Deriving the Shell Volume Formula

Why a Shell Formula Leads to an Integral

A solid formed by rotating a region around an axis can be viewed as a stack of thin cylindrical layers. Each layer has a radius, a height, and a small thickness. Its volume is approximately the area around the cylinder multiplied by its height and thickness.

The key word is approximately. A layer with nonzero thickness is an annular cylinder, not a zero-thickness surface. Its exact volume can be found by subtracting the volume of its inner cylinder from the volume of its outer cylinder. As the layers become thinner, the resulting sum approaches a definite integral.

This derivation explains both the shell formula and the reason it gives the volume of the solid. We will use \(r\) for distance from the axis of rotation, \(h(r)\) for the height of a shell at radius \(r\), and \(dr\) for an infinitesimal change in radius.

Shell volume formula: When a solid is represented by cylindrical shells with radius \(r\) and height \(h(r)\), its volume is

V=\int_a^b 2\pi r h(r)\,dr.

Here, \(a\) and \(b\) are the smallest and largest shell radii. The formula applies when the shells fill the solid without overlapping in their interiors and their heights describe the solid throughout that radial interval.

Start with One Annular Shell

Consider a cylindrical layer with constant height \(h\), inner radius \(r\), and thickness \(\Delta r\). Its outer radius is \(r+\Delta r\). The layer is the outer cylinder with the inner cylinder removed.

A cylinder of radius \(R\) and height \(h\) has volume \(\pi R^2h\). Therefore, the annular layer has exact volume

\begin{aligned} \Delta V &=\pi h(r+\Delta r)^2-\pi h r^2\\ &=\pi h\left(r^2+2r\Delta r+(\Delta r)^2-r^2\right)\\ &=2\pi r h\,\Delta r+\pi h(\Delta r)^2. \end{aligned}

The first term, \(2\pi r h\,\Delta r\), is the familiar thin-shell expression: circumference \(2\pi r\), multiplied by height \(h\), multiplied by thickness \(\Delta r\). The second term, \(\pi h(\Delta r)^2\), accounts for the fact that the layer has a nonzero thickness.

There is also a useful exact interpretation of the first expression. If \(\bar r=r+\frac12\Delta r\) is the middle radius of the layer, then

\begin{aligned} \pi h\left((r+\Delta r)^2-r^2\right) &=\pi h(2r\Delta r+(\Delta r)^2)\\ &=2\pi \bar r h\,\Delta r. \end{aligned}

So a shell of constant height has exactly the volume \(2\pi\bar r h\,\Delta r\) when its middle radius is used. If instead we use the inner radius \(r\), the missing correction term becomes negligible as the thickness shrinks.

From Many Shells to a Definite Integral

Now suppose the shell height changes with radius according to a continuous function \(h(r)\) on the interval \([a,b]\). Divide that interval into \(n\) subintervals. Let the \(i\)-th subinterval have width \(\Delta r_i\), and choose a representative radius \(r_i^*\) within it.

Across a sufficiently narrow subinterval, the shell heights are close to \(h(r_i^*)\). The layer’s annular cross-section has area \(2\pi r_i^*\Delta r_i\) when \(r_i^*\) is the middle radius. Multiplying by the representative height gives the shell-volume estimate

\Delta V_i\approx 2\pi r_i^*h(r_i^*)\,\Delta r_i.

Adding the estimates gives an approximation to the whole volume:

V\approx \sum_{i=1}^{n}2\pi r_i^*h(r_i^*)\,\Delta r_i.

This is a Riemann sum for the function \(2\pi r h(r)\). Because \(h\) is continuous, the product \(2\pi r h(r)\) is continuous on the finite interval \([a,b]\), so it is integrable there. As the largest subinterval width approaches zero, the Riemann sums converge to the definite integral.

Partition the radial interval.
Split \([a,b]\) into thin intervals of width \(\Delta r_i\).

Estimate each layer.
Use its representative radius and height to estimate its volume by \(2\pi r_i^*h(r_i^*)\Delta r_i\).

Add the layer estimates.
The sum accounts for the volume of all the layers in the partition.

Take the limit.
As the widths shrink, the Riemann sums converge to \(\int_a^b 2\pi r h(r)\,dr\).

Why the integral represents volume: Each summand estimates the volume of one thin layer. A definite integral is the limit of these sums as the layers become arbitrarily thin. Therefore, when the shell heights describe the solid and the shells fill it without interior overlap, the integral gives the solid’s volume.

Worked Examples

Worked Example: A Solid with Constant Shell Height

Suppose shells extend from radius \(r=1\) to radius \(r=3\), and each shell has height \(4\). Find the volume using the shell formula, then check it using the exact annular-cylinder calculation.

Here \(h(r)=4\), so

\begin{aligned} V &=\int_1^3 2\pi r(4)\,dr\\ &=8\pi\int_1^3 r\,dr\\ &=8\pi\left[\frac{r^2}{2}\right]_1^3\\ &=4\pi(3^2-1^2)\\ &=4\pi(9-1)\\ &=32\pi. \end{aligned}

For an independent check, the solid is an annular cylinder of height \(4\), outer radius \(3\), and inner radius \(1\). Its volume is

\pi(4)(3^2-1^2)=4\pi(9-1)=32\pi.

The two calculations agree. The antiderivative also checks correctly because \(\frac{d}{dr}(4\pi r^2)=8\pi r\), the integrand.

Worked Example: Shell Height Decreases with Radius

Suppose the shell radii run from \(0\) to \(2\), with height \(h(r)=3-r\). The height is nonnegative on this interval. Find the volume.

