Deriving the Cylindrical Shell Volume Formula

From a Rotating Slice to a Cylindrical Layer

A volume integral can be understood by dividing a solid into many small pieces, estimating the volume of each piece, and adding those estimates. The disk and washer methods build a solid from slices perpendicular to the axis of rotation. The shell method uses a different kind of piece: a thin cylindrical layer whose length runs around the axis.

To see where the method comes from, consider a region between $x=a$ and $x=b$, with vertical slices of height $H(x)$, where $H(x)\ge 0$. Suppose $0\le a<b$, and rotate the region about the $y$-axis. A narrow vertical strip at $x$ sweeps out a thin cylindrical layer. Its radius is approximately $x$, its height is approximately $H(x)$, and its thickness is a small change in $x$.

The key geometric idea is that the layer’s volume is approximately its circumference multiplied by its height and its thickness. The derivation below explains why this product appears and when the approximation becomes an exact integral.

Central idea: A thin layer at radius $r$, height $h$, and thickness $\Delta r$ has approximate volume $2\pi r h\,\Delta r$. Adding such contributions and taking a limit leads to the shell integral.

The Volume of One Cylindrical Layer

Begin with an actual cylindrical layer of constant height $h$, inner radius $r$, and thickness $\Delta r$. Its outer radius is $r+\Delta r$. The layer is the difference between two cylinders, so its exact volume is the volume of the outer cylinder minus that of the inner cylinder.

V_{\text{layer}} =\pi(r+\Delta r)^2h-\pi r^2h =\pi\bigl((r+\Delta r)^2-r^2\bigr)h.

Expanding the square gives an exact expression:

V_{\text{layer}} =\pi(2r\Delta r+(\Delta r)^2)h =2\pi\left(r+\frac{\Delta r}{2}\right)h\,\Delta r.

This identity reveals two useful facts. First, the exact volume equals the circumference at the layer’s mean radius, $r+\frac{\Delta r}{2}$, multiplied by height and thickness. Second, if the thickness is small, the term involving $(\Delta r)^2$ is much smaller than the term involving $\Delta r$. The thin-layer approximation is therefore circumference at radius $r$, times height, times thickness.

Important distinction: For a layer of finite thickness, $2\pi r h\,\Delta r$ using the inner radius is generally only an approximation. The exact formula is $2\pi\left(r+\frac{\Delta r}{2}\right)h\,\Delta r$, using the mean radius. In the shell integral, the thickness tends to zero, so the thin-layer contribution leads to the correct limit.

Deriving the Shell Integral

Now divide the interval $[a,b]$ into $n$ subintervals. Write their widths as $\Delta x_i$, and choose a sample point $x_i^*$ in each subinterval. The vertical strip at $x_i^*$ has height approximately $H(x_i^*)$ and thickness $\Delta x_i$. When revolved about the $y$-axis, it is approximated by a cylindrical layer with radius $x_i^*$.

The circumference of that layer is $2\pi x_i^*$. Multiplying circumference, height, and thickness gives the approximate contribution

2\pi x_i^* H(x_i^*)\,\Delta x_i.

Adding the contributions from all the subintervals gives a Riemann sum for the solid’s volume. If $H$ is Riemann integrable on $[a,b]$, the sums approach the integral as the largest subinterval width approaches zero. The exact mean-radius formula for each layer also supports this limiting argument: the difference between using a layer’s inner radius and its mean radius involves a term proportional to $(\Delta x_i)^2$, which vanishes in the limit.

Partition the interval.
Divide $[a,b]$ into small widths $\Delta x_i$, and select a sample point $x_i^*$ in each one.

Approximate each layer.
Use radius $x_i^*$, height $H(x_i^*)$, and thickness $\Delta x_i$. Its approximate volume is $2\pi x_i^*H(x_i^*)\Delta x_i$.

Add the contributions.
The sum of the layer estimates is a Riemann sum for the volume.

Take the limit.
As the widths shrink to zero, the Riemann sum becomes the definite integral.

Thus the shell formula for a region with vertical-slice height $H(x)$, rotated about the $y$-axis, is

V=\int_a^b 2\pi xH(x)\,dx.

Theorem: Suppose $0\le a<b$, the region lies between $x=a$ and $x=b$, and its vertical slices have nonnegative height $H(x)$. If $H$ is Riemann integrable on $[a,b]$, revolving the region about the $y$-axis gives volume $$ V=\int_a^b 2\pi xH(x)\,dx. $$

The restriction $0\le a<b$ ensures that the radius of a vertical slice is $x$. If the entire region instead lies to the left of the $y$-axis, the radius is $|x|$, not $x$; the corresponding contribution uses $2\pi|x|H(x)$. The theorem as stated does not cover a region that crosses the axis: in that situation, different strips can rotate into overlapping parts of the solid, so the simple sum of their shell volumes need not represent the volume correctly.

Worked Examples

Worked Example: A Rectangular Region

Rotate the rectangle $1\le x\le 3$, $0\le y\le 2$, about the $y$-axis. Here each vertical slice has height $H(x)=2$, and all radii are nonnegative. The shell integral is

\begin{aligned} V &=\int_1^3 2\pi x(2)\,dx\\ &=4\pi\int_1^3 x\,dx\\ &=4\pi\left[\frac{x^2}{2}\right]_1^3\\ &=2\pi(3^2-1^2)\\ &=2\pi(9-1)\\ &=16\pi. \end{aligned}

Check by treating the solid as a hollow cylinder of height $2$, outer radius $3$, and inner radius $1$:

V=\pi(3^2-1^2)(2)=\pi(9-1)(2)=16\pi.

