Quick Reference
Axis of Rotation Radius Strip Direction Integrate $y$-axis ($x = 0$) $x$ Vertical $dx$ $x = k$ (vertical) $\lvert x - k \rvert$ Vertical $dx$ $x$-axis ($y = 0$) $y$ Horizontal $dy$ $y = k$ (horizontal) $\lvert y - k \rvert$ Horizontal $dy$ Universal rule: Radius = distance from shell to axis
For $y = f(x)$ rotated about the $y$-axis: $$V = \int_a^b 2\pi x f(x) \, dx$$
If this is unclear, master Shell Method: y-axis first.
Answer: $\vert 3 - 7\vert = \vert -4\vert = 4$
The distance between a point and a vertical line is always positive. This concept is essential for finding shell radii.
Answer: Square both sides: $y^2 = x$, so $x = y^2$.
For rotation about horizontal axes, we often need to express boundaries as functions of $y$.
Real problems don't always rotate about the $y$-axis. The axis might be $x = 5$, $x = -2$, or even a horizontal line like $y = 3$. The shell method adapts to all these cases with one key insight:
The radius is always the distance from the shell to the axis of rotation.
Once you internalize this rule, any axis becomes manageable.
| Property | Value |
|---|---|
| Chapter | 5 - Applications of Integration |
| Section | 5.3 |
| Difficulty | Intermediate |
| Time | ~25 minutes |
For any axis of rotation:
$$\text{Radius} = \text{distance from shell to axis}$$
This is always positive. Use absolute value when the region might span both sides of the axis.
When rotating about the vertical line $x = k$:
Step 1: Use vertical strips (integrate with $dx$).
Step 2: Find the radius = distance from strip at $x$ to the line $x = k$:
$$\text{radius} = \lvert x - k \rvert$$
Step 3: Simplify based on where the region lies:
| Region Position | Radius Formula |
|---|---|
| Entirely right of axis ($x > k$ for all $x$) | $x - k$ |
| Entirely left of axis ($x < k$ for all $x$) | $k - x$ |
| Spans across axis | Split integral or use $\lvert x - k \rvert$ |
Step 4: Set up the integral: $$V = \int_a^b 2\pi \cdot (\text{radius}) \cdot (\text{height}) \, dx$$
Axis x = k
│
│
←─────────┼──────────→
k - x │ x - k
(radius │ (radius
if x < k) │ if x > k)
│
Example: For axis x = 2 and strip at x = 5:
radius = 5 - 2 = 3
Example: For axis x = 2 and strip at x = -1:
radius = 2 - (-1) = 3When rotating about the horizontal line $y = k$:
Step 1: Use horizontal strips (integrate with $dy$).
Step 2: Express boundaries as functions of $y$ (solve $y = f(x)$ for $x$).
Step 3: Find the radius = distance from strip at height $y$ to the line $y = k$:
$$\text{radius} = \lvert y - k \rvert$$
Step 4: Identify the "width" of each strip: $$\text{width} = (\text{right boundary}) - (\text{left boundary})$$
Step 5: Set up the integral: $$V = \int_c^d 2\pi \cdot (\text{radius}) \cdot (\text{width}) \, dy$$
This is the special case of $y = k$ with $k = 0$:
$$V = \int_c^d 2\pi y \cdot (\text{width}) \, dy$$
The radius is simply $y$ (the height of the strip).
| Axis | Strip Direction | Variable | Radius | Typical Setup |
|---|---|---|---|---|
| $y$-axis ($x = 0$) | Vertical | $dx$ | $x$ | $\int 2\pi x f(x) \, dx$ |
| $x = k$ | Vertical | $dx$ | $\lvert x - k \rvert$ | $\int 2\pi \lvert x-k \rvert f(x) \, dx$ |
| $x$-axis ($y = 0$) | Horizontal | $dy$ | $y$ | $\int 2\pi y \cdot \text{width} \, dy$ |
| $y = k$ | Horizontal | $dy$ | $\lvert y - k \rvert$ | $\int 2\pi \lvert y-k \rvert \cdot \text{width} \, dy$ |
Any line in the plane is either vertical ($x = k$) or has some other slope. For volumes of revolution in calculus courses, we only rotate about vertical or horizontal axes because:
Rotation about slanted axes (like $y = x$) is possible but requires more advanced techniques (typically covered in multivariable calculus using different coordinate systems).
For each situation, determine the radius of a shell at position $x$:
Find the volume of the solid obtained by rotating the region bounded by $y = x - x^2$ and $y = 0$ about the line $x = 2$.
Use cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$ about the $x$-axis.
Find the volume of the solid obtained by rotating the region bounded by $y = x^2$ and $y = 2x$ about the line $x = -1$.
The region bounded by $y = 4 - x^2$ and $y = 0$ is rotated about the line $x = 1$.
| Mistake | Why It Happens | Correction |
|---|---|---|
| Using $x - k$ when $x < k$ | Not checking which side of axis | Always sketch! If region is left of axis, use $k - x$. |
| Forgetting to change strip direction for horizontal axes | Habit from $y$-axis problems | For horizontal axes, use horizontal strips and integrate with $dy$. |
| Wrong integration variable | Confusion between cases | Vertical axis → $dx$; horizontal axis → $dy$. |
| Not splitting when region spans axis | Treating $\lvert x - k \rvert$ as $x - k$ | If region is on both sides of axis, split the integral at the axis. |
| Forgetting to convert bounds for $y$ | Using $x$-bounds when integrating $dy$ | Re-express bounds in terms of the new variable. |
You now have the tools to use shells with any axis. The final skill brings everything together:
The Distance Rule:
No matter where the axis is, ask yourself: "How far is my shell from the axis?" That distance is the radius.
Think of it like measuring distance to a wall:
Draw the axis, draw your strip, and measure the gap.
| Previous | Up | Next |
|---|---|---|
| Shell Method: y-axis | Section 5.3 | Shells vs Washers |
Last updated: 2026-01-23