Substitute the height function into the formula:

\begin{aligned} V &=\int_0^2 2\pi r(3-r)\,dr\\ &=2\pi\int_0^2(3r-r^2)\,dr\\ &=2\pi\left[\frac{3r^2}{2}-\frac{r^3}{3}\right]_0^2\\ &=2\pi\left(\frac{3(2^2)}{2}-\frac{2^3}{3}-0\right)\\ &=2\pi\left(6-\frac{8}{3}\right)\\ &=2\pi\left(\frac{18-8}{3}\right)\\ &=\frac{20\pi}{3}. \end{aligned}

To check, differentiate the antiderivative \(2\pi(\frac{3r^2}{2}-\frac{r^3}{3})\). Its derivative is \(2\pi(3r-r^2)=2\pi r(3-r)\), exactly the integrand. Evaluating at the endpoints gives \(\frac{20\pi}{3}-0=\frac{20\pi}{3}\), as required.

Worked Example: Comparing Finite-Shell Estimates with the Integral

Let \(h(r)=r\) for \(0\le r\le 2\). First calculate the exact volume. Then estimate it using midpoint shells with two subintervals and with four subintervals.

The exact volume is

\begin{aligned} V &=\int_0^2 2\pi r(r)\,dr\\ &=2\pi\left[\frac{r^3}{3}\right]_0^2\\ &=2\pi\left(\frac{8}{3}\right)\\ &=\frac{16\pi}{3}. \end{aligned}

With two equal subintervals, \([0,1]\) and \([1,2]\), the width is \(\Delta r=1\), and the midpoint radii are \(\frac12\) and \(\frac32\). The midpoint shell sum is

\begin{aligned} V_2 &=2\pi\left(\frac12\right)h\left(\frac12\right)(1) +2\pi\left(\frac32\right)h\left(\frac32\right)(1)\\ &=2\pi\left(\frac12\right)\left(\frac12\right) +2\pi\left(\frac32\right)\left(\frac32\right)\\ &=2\pi\left(\frac14+\frac94\right)\\ &=5\pi. \end{aligned}

With four equal subintervals, the width is \(\Delta r=\frac12\), and the midpoint radii are \(\frac14,\frac34,\frac54,\frac74\). Since \(h(r)=r\), each term contains the square of its midpoint radius:

\begin{aligned} V_4 &=2\pi\left(\frac12\right) \left[ \left(\frac14\right)^2+ \left(\frac34\right)^2+ \left(\frac54\right)^2+ \left(\frac74\right)^2 \right]\\ &=\pi\left(\frac{1+9+25+49}{16}\right)\\ &=\pi\left(\frac{84}{16}\right)\\ &=\frac{21\pi}{4}. \end{aligned}

The estimates are \(5\pi\) and \(\frac{21\pi}{4}\), while the exact volume is \(\frac{16\pi}{3}\). Their differences from the exact value are, respectively,

\frac{16\pi}{3}-5\pi=\frac{\pi}{3}, \qquad \frac{16\pi}{3}-\frac{21\pi}{4}=\frac{\pi}{12}.

The four-shell estimate is closer. This illustrates the limiting idea: finer partitions give more detailed approximations, and the integral is the limit of the corresponding Riemann sums. Differentiating \(2\pi r^3/3\) gives \(2\pi r^2\), confirming the integrand used for the exact volume.

A Common Mistake: Using Only the Circumference

A frequent error is to multiply the circumference of a shell by its height but forget the shell’s thickness. That gives a surface area, not a volume. For example, a shell of radius \(2\) and height \(3\) has lateral surface area \(2\pi(2)(3)=12\pi\), but this is not a volume. If its thickness is \(\Delta r\), its thin-shell volume is approximately \(12\pi\Delta r\).

The exact annular volume, with inner radius \(2\), outer radius \(2+\Delta r\), and height \(3\), is

\begin{aligned} \Delta V &=3\pi\left((2+\Delta r)^2-2^2\right)\\ &=3\pi\left(4\Delta r+(\Delta r)^2\right)\\ &=12\pi\Delta r+3\pi(\Delta r)^2. \end{aligned}

For a thin layer, the quadratic term is small, so \(12\pi\Delta r\) is a good approximation. The factor \(\Delta r\) is essential: without it, the expression measures area rather than volume.

When the Derivation Applies

The derivation relies on the shell layers representing the solid correctly. In particular, the radial interval must cover the shells being counted, the height function must describe each shell, and the layers must not count the same interior volume more than once. Continuity of \(h(r)\) on a finite interval is a convenient sufficient condition for the Riemann-sum argument above. Piecewise continuous heights can also be handled by splitting the interval at their discontinuities and integrating each piece.

The formula does not claim that every volume is automatically a shell integral. Its usefulness depends on whether the solid can be described by shells with a suitable radius and height. Once that description is valid, the integral follows from the same principle used throughout the derivation: add the volumes of thin pieces, then take the limit as their thicknesses shrink.

Check Your Understanding

Use the derivation and examples to answer these questions.

Write the exact volume of an annular shell with inner radius \(r\), thickness \(\Delta r\), and constant height \(h\).
Why does \(2\pi r h\,\Delta r\) approximate a thin-shell volume?
What condition on the shell heights makes the Riemann-sum argument straightforward?
For \(h(r)=2\) on \(1\le r\le 4\), write the shell integral and evaluate it.
Explain why omitting \(\Delta r\) from a shell-volume estimate gives the wrong type of quantity.