The geometric volume agrees with the shell integral.

Worked Example: A Triangular Region

Rotate the region $0\le x\le 2$, $0\le y\le x$, about the $y$-axis. Its vertical-slice height is $H(x)=x$. The shell formula gives

\begin{aligned} V &=\int_0^2 2\pi x(x)\,dx\\ &=2\pi\int_0^2 x^2\,dx\\ &=2\pi\left[\frac{x^3}{3}\right]_0^2\\ &=2\pi\left(\frac{8}{3}-0\right)\\ &=\frac{16\pi}{3}. \end{aligned}

For a check, at a fixed height $y$ with $0\le y\le 2$, the region extends horizontally from $x=y$ to $x=2$. Its rotation gives a washer with outer radius $2$ and inner radius $y$. Integrating those cross-sectional areas gives

\begin{aligned} V &=\int_0^2 \pi(2^2-y^2)\,dy\\ &=\pi\left[4y-\frac{y^3}{3}\right]_0^2\\ &=\pi\left(8-\frac{8}{3}\right)\\ &=\frac{16\pi}{3}. \end{aligned}

This independent calculation confirms the shell result.

Worked Example: A Region with Decreasing Height

Rotate the region $0\le x\le 2$, $0\le y\le 2-x$, about the $y$-axis. The slice height is $H(x)=2-x$, so

\begin{aligned} V &=\int_0^2 2\pi x(2-x)\,dx\\ &=2\pi\int_0^2 (2x-x^2)\,dx\\ &=2\pi\left[x^2-\frac{x^3}{3}\right]_0^2\\ &=2\pi\left(4-\frac{8}{3}\right)\\ &=2\pi\left(\frac{12}{3}-\frac{8}{3}\right)\\ &=\frac{8\pi}{3}. \end{aligned}

For a check, a horizontal slice at height $y$, where $0\le y\le 2$, runs from the $y$-axis to $x=2-y$. It forms a disk of radius $2-y$, so

\begin{aligned} V &=\int_0^2 \pi(2-y)^2\,dy\\ &=\pi\int_0^2 (4-4y+y^2)\,dy\\ &=\pi\left[4y-2y^2+\frac{y^3}{3}\right]_0^2\\ &=\pi\left(8-8+\frac{8}{3}\right)\\ &=\frac{8\pi}{3}. \end{aligned}

The two methods agree.

Why the Formula Works

The derivation is a direct application of the definition of a definite integral as a limit of sums. A small vertical strip becomes a thin cylindrical layer; the layer’s circumference accounts for the distance around the axis, its height accounts for the vertical extent of the region, and its thickness accounts for the small horizontal width of the original strip.

The factor $2\pi x$ is therefore not an arbitrary addition to the integral. It is the circumference of a circle of radius $x$. The factor $H(x)$ is the height of the layer, and $dx$ represents its infinitesimal thickness. Multiplication produces the small volume contribution $2\pi xH(x)\,dx$. Summing all contributions over the interval produces the volume.

A Common Pitfall: Treating a Finite Layer as Infinitesimal

Suppose a cylindrical layer has inner radius $2$, thickness $0.1$, and height $3$. Using the inner radius in the thin-shell approximation gives

2\pi(2)(3)(0.1)=1.2\pi.

This is an approximation, not the exact volume of this finite-thickness layer. Its outer radius is $2.1$, so its exact volume is

\begin{aligned} V &=\pi(2.1^2-2^2)(3)\\ &=3\pi(4.41-4)\\ &=3\pi(0.41)\\ &=1.23\pi. \end{aligned}

Using the mean radius $\frac{2+2.1}{2}=2.05$ gives the exact same result:

2\pi(2.05)(3)(0.1)=1.23\pi.

The inner-radius estimate is reasonable for a thin layer, but it should not be called exact when the thickness is finite. The integral formula becomes exact because it is formed from a limit in which the layer widths approach zero.

When the Shell Method Is Useful

The shell method is useful when the solid can be divided into layers parallel to the axis of rotation and the resulting radius and height are straightforward to describe. Its derivation also provides a practical check on a setup: the integrand must represent circumference times height, and the variable of integration must represent shell thickness. If an expression does not have that geometric meaning, revisit the setup before evaluating it.

Takeaway: For vertical slices rotated about the $y$-axis, with $0\le a<b$ and integrable nonnegative height $H(x)$, each thin layer contributes approximately $2\pi xH(x)\,dx$. The limiting sum is $$ V=\int_a^b 2\pi xH(x)\,dx. $$

Check Your Understanding

Use the geometric meaning of each factor, not just the memorized formula, to answer these questions.

For a thin layer of radius $r$, height $h$, and thickness $\Delta r$, what three quantities are multiplied to form its thin-layer volume estimate?
Why is $2\pi(r+\frac{\Delta r}{2})h\,\Delta r$ the exact volume of a cylindrical layer with inner radius $r$ and thickness $\Delta r$?
What conditions on $a$, $b$, and $H$ allow the stated shell formula about the $y$-axis to apply?
For a region entirely to the left of the $y$-axis, why is the radius $|x|$ rather than $x$?
For the region $0\le x\le 1$, $0\le y\le 1$, rotated about the $y$-axis, write the shell integral but do not evaluate it.

Deriving the Cylindrical Shell Volume Formula

What You'll Learn

From a Rotating Slice to a Cylindrical Layer

The Volume of One Cylindrical Layer

Deriving the Shell Integral