Calculus · Step-by-step guide

Disk vs. Washer Method

Set up volumes of revolution with disks and washers, including correct radii and bounds.

By Tyler Blovat··9 min read
The short answer

Start with the central idea

A disk has no hole: V=π∫R2\displaystyle V=\pi\int R^2. A washer has an outer and inner radius: V=π∫(R2−r2)\displaystyle V=\pi\int(R^2-r^2). Both radii are distances from the axis of rotation.

V=π∫ab(R(x)2−r(x)2)dx\displaystyle V=\pi\int_a^b\left(R(x)^2-r(x)^2\right)dx
01

How it works

Build the method from meaning before memorizing the moves.

These are the decisions that make the procedure reliable. Read them once, then look for each idea in the worked examples below.

1

Slice perpendicular to the axis

Disk and washer cross-sections come from slices perpendicular to the line of rotation.

2

Measure distances

A radius is top minus axis, axis minus bottom, right minus axis, or axis minus left—not automatically the function itself.

3

Outer minus inner

Square each radius separately, then subtract inner cross-sectional area from outer area.

02

See the structure

A picture makes the relationships easier to remember.

DISKRdxsolid circular sliceWASHERRrdxouter disk − inner hole
A disk slice is solid; a washer slice subtracts the circular hole with area π(R² − r²). Adding slices of thickness dx builds the 3D volume.
03

Worked examples

Follow the reason for each line, then try to reproduce it without looking.

Example 1Disk about x-axis

Rotate y=x\displaystyle y=x from x = 0 to 2 about the x-axis.

  1. There is no gap, so r = 0.
  2. Outer radius R = x.
  3. V=π∫02x2dx\displaystyle V=\pi\int_0^2x^2dx.
Answer8π/3\displaystyle 8\pi/3
Example 2Washer

Rotate the region between y = 4 and y = x², −2 ≤ x ≤ 2, about the x-axis.

  1. Outer radius is 4; inner radius is x².
  2. V=π∫−22(16−x4)dx\displaystyle V=\pi\int_{-2}^2(16-x^4)dx.
  3. Use symmetry or evaluate directly.
Answer256π/5\displaystyle 256\pi/5
Example 3Shifted axis

Rotate the region from y = x to y = 3 about y = 5.

  1. Distances are measured downward from y = 5.
  2. Outer radius to y = x is 5−x\displaystyle 5-x.
  3. Inner radius to y = 3 is 2.
AnswerIntegrand π((5−x)2−4)\displaystyle \pi((5-x)^2-4)
04

Common mistakes—and how to avoid them

Accuracy improves fastest when you know what to check.

  • Using inner minus outer.

  • Forgetting to measure radius from a shifted axis.

  • Using disk/washer slices parallel to the rotation axis.

05

Turn the explanation into a skill

Reading creates recognition. Independent practice creates recall.

Free printable worksheet · Answer key included

Basic Integration

Strengthen the definite-integration skills used to evaluate disk and washer volumes.

Preview the worksheet

A strong practice loop: solve one example with the guide open, solve a similar problem without it, explain the method aloud, and return the next day for a short mixed review.

06

Practice problems with worked solutions

Solve each problem on paper, use the hint only if needed, then compare every step.

1

Rotate y=x\displaystyle y=x from x = 0 to 2 about the x-axis. Find the volume.

Show hint

There is no gap, so use disks.

Show worked solution
  1. Radius R=x\displaystyle R=x.
  2. V=π∫02x2dx\displaystyle V=\pi\int_0^2x^2dx.
  3. V=π[x3/3]02\displaystyle V=\pi[x^3/3]_0^2.

Answer: 8π/3\displaystyle 8\pi/3

2

Rotate the region between y=3\displaystyle y=3 and the x-axis from x = 0 to 4 about the x-axis.

Show hint

Use constant-radius disks.

Show worked solution
  1. R=3\displaystyle R=3.
  2. V=π∫049dx\displaystyle V=\pi\int_0^4 9dx.
  3. V=9π(4)\displaystyle V=9\pi(4).

Answer: 36π\displaystyle 36\pi

3

Rotate the region between y=x\displaystyle y=\sqrt{x} and y=1\displaystyle y=1 from x = 1 to 4 about the x-axis.

Show hint

Outer radius is the upper curve.

Show worked solution
  1. R=x\displaystyle R=\sqrt{x} and r=1\displaystyle r=1.
  2. V=π∫14(x−1)dx\displaystyle V=\pi\int_1^4(x-1)dx.
  3. Evaluate π[x2/2−x]14\displaystyle \pi[x^2/2-x]_1^4.

Answer: 9π/2\displaystyle 9\pi/2

4

Rotate the region between y=4\displaystyle y=4 and y=x2\displaystyle y=x^2 from x = 0 to 2 about the x-axis.

Show hint

Use outer radius 4 and inner radius x².

Show worked solution
  1. V=π∫02(42−(x2)2)dx\displaystyle V=\pi\int_0^2(4^2-(x^2)^2)dx.
  2. V=π[16x−x5/5]02\displaystyle V=\pi[16x-x^5/5]_0^2.

Answer: 128π/5\displaystyle 128\pi/5

5

A rotated region has outer radius 5, inner radius 2, and thickness 7. Find its volume.

Show hint

Use washer area times thickness.

Show worked solution
  1. Washer area =π(52−22)=21π\displaystyle =\pi(5^2-2^2)=21\pi.
  2. Multiply by thickness 7.

Answer: 147π\displaystyle 147\pi cubic units.

07

Frequently asked questions

Quick answers before you move on.

What should I remember first?

A disk has no hole: V=π∫R2\displaystyle V=\pi\int R^2. A washer has an outer and inner radius: V=π∫(R2−r2)\displaystyle V=\pi\int(R^2-r^2). Both radii are distances from the axis of rotation.

How do I know whether I understand this topic?

You should be able to name the method, explain why each step is valid, complete a new example without copying, and check whether your answer is reasonable.

What should I do if I keep making the same mistake?

Write the error as a specific checkpoint—for example, “I will identify the hypotenuse before substituting.” Then use that checkpoint on three short problems rather than repeating a full page without feedback.

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