Shell Method: Rotation About the y-axis
Quick Reference $$\boxed{V = \int_a^b 2\pi x f(x) \, dx}$$
Component Meaning $x$ Radius (distance from strip to $y$-axis) $f(x)$ Height (length of vertical strip) $dx$ Thickness $2\pi x$ Circumference of the shell
Before You Start
1. Can you identify the radius, height, and thickness from a shell formula setup?
In $V = 2\pi r h \Delta r$:
- $r$ = radius (distance to axis)
- $h$ = height (length of strip)
- $\Delta r$ = thickness
If this is unclear, review Shell Method Formula.
2. Can you find where $y = x^2$ and $y = 2x$ intersect?
Solution: Set $x^2 = 2x$, so $x^2 - 2x = 0$, giving $x(x - 2) = 0$. Thus $x = 0$ or $x = 2$.
If this was difficult, you’ll struggle with bounds for shell integrals. Review solving systems of equations.
3. Can you evaluate $\int_0^2 x^{3/2} \, dx$?
Solution: $\left[\frac{x^{5/2}}{5/2}\right]_0^2 = \frac{2}{5} \cdot 2^{5/2} = \frac{2}{5} \cdot 4\sqrt{2} = \frac{8\sqrt{2}}{5}$
If fractional exponents feel unfamiliar, review Power Rule Integration.
The Standard Case
Rotation about the $y$-axis is the natural habitat for the shell method. When a region defined by $y = f(x)$ rotates about the $y$-axis, vertical strips become cylindrical shells in a direct way:
- The strip’s distance from the $y$-axis becomes the radius
- The strip’s length becomes the height
- The strip’s width becomes the thickness
This is often simpler than the disk/washer method, which would require solving $y = f(x)$ for $x$: impossible or painful when $f$ is a cubic or worse.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Chapter | 5 - Applications of Integration |
| Section | 5.3 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
| Scenario | Shell Formula |
|---|---|
| Region under $y = f(x)$ | $V = \int_a^b 2\pi x f(x) \, dx$ |
| Region between $y = f(x)$ and $y = g(x)$ | $V = \int_a^b 2\pi x [f(x) - g(x)] \, dx$ |
Key Concepts
The Standard Formula
For a region bounded by $y = f(x)$ (where $f(x) \geq 0$), $y = 0$, $x = a$, and $x = b$ (with $0 \leq a < b$), rotated about the y-axis:
$$\boxed{V = \int_a^b 2\pi x f(x) \, dx}$$
Understanding Each Part
| Component | What It Represents | In the Formula |
|---|---|---|
| Radius | Distance from $y$-axis to strip | $x$ |
| Height | Length of vertical strip | $f(x)$ |
| Thickness | Width of strip | $dx$ |
| Circumference | Perimeter of circular cross-section | $2\pi x$ |
Step-by-Step Process
Step 1: Sketch the region and identify the bounds on $x$.
Step 2: Draw a typical vertical strip at position $x$.
Step 3: Identify the shell components:
- Radius = $x$ (distance to $y$-axis)
- Height = $f(x)$ (or top curve minus bottom curve)
- Thickness = $dx$
Step 4: Write the integral: $$V = \int_a^b 2\pi \cdot (\text{radius}) \cdot (\text{height}) \, dx$$
Step 5: Simplify the integrand before integrating.
Step 6: Evaluate the definite integral.
Step 7: Verify (optional but recommended): Check dimensions or compare with another method.
Visualizing the Setup
y
│ ╭───────╮
│ ╱ │
│ ╱ f(x) │ ← height of shell = f(x)
│ ╱ │
│ ╱───────────────┤
├─────┬───────────┼─── x
0 a x b
↑
└── radius of shell = x (distance to y-axis)
When this strip rotates around the y-axis,
it sweeps out a cylindrical shell.Region Between Two Curves
When the region is between $y = g(x)$ (bottom) and $y = f(x)$ (top):
$$V = \int_a^b 2\pi x [f(x) - g(x)] \, dx$$
The height of each shell is the vertical distance between the curves.
y
│ ╭─── f(x) (top curve)
│ ╱
│ ╱ ↕ height = f(x) - g(x)
│ ╱
│ ╲─────── g(x) (bottom curve)
├─────────────────── x
0 a x b📜 Historical Note: Why Shells Matter
The shell method was developed because many natural shapes, like solids formed by rotating cubics such as $y = x^3 - x^2 + x$, lead to equations that are impractical to solve when using the disk method. Before computer algebra systems, mathematicians needed methods that worked with the “natural” description of a region. The shell method lets you work with $y = f(x)$ directly, without solving for $x = f^{-1}(y)$.
