Washer method around the x- or y-axis

AP Calculus AB· AP Calculus AB CED — Applications of Integration· 14 min read

1. What Is the Washer Method?★☆☆☆☆⏱ 2 min

The washer method is an integration technique used to calculate the volume of a solid of revolution that has a hollow core, formed when rotating a region between two curves around an axis. Each cross-section perpendicular to the axis forms a "washer" (a flat disk with a smaller circular hole cut out of the center), hence the name, and it is an extension of the disk method for solid solids of revolution.

Per the AP Calculus AB Course and Exam Description, this topic is part of Unit 8, which accounts for 10–15% of the total AP exam score. Washer method problems appear on both multiple-choice and free-response sections, and you will often be asked to set up or evaluate integrals for volume on FRQs, making this a high-weight topic.

📘 Definition

Solid of Revolution

A 3D solid formed by rotating a 2D region around a fixed axis (the axis of rotation)

Example:

Rotating a region bounded by two curves around the x-axis

2. Revolution Around the x-Axis (Horizontal Axes)★★☆☆☆⏱ 4 min

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When rotating around any horizontal axis (of the form $y = k$ , including the x-axis $y = 0$ ), cross-sections perpendicular to the axis are vertical slices, so we integrate with respect to $x$ . The area of one washer is the area of the outer circle minus the area of the inner circle:

A = \pi R^2 - \pi r^2 = \pi\left(R^2 - r^2\right)

where $R (x)$ is the outer radius (distance from the axis to the farther curve from the axis), and $r (x)$ is the inner radius (distance from the axis to the closer curve from the axis). Total volume is the integral of all cross-sectional areas between the bounds of the region:

V = \pi \int_a^b \left(\left[R(x)\right]^2 - \left[r(x)\right]^2\right)dx

For rotation around the x-axis ( $y = 0$ ), this simplifies to $R (x) = y_{upper} (x)$ and $r (x) = y_{lower} (x)$ , since the distance from $y = 0$ to a curve is just the $y$ -value of the curve.

📐 Worked Example

Find the volume of the solid formed when the region bounded by $y = 2 x$ , $y = x^{2}$ , $0 \leq x \leq 2$ is rotated around the x-axis.

1
Confirm the axis is horizontal ( $y = 0$ ), so we integrate with respect to $x$ with bounds $a = 0$ , $b = 2$ .
2
Identify radii: Between $0$ and $2$ , $2 x \geq x^{2}$ , so the farther curve from the x-axis is $y = 2 x$ , and the closer curve is $y = x^{2}$ . Thus:
3
$R(x) = 2x, \quad r(x) = x^2$
4
Set up the volume integral:
5
$V = \pi \int_0^2 \left((2x)^2 - (x^2)^2\right)dx = \pi \int_0^2 \left(4x^2 - x^4\right)dx$
6
Evaluate the definite integral:
7
$V = \pi \left[\frac{4x^3}{3} - \frac{x^5}{5}\right]_0^2 = \pi \left(\frac{32}{3} - \frac{32}{5}\right) = \frac{64\pi}{15}$

Exam tip:

Always square each radius separately before subtracting. Never subtract first then square, as this will always give the wrong volume.

3. Revolution Around the y-Axis (Vertical Axes)★★☆☆☆⏱ 4 min

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When rotating around any vertical axis (of the form $x = k$ , including the y-axis $x = 0$ ), cross-sections perpendicular to the axis are horizontal slices, so we integrate with respect to $y$ . This means we must first rewrite all bounding curves as functions of $y$ , instead of $y$ as a function of $x$ . The logic for radii and area is identical to the horizontal case:

V = \pi \int_c^d \left(\left[R(y)\right]^2 - \left[r(y)\right]^2\right)dy

For rotation around the y-axis ( $x = 0$ ), $R (y) = x_{right} (y)$ (farther from the axis) and $r (y) = x_{left} (y)$ (closer to the axis).

📐 Worked Example

Find the volume of the solid formed when the region bounded by $y = 2 x$ , $y = x^{2}$ , $0 \leq y \leq 4$ is rotated around the y-axis.

1
Confirm the axis is vertical ( $x = 0$ ), so we integrate with respect to $y$ with bounds $c = 0$ , $d = 4$ . Rewrite both curves as $x$ in terms of $y$ .
2
Rewrite the functions:
3
$y = 2x \implies x = \frac{y}{2}; \quad y = x^2 \implies x = \sqrt{y} \quad (\text{positive root for this region})$
4
Identify radii: Between $0$ and $4$ , $y \geq \frac{y}{2}$ , so $R (y) = y$ (farther from y-axis), $r (y) = \frac{y}{2}$ (closer to y-axis).
5
Set up and evaluate the integral:
6
$V = \pi \int_0^4 \left((\sqrt{y})^2 - \left(\frac{y}{2}\right)^2\right)dy = \pi \int_0^4 \left(y - \frac{y^2}{4}\right)dy$
7
$V = \pi \left[\frac{y^2}{2} - \frac{y^3}{12}\right]_0^4 = \pi \left(8 - \frac{64}{12}\right) = \frac{8\pi}{3}$

Exam tip:

Always rewrite all functions in terms of the integration variable before calculating radii. Mixing $x$ and $y$ in the same integral is an immediate point loss on FRQs.

