What happens when you rotate a region around an axis it doesn't touch? You get a solid with a hollow core—like a donut, a pipe, or a vase. The cross-sections are no longer solid disks but washers (annular rings).
A washer is the region between two concentric circles. Its area is simply the outer circle's area minus the inner circle's area: $\pi R^2 - \pi r^2$.
The key insight: Two radii matter—the outer radius (to the farther curve) and the inner radius (to the closer curve). Subtract the inner from the outer.
Self-check: Can you answer these questions? If not, review the linked prerequisites first.
| Question | If you struggle... |
|---|---|
| Find the volume when $y = x^2$ (from $x=0$ to $x=1$) is rotated about the $x$-axis | Review Disk Method |
| What is the area of the region between two concentric circles with radii 3 and 5? | This is a washer! Answer: $\pi(5^2) - \pi(3^2) = 16\pi$ |
| If a region is bounded by $y = 2x$ and $y = x^2$, which curve is on top for $0 < x < 2$? | Sketching regions is essential for identifying radii |
When rotating about the $x$-axis with radius $r = f(x)$:
$$V = \pi\int_a^b [f(x)]^2\,dx$$
The washer method extends this by subtracting the volume of the "hole."
To find outer and inner radii:
Key: The radius is a distance, so it's always positive. Use absolute value thinking if curves are on opposite sides of the axis.
| Property | Value |
|---|---|
| Concept | Applications of Integration |
| Course | MATH162 |
| Section | Stewart 5.2 |
| Difficulty | Intermediate |
| Time | ~25 minutes |
When a region between two curves is rotated about an axis, the cross-sections are washers:
$$\boxed{V = \pi\int_a^b \left([R(x)]^2 - [r(x)]^2\right)\,dx}$$
where:
Outer curve
y ↓
| ___________
| / \
| ____/ Region \_____ ← Inner curve
| to rotate
+---------------------------→ x
axis of rotation
Cross-section (perpendicular to axis):
┌─────────────┐
│ ┌─────┐ │
│ │ │ │ Washer = outer disk − inner disk
│ │ ○ │ │ Area = πR² − πr²
│ │ │ │
│ └─────┘ │
└─────────────┘
↑ ↑
r RHow do you know if you need washers or disks? Ask yourself: Does the region touch the axis of rotation?
| Situation | Method |
|---|---|
| Region touches the axis | Disk (no hole) |
| Region does NOT touch the axis | Washer (has a hole) |
| Region between two curves, axis inside | Washer |
| Region between two curves, axis outside | Washer |
Simple test: If there's empty space between the region and the axis, you need washers.
For region between $y = f(x)$ (top) and $y = g(x)$ (bottom), rotating about $x$-axis:
$$V = \pi\int_a^b \left([f(x)]^2 - [g(x)]^2\right)\,dx$$
About $y = k$ (horizontal line):
$$R = \vert f(x) - k\vert , \quad r = \vert g(x) - k\vert $$
About $x = h$ (vertical line): Switch to integrating with respect to $y$:
$$V = \pi\int_c^d \left([R(y)]^2 - [r(y)]^2\right)\,dy$$
| Mistake | Why It's Wrong | Correction |
|---|---|---|
| Writing $(R - r)^2$ instead of $R^2 - r^2$ | These are NOT equal: $(R-r)^2 = R^2 - 2Rr + r^2$ | Area is $\pi R^2 - \pi r^2$, keep separate |
| Swapping inner and outer radii | Gives negative volume or wrong answer | Outer is farther from axis, inner is closer |
| Using disks when washers needed | Missing the hollow core | If region doesn't touch axis, you need washers |
| Forgetting which curve is which | When axis is above/below region, radii flip | Draw the cross-section and measure distances carefully |
The region bounded by $y = 2x$ and $y = x^2$ is rotated about the $x$-axis.
Find the volume of the solid obtained by rotating the region bounded by $y = 2x$ and $y = x^2$ about the $x$-axis.
Find the volume of the solid obtained by rotating the region bounded by $y = 2x$ and $y = x^2$ about the line $y = 5$.
Find the volume of the solid obtained by rotating the region bounded by $y = 2x$ and $y = x^2$ about the line $x = -2$.
A torus is the donut-shaped solid formed by rotating a circle of radius $r$ about a line at distance $R$ from the center of the circle (where $R > r$).
The Pipe Cross-Section: Think of cutting a pipe perpendicular to its length. You see a ring (washer) with an outer radius and an inner radius. The area of material is the big circle minus the hole: $\pi R^2 - \pi r^2$. When building a solid of revolution, you're stacking infinitely many of these washer-shaped slices.
Don't simplify $R^2 - r^2$ to $(R - r)^2$!
$$\pi R^2 - \pi r^2 \neq \pi(R - r)^2$$
The correct factorization (if useful) is: $$\pi(R^2 - r^2) = \pi(R + r)(R - r)$$
Looking back:
Looking ahead:
When to use which method:
| Previous | Up | Next |
|---|---|---|
| Disk Method | Skills Index | Known Cross-Sections |
Last updated: 2026-01-23