Practice Problems
Find the volume of the solid obtained by rotating the region under $y = x^2$ from $x = 0$ to $x = 3$ about the $y$-axis.
Find the volume of the solid obtained by rotating the region bounded by $y = 3x - x^3$ and $y = 0$ (for $x \geq 0$) about the $y$-axis.
Find the volume of the solid obtained by rotating about the $y$-axis the region bounded by $y = 2x$ and $y = x^2$.
Find the volume of the solid obtained by rotating about the $y$-axis the region bounded by $y = 3x^2 - x^3$ and $y = 0$.
Consider the region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$, rotated about the $y$-axis.
- Set up the volume integral using the shell method.
- Set up the volume integral using the disk/washer method.
- Evaluate both integrals and verify they give the same answer.
- Discuss which method required fewer steps for this problem.
Common Mistakes
| Mistake | Why It Happens | Correction |
|---|---|---|
| Using $2\pi f(x)$ as radius | Confusing radius with height | Radius is $x$ (distance to axis), height is $f(x)$. |
| Forgetting the $2\pi$ | Rushed setup or mixing with disk method | Always write out: circumference × height × thickness = $2\pi x \cdot f(x) \cdot dx$. |
| Wrong bounds (using $y$-values) | Mixing up integration variables | When integrating with $dx$, bounds are $x$-values. |
| Height as $(f(x))^2$ | Confusing with the disk method | The disk method squares the radius; the shell method doesn’t square the height. |
| Subtracting curves in wrong order | Not checking which is “on top” | Test a point between intersections to see which $y$-value is larger. |
Still Confused?
- Formula components unclear? → Review Shell Method Formula
- Integration of powers giving trouble? → Review Power Rule Integration
- Finding intersections is hard? → Practice solving $f(x) = g(x)$ by setting to zero and factoring
- Not sure when shells beat washers? → Preview Shells vs. Washers
Common Misconceptions
in the shell formula $V = \int 2\pi x f(x)\,dx$, squaring $f(x)$ is necessary because volumes involve squared quantities.
This is the concept-image-conflicts-definition error arising from confusing the shell method with the disk method. The disk method squares the radius because cross-sectional area is $\pi r^2$. In the shell method, the integrand $2\pi x \cdot f(x)$ represents circumference times height, and neither factor is squared: the height $f(x)$ enters linearly. For the region under $y = x^2$ from $x = 0$ to $x = 3$ rotated about the $y$-axis, the correct integral is $2\pi \int_0^3 x \cdot x^2\,dx = 2\pi \int_0^3 x^3\,dx$, not $2\pi \int_0^3 x \cdot (x^2)^2\,dx$.
Mastery Checklist
Looking Ahead
You’ve mastered the standard case. Next, we generalize:
- Shell Method: Other Axes: what changes when the axis is $x = 3$ instead of $x = 0$?
- Shells vs. Washers: how to decide which method to use in any situation
Mental Model
The Vertical Strip Wrap:
Picture taking a tall, thin strip of paper and wrapping it around a pole (the $y$-axis). The distance from the strip to the pole is the radius, the strip’s length is the height, and its width is the thickness. Sum up infinitely many such wrapped strips to get the solid.
When $y = f(x)$ is complicated, this “wrap around the axis” approach beats having to unwrap and re-describe the curves in terms of $y$.
| Previous | Up | Next |
|---|---|---|
| Shell Method Formula | Section 5.3 | Shell Method: Other Axes |
Last updated: 2026-01-23