Quick Reference $$\boxed{V = 2\pi r h \Delta r = \text{(circumference)} \times \text{(height)} \times \text{(thickness)}}$$
For continuous shells: $V = \int 2\pi \cdot (\text{radius}) \cdot (\text{height}) \, d(\text{radius})$
Test yourself on these prerequisite skills. If any feel unfamiliar, follow the review link before continuing.
Answer: $\left[\frac{x^4}{4}\right]_0^2 = \frac{16}{4} = 4$
If this was difficult, review Power Rule Integration.
Answer: $C = 2\pi r$
This is essential for the shell method. The factor $2\pi r$ appears in every shell formula.
Answer: $V = \int_0^1 \pi (x^2)^2 \, dx = \pi \int_0^1 x^4 \, dx$
If this was difficult, review Disk/Washer Method.
Some volume problems become nightmares with disks. Consider rotating the region under $y = x^3 - 3x + 2$ about the $y$-axis. To use washers, you would need to solve this cubic for $x$—typically requiring the cubic formula, which is messy and error-prone.
The shell method offers an elegant alternative. Instead of slicing perpendicular to the axis (creating disks), we slice parallel to it. Each slice, when rotated, forms a thin cylindrical tube called a shell.
Think of peeling an onion: each thin layer is a cylindrical shell. The shell method builds a solid by summing infinitely many such layers.
| Property | Value |
|---|---|
| Chapter | 5 - Applications of Integration |
| Section | 5.3 |
| Difficulty | Beginner |
| Time | ~15 minutes |
| Formula | When to Use |
|---|---|
| $V = 2\pi r h \Delta r$ | Single shell with finite thickness |
| $V = \int 2\pi r h \, dr$ | Continuous shells (infinitesimal thickness) |
A cylindrical shell is the region between two concentric cylinders—like a pipe or a toilet paper tube.
╭─────────────────╮
/ \
/ ╭───────╮ \
│ │ │ │
│ │ │ │ height h
│ │ │ │
│ └───────┘ │
\ /
\___________________/
←─── thickness Δr ───→
Inner radius: r₁
Outer radius: r₂ = r₁ + Δr
Average radius: r = (r₁ + r₂)/2Step 1: Start with the volumes of the outer and inner cylinders.
$$V_{\text{outer}} = \pi r_2^2 h \qquad V_{\text{inner}} = \pi r_1^2 h$$
Step 2: Subtract to get the shell volume.
$$V = \pi r_2^2 h - \pi r_1^2 h = \pi(r_2^2 - r_1^2)h$$
Step 3: Factor the difference of squares.
$$V = \pi(r_2 + r_1)(r_2 - r_1)h$$
Step 4: Introduce average radius $r = \frac{r_1 + r_2}{2}$ and thickness $\Delta r = r_2 - r_1$.
Note that $r_2 + r_1 = 2r$, so:
$$\boxed{V = 2\pi r h \Delta r}$$
Step 5: Interpret geometrically.
$$V = \underbrace{2\pi r}_{\text{circumference}} \times \underbrace{h}_{\text{height}} \times \underbrace{\Delta r}_{\text{thickness}}$$
The Tin Can Label: Peel the label off a tin can. You get a rectangle with width = circumference = $2\pi r$ and height = $h$. The shell's volume equals this rectangle's area times the shell's thickness.
Cut and
Cylindrical Shell unfold Rectangular Slab
───────────────── ──────► ─────────────────
╭──────╮ ┌──────────────────┐
/ \ │ │
│ h │ │ h │
│ │ ===► │ │
\ / └──────────────────┘
╰──────╯ 2πr
radius r, thickness Δr Area = 2πr × h
Volume = 2πrh × ΔrWhen building a solid from infinitely thin shells:
$$V = \int 2\pi \cdot (\text{radius}) \cdot (\text{height}) \, d(\text{radius})$$
The specific form depends on which axis you rotate around and which variable you integrate with respect to. The next skill pages cover these cases.
The shell method emerged as mathematicians sought efficient ways to compute volumes of revolution. While Cavalieri's principle (1635) and the disk method handle many cases elegantly, some shapes—particularly those defined by functions difficult to invert—motivated the development of this "parallel slicing" approach. The shell method is sometimes called the "method of cylindrical shells" or "tube method."
A cylindrical shell has inner radius 3, outer radius 3.5, and height 4. Find its volume.
The region under $y = x^2$ from $x = 0$ to $x = 2$ is rotated about the $y$-axis. A thin vertical strip at position $x$ with width $\Delta x$ generates a shell. Identify:
The region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$ is rotated about the $y$-axis.
The region bounded by $y = 3x$ and $y = x^2$ (in the first quadrant) is rotated about the $y$-axis.
| Mistake | Why It Happens | Correction |
|---|---|---|
| Using $\pi r^2 h$ instead of $2\pi r h \Delta r$ | Confusing shell volume with cylinder volume | A shell is a hollow tube, not a solid cylinder. The $2\pi r$ is circumference, not $\pi r^2$ area. |
| Forgetting the $2\pi$ | Rushed setup | Always write "circumference × height × thickness" and remember circumference = $2\pi r$. |
| Using wrong radius | Not visualizing the setup | Draw the region and axis. Radius = distance from strip to axis. |
| Treating the formula as approximate | Misunderstanding the derivation | The formula $2\pi r h \Delta r$ is exactly equal to $\pi(r_2^2 - r_1^2)h$, not an approximation. |
Now that you understand what a shell is and how to compute its volume, the next skills apply this to actual integration problems:
The Tin Can Label:
Peel the label off a tin can. You get a rectangle. The rectangle's area is the can's circumference ($2\pi r$) times its height ($h$). The shell method treats infinitely thin labels wrapped around an axis—each "label" has area $2\pi r \times h$, and multiplying by the tiny thickness $dr$ gives the shell's volume.
| Previous | Up | Next |
|---|---|---|
| Disk/Washer Method | Section 5.3 | Shell Method: y-axis |
Last updated: 2026-01-23