4. Revolution Around Non-Coordinate Axes★★★☆☆⏱ 4 min

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AP Calculus AB exam questions frequently ask for rotation around an axis that is not the x or y axis (e.g. $y = 2$ , $x = - 1$ ). The core logic stays the same, but you must correctly calculate the radius as the distance between the curve and the shifted axis, not just the value of the curve.

📐 Worked Example

Find the volume of the solid formed when the region bounded by $y = 2 x$ , $y = x^{2}$ , $0 \leq x \leq 2$ is rotated around the horizontal line $y = 5$ .

1
Axis is horizontal ( $y = 5$ ), so integrate with respect to $x$ , bounds $0$ to $2$ . The entire region lies below $y = 5$ , so all distances are $5 - y$ .
2
Identify radii: The lower curve $y = x^{2}$ is farther from $y = 5$ , so it gives the outer radius, and the upper curve $y = 2 x$ is closer to $y = 5$ , so it gives the inner radius:
3
$R(x) = 5 - x^2, \quad r(x) = 5 - 2x$
4
Set up the integral:
5
$V = \pi \int_0^2 \left((5 - x^2)^2 - (5 - 2x)^2\right)dx$
6
Simplify and evaluate the integral:
7
$(25 - 10x^2 + x^4) - (25 - 20x + 4x^2) = x^4 - 14x^2 + 20x$
8
$V = \pi \int_0^2 (x^4 - 14x^2 + 20x)dx = \pi \left[\frac{x^5}{5} - \frac{14x^3}{3} + 10x^2\right]_0^2 = \frac{136\pi}{15}$

Exam tip:

Test a sample point in your interval to confirm $R > r$ : if your integrand is negative, you swapped outer and inner radii.

5. AP-Style Concept Check★★★☆☆⏱ 3 min

✓ Quick check

Test your understanding with these AP-style multiple choice questions:

The region $R$ is bounded by $y = x$ , $y = 2$ , and $x = 0$ . Which of the following gives the volume of the solid formed when $R$ is rotated around the y-axis?
- A) $π \int_{0}^{4} ((y^{2})^{2} - 0^{2}) d y$
- B) $π \int_{0}^{2} (2^{2} - (x)^{2}) d x$
- C) $π \int_{0}^{2} ((y^{2})^{2} - 0^{2}) d y$
- D) $π \int_{0}^{4} (y - 4) d x$
Reveal answer
2 —
Correct. Rotation around the vertical y-axis requires integrating with respect to $y$ , with bounds from $y = 0$ to $y = 2$ , matching this option. A has wrong bounds, B uses the wrong integration variable, D is incorrect.
Let $R$ be the region bounded by $y = x^{2} + 1$ , $y = x + 3$ , $x \geq 0$ . What is the correct integral for the volume when $R$ is rotated around the horizontal line $y = - 1$ using the washer method?
- A) $π \int_{0}^{2} ((x + 3)^{2} - (x^{2} + 1)^{2}) d x$
- B) $π \int_{- 1}^{2} ((x + 4)^{2} - (x^{2} + 2)^{2}) d x$
- C) $π \int_{0}^{2} ((x + 4)^{2} - (x^{2} + 2)^{2}) d x$
- D) $π \int_{0}^{2} ((x + 2)^{2} - (x^{2} + 1)^{2}) d x$
Reveal answer
2 —
Correct. Intersection of curves is at $x = 2$ , so bounds are $x = 0$ to $x = 2$ . Radii are distance from $y = - 1$ : outer radius is $(x + 3) - (- 1) = x + 4$ , inner radius is $(x^{2} + 1) - (- 1) = x^{2} + 2$ .

6. Common Pitfalls

Wrong move:

Writing $(R - r)^{2}$ instead of $R^{2} - r^{2}$ , subtracting radii before squaring.

Why:

Students confuse factoring rules with the circle area formula, or rush through setup.

Correct move:

Always square each radius separately first, then subtract the squared inner radius from the squared outer radius.

Wrong move:

Integrating with respect to $x$ when rotating around a vertical axis, or with respect to $y$ when rotating around a horizontal axis.

Why:

Students forget cross-sections must be perpendicular to the axis of rotation, which dictates the integration variable.

Correct move:

For any horizontal axis, integrate with respect to $x$ ; for any vertical axis, integrate with respect to $y$ , and rewrite all functions first.

Wrong move:

Calculating radius for a non-coordinate axis as just the function value, e.g. for rotation around $y = 2$ , writing $R (x) = x^{2}$ instead of $∣2 - x^{2} ∣$ .

Why:

Students get used to rotating around the x-axis ( $y = 0$ ) and forget to adjust radii for a shifted axis.

Correct move:

Always calculate radius as the absolute difference between the curve and the axis: $∣ c u r v e - a x i s ∣$ .

Wrong move:

Swapping outer and inner radii, resulting in a negative integrand.

Why:

Students mix up which curve is farther from the axis, especially when the axis is above or to the left of the region.

Correct move:

After identifying radii, test one point in the interval to confirm $R > r$ and the integrand is positive.

Wrong move:

Forgetting to multiply the entire integral by $π$ , even when asked for