Volume with Washer Method: Revolving Around the x- or y-Axis

Identify nonnegative outer and inner radii and subtract their squared disc areas.

This is shared AB/BC instruction. Open the AB edition of Topic 8.11. Your BC practice and progress stay in this book.

College Board Topic 8.11 · CED p. 157 (PDF p. 162) · CED effective Fall 2020; project course outline 2026–27.

A = π (R^{2} - r^{2})

A slice that stops short of the rotation axis creates a washer: outer disc minus inner disc. Its area is π(R²−r²), with R≥r≥0. It is not π(R−r)². About the x-axis measure both vertical radii from zero and use dx; about the y-axis measure horizontal radii and use dy.

Measure both radii from the same rotation axis. The hole has its own area.

For x≤y≤2 on 0≤x≤2 rotated about the x-axis, outer radius R=2 and inner radius r=x. The volume is π∫₀²(4−x²)dx=16π/3. If the identity of the farther boundary changes, split the interval. A region crossing the axis can fill the center instead of leaving a hole; inspect the geometry rather than automatically subtracting two squared functions.

Let's work through it

For y²≤x≤y on 0≤y≤1 rotated about the y-axis, outer radius is y and inner radius is y². Reason: Horizontal distances are measured from x=0.
V=π∫₀¹(y²−y⁴)dy=2π/15. Reason: Subtract the two disc areas, then integrate.

Inspect the axis and the hole before squaring

A perpendicular segment rotated around an axis forms a washer only if it leaves a gap from that axis. Measure the farther boundary distance R and nearer boundary distance r from the same axis, with R ≥ r ≥ 0. The rotated slice has area πR² − πr², not π(R − r)². About the x-axis, vertical segments produce dx slices; about the y-axis, horizontal segments produce dy slices. Changing the axis can change both the radius expressions and the bounds.

For the region 1 ≤ x ≤ 4 and √x ≤ y ≤ 2, rotation about the x-axis gives R = 2 and r = √x. Rotation about the y-axis uses the equivalent horizontal description 1 ≤ y ≤ 2 and 1 ≤ x ≤ y², giving R = y² and r = 1. The completed graphical keys show both base slices and actual ring sizes. The full volumes are π∫₁⁴(4 − x) dx = 9π/2 cm³ and π∫₁²(y⁴ − 1) dy = 26π/5 cm³. If a segment crosses an axis, the rotation fills the center; inspect the farthest distance and split the integral when the geometry changes.

The base region is 1 to 4 centimeters in x, between y equals square root of x and y equals 2. About the x-axis a vertical slice rotates to a washer. The representative x equals 2.25 slice has outer radius 2 centimeters and inner radius 1.5 centimeters, shown by concentric circles at their true ratio. The integral pi from 1 to 4 of (4 minus x) dx evaluates to 9 pi/2 cubic centimeters. — Coordinate-axis washer geometry and x-axis completed key

The base region is 1 to 4 centimeters in x, between y equals square root of x and y equals 2. About the y-axis a horizontal slice runs from x equals 1 to x equals y squared. At y equals 1.5 its outer radius is 2.25 centimeters and inner radius 1 centimeter, shown at their true ratio. The integral pi from 1 to 2 of (y to the fourth minus 1) dy is 26 pi/5 cubic centimeters. — Coordinate-axis washer geometry and y-axis completed key

CHA-5.C CHA-5.C.3

Integrate the ring, not its thickness

Rotate the region 0 ≤ x ≤ 1 cm and 1 ≤ y ≤ 2 − x cm about the x-axis.

A vertical slice is perpendicular to the rotation axis and uses dx.
The outer radius is R = 2 − x and the inner radius is r = 1. The slice stops short of the axis except where the two boundaries meet.
Set V = π∫₀¹[(2 − x)² − 1] dx = π∫₀¹(3 − 4x + x²) dx.
Evaluate π[3x − 2x² + x³/3]₀¹ = 4π/3 cm³.

Answer and check

V = 4π/3 cm³. Squaring the thickness 1 − x would instead describe a filled disc of a different radius, not this ring.

CHA-5.C CHA-5.C.3

Try it from memory

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Try it before you peek.

Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.

Question 1 of 20 self-assessed

Question 1

Why is washer area not π(R−r)²?

Your answer and reasoning

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Question 1: prompt and worked reasoning

Why is washer area not π(R−r)²?

