0^{2\pi}\int_0^\pi\int_a^b \rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=(2\pi)(2)\left[\frac{\rho^3}{3}\right]_a^b\\ &=4\pi\left(\frac{b^3-a^3}{3}\right) =\frac{4\pi}{3}(b^3-a^3). \end{aligned}

This argument also explains the difference between a shell and a ball. Setting \(a=0\) gives \(V=\frac{4\pi}{3}b^3\), the ball formula. For \(a>0\), the term involving \(a^3\) removes the volume of the hollow interior. The formula depends on the difference of the cubes of the radii, not simply on the difference \(b-a\).

Integrating a Radial Function

A radial function is a function whose value depends only on distance from the origin. Write it as \(f(\rho)\). Because it has no dependence on either angle, its integral over a full spherical shell separates into an angular factor and a radial integral.

Theorem: Suppose \(f(\rho)\) is continuous for \(a\leq\rho\leq b\), where \(0\leq a<b\). Over the full spherical shell, \[ \iiint_{\text{shell}} f(\rho)\,dV =4\pi\int_a^b f(\rho)\rho^2\,d\rho. \] If \(a=0\) and \(f\) is not continuous at the origin, the corresponding improper integral must converge for this formula to give a finite integral.

The factor \(\rho^2\) is essential: it comes from the spherical volume element, not from the function \(f\). The formula follows by inserting \(f(\rho)\) into the triple integral and evaluating the angular integrals:

\begin{aligned} \iiint_{\text{shell}}f(\rho)\,dV &=\int_0^{2\pi}\int_0^\pi\int_a^b f(\rho)\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=\left(\int_a^b f(\rho)\rho^2\,d\rho\right) \left(\int_0^\pi\sin\phi\,d\phi\right) \left(\int_0^{2\pi}d\theta\right)\\ &=4\pi\int_a^b f(\rho)\rho^2\,d\rho. \end{aligned}

For example, if \(f\) is a mass density depending only on distance from the center, this integral gives the total mass of the shell. The same calculation applies to any integrable radial quantity, with the physical interpretation determined by the function and its units.

Worked Example: Mass with Density Proportional to Radius

Suppose a shell has inner radius \(1\), outer radius \(3\), and density \(\delta(\rho)=\rho\) kilograms per cubic metre when \(\rho\) is measured in metres. Its mass is the integral of density over the shell:

\begin{aligned} M &=4\pi\int_1^3 \delta(\rho)\rho^2\,d\rho\\ &=4\pi\int_1^3 \rho\cdot\rho^2\,d\rho\\ &=4\pi\left[\frac{\rho^4}{4}\right]_1^3\\ &=\pi(3^4-1^4) =\pi(81-1) =80\pi\text{ kg}. \end{aligned}

To check the radial antiderivative, differentiate \(\rho^4/4\) to obtain \(\rho^3\), the integrand after including the volume factor \(\rho^2\). The bounds give \(3^4/4-1^4/4=20\), and multiplying by \(4\pi\) gives \(80\pi\) kilograms.

Worked Example: Integrating an Inverse-Square Radial Function

Evaluate the integral of \(f(\rho)=1/\rho^2\) over the shell \(1\leq\rho\leq4\). The function is continuous throughout this shell because it does not include \(\rho=0\). Use the radial formula:

\begin{aligned} \iiint_{\text{shell}}\frac{1}{\rho^2}\,dV &=4\pi\int_1^4\frac{1}{\rho^2}\rho^2\,d\rho\\ &=4\pi\int_1^4 1\,d\rho\\ &=4\pi[\rho]_1^4\\ &=4\pi(4-1)=12\pi. \end{aligned}

For a check, the factors \(1/\rho^2\) and \(\rho^2\) cancel at every point in the radial integral, leaving an interval of length \(4-1=3\). Multiplying by the full angular factor \(4\pi\) gives \(12\pi\), as computed.

A Common Error: Omitting the Inner Radius

A hollow ball is not the same region as the solid ball out to its outer radius. A common mistake is to use \(0\leq\rho\leq b\) even when the region begins at a positive inner radius. For instance, the incorrect calculation for the shell \(2\leq\rho\leq5\) would be

\frac{4\pi}{3}(5^3)=\frac{500\pi}{3}.

This is the volume of the entire ball of radius \(5\); it includes the hollow interior. The correction is to subtract the inner ball, or equivalently to use both radial limits:

\frac{4\pi}{3}(5^3)-\frac{4\pi}{3}(2^3) =\frac{4\pi}{3}(125-8) =156\pi.

Before integrating, check that the lower radial bound matches the inner spherical boundary. Also keep the Jacobian factor \(\rho^2\): replacing the radial volume integral by an unweighted integral of \(f(\rho)\) generally gives the wrong result.

Why Radial Bounds Are Useful

For a concentric shell, the geometry is encoded by just two radial numbers. This reduces volume and radial-density calculations to one-dimensional integration after the angular factor has been evaluated. It also makes a geometric check immediate: the shell volume must equal the volume of the outer ball minus that of the inner ball, and must be positive whenever \(b>a\).

These bounds should not be used blindly for every region that lies between spherical surfaces. They describe a complete shell only when the same inner and outer radii apply in every direction. For the region defined here, the radial interval is constant and the full angular ranges are appropriate.

Check Your Understanding

Use the radial bounds and formula from this tutorial to answer the following.

Write the spherical-coordinate bounds for the shell with inner radius \(3\) and outer radius \(6\).
Find the volume of that shell, and check it by subtracting the inner-ball volume from the outer-ball volume.
Set up, but do not evaluate, the integral of a continuous radial density \(\delta(\rho)\) over a shell with radii \(a\) and \(b\).
Why does the radial integral of \(f(\rho)\) include a factor of \(\rho^2\)?
What region would be described incorrectly by using \(0\leq\rho\leq b\) when the intended inner radius is positive?