Washer Method
Solids with Holes
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.
Prerequisite Map
Before You Start
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 |
Refresh: The Disk Method
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.”
Refresh: Sketching to Identify Radii
To find outer and inner radii:
- Draw the region in the $xy$-plane
- Draw the axis of rotation
- At a typical $x$-value, draw a line from the axis to each curve
- The longer distance is the outer radius $R$
- The shorter distance is the inner radius $r$
Key: The radius is a distance, so it’s always positive. Use absolute value thinking if curves are on opposite sides of the axis.
Quick Reference
| Property | Value |
|---|---|
| Concept | Applications of Integration |
| Course | MATH162 |
| Section | Stewart 5.2 |
| Difficulty | Intermediate |
| Time | ~25 minutes |
Key Concepts
The Washer Method Formula
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:
- $R(x)$ = outer radius (distance from axis to the farther curve)
- $r(x)$ = inner radius (distance from axis to the closer curve)
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 RWhen to Use Washers vs. Disks
How 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.
Identifying Inner and Outer Radii
- Draw a cross-section perpendicular to the axis of rotation
- Mark both curves where they intersect this cross-section
- Measure distances from the axis:
- The curve farther from the axis gives $R$ (outer)
- The curve closer to the axis gives $r$ (inner)
Rotation About the $x$-Axis
For region between $y = f(x)$ (top) and $y = g(x)$ (bottom), rotating about $x$-axis:
- If both functions are positive: $R = f(x)$, $r = g(x)$
- If both functions are negative: $R = \vert g(x)\vert $, $r = \vert f(x)\vert $
$$V = \pi\int_a^b \left([f(x)]^2 - [g(x)]^2\right)\,dx$$
Rotation About Other Lines
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 $y$:
$$V = \pi\int_c^d \left([R(y)]^2 - [r(y)]^2\right)\,dy$$
The Algorithm
- Sketch the region and axis of rotation
- Determine if you need washers (region doesn’t touch axis)
- Identify outer and inner radii in terms of the integration variable
- Find limits of integration
- Set up and evaluate: $V = \pi\int (R^2 - r^2)\,d(\text{variable})$
Common Pitfalls
| 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 |
Practice Problems
The region bounded by $y = 2x$ and $y = x^2$ is rotated about the $x$-axis.
- Find the intersection points and sketch the region.
- Identify which curve gives the outer radius and which gives the inner radius.
- Set up (but do not evaluate) the integral for the volume.
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 $R > r$).
- Set up the washer integral for the volume of a torus.
- Evaluate the integral (hint: recognize a known area).
- Verify your answer makes sense when $r$ is small compared to $R$.
Common Misconceptions
the washer area is $\pi(R - r)^2$ because subtracting the inner radius from the outer radius gives the “width” of the washer.
This is the multiplicative-not-additive error. The washer area is $\pi R^2 - \pi r^2$, which factors as $\pi(R+r)(R-r)$, not $\pi(R-r)^2$. For a washer with outer radius 5 and inner radius 3, the correct area is $\pi(25 - 9) = 16\pi$, whereas $\pi(5-3)^2 = 4\pi$ is less than the area of a disk of radius 2 and clearly too small. The two radii contribute their squares separately because area scales with the square of radius, and the square of a difference is not the difference of squares.
Mastery Checklist
Mental Model
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.
Common Pitfall
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: $$\pi(R^2 - r^2) = \pi(R + r)(R - r)$$
Connections
Looking back:
- The Disk Method is the special case where $r = 0$ (no hole)
- Volume by Slicing provides the general framework
Looking ahead:
- The Shell Method offers an alternative that’s sometimes easier, especially for rotation about the $y$-axis
When to use which method:
- Washers/Disks: Cross-sections are perpendicular to the axis of rotation
- Shells: Cross-sections are parallel to the axis of rotation
| Previous | Up | Next |
|---|---|---|
| Disk Method | Skills Index | Known Cross-Sections |
Last updated: 2026-01-23