The hole area is πr², subtracted from πR².

Thickness is not the radius of a filled disc.

Question 2: prompt and worked reasoning

Can a negative x-coordinate be an inner radius directly?

No.

Radius is a nonnegative distance to the axis.

See every recall question with its answer

These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.

Retrieval 1

Why is washer area not π(R−r)²?

Worked answer and reason

The hole area is πr², subtracted from πR².

Thickness is not the radius of a filled disc.

Retrieval 2

Can a negative x-coordinate be an inner radius directly?

Worked answer and reason

No.

Radius is a nonnegative distance to the axis.

Catch the tempting mistake

The upper curve is always the outer radius.: Outer means farther from the axis. Coordinate ordering alone may not determine distance, especially below or left of the axis.

Calculator workspace and limits

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Put it to work

Unit 8 problem 22 · No calculator · analytic

Explain and apply the mathematics

Try these problems before opening the solutions. Show your setup, the hypotheses behind your method, and what your answer means.

Rotate one actual base about two axes

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The base region is 1 ≤ x ≤ 4 and √x ≤ y ≤ 2. Coordinates are in centimeters. Compare solids made by rotating this actual region about the x-axis and about the y-axis; use washers perpendicular to each selected axis.

The base region is 1 to 4 centimeters in x, between y equals square root of x and y equals 2. Both coordinate axes are visible; the task asks for perpendicular slices and washers about each. — Coordinate-axis washer geometry

(a) For rotation about the x-axis, sketch a representative vertical base slice and the resulting washer at x = 2.25. Label the rotation axis, outer and inner radii and slice thickness, then give R(x), r(x) and the area function. Explain why π(R − r)² is not this washer’s area.

CHA-5.C CHA-5.C.3

(b) Construct and evaluate the full volume integral for rotation about the x-axis. Include the actual bounds and units.

CHA-5.C CHA-5.C.3

(c) For rotation about the y-axis, rewrite the same base with horizontal slices and actual y-bounds. Sketch the slice and its washer at y = 1.5, label both radii and thickness, then construct and evaluate the full volume integral. Explain why the x-axis integral cannot simply be reused.

CHA-5.C CHA-5.C.3

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Worked solution and rubric →Revisit the explanation →Revisit the explanation →

Does the rotating segment actually leave a hole?

No calculator

The supplied figure shows the region 0 ≤ x ≤ 2 and x − 1 ≤ y ≤ x + 1. Coordinates are in centimeters, and the rotation axis is y = 0. Inspect the actual geometry before choosing a formula.

The base lies from x equals 0 to 2 centimeters between y equals x minus 1 and y equals x plus 1, rotated about the x-axis. The lower boundary crosses the rotation axis at x equals 1. — Axis-crossing geometry

(a) Rotate 0 ≤ x ≤ 2 and x − 1 ≤ y ≤ x + 1 about the x-axis. Which complete volume setup and value correctly account for the slice crossing the axis?

CHA-5.C CHA-5.C.1 CHA-5.C.3

Worked solution and rubric →Revisit the explanation →

For the teacher

Use the accessible ring diagram and a text-only radius table. Ask students to identify R and r before seeing the formula.

Unit planning and formative guidance →

Glossary and deeper reading · 2 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Disc method
The disc method finds a solid’s volume using circular slices perpendicular to the rotation axis. A slice with radius R has area πR². For example, rotating the region under y=x on [0,1] around the horizontal axis gives volume equal to the integral of πx². If the region leaves a hole around the axis, a washer rather than a full disc is needed.
Read more: 8.9 · Disc method →
Integral
Integration can mean accumulating a function over an interval or finding an antiderivative. A definite integral gives a number, while an indefinite integral represents a family of antiderivatives. For example, integrating 2x indefinitely gives x²+C, and integrating it from 0 to 2 gives 4. The Fundamental Theorem of Calculus connects these two operations under its hypotheses.
Read more: 6.7 · Integral →

Let's work through it

Inspect the axis and the hole before squaring

Integrate the ring, not its thickness

Answer and check

Try it from memory

Try it before you peek.

Question 1

Retrieval 1

Retrieval 2

Catch the tempting mistake

Put it to work

Explain and apply the mathematics

Rotate one actual base about two axes

Does the rotating segment actually leave a hole?

Disc method

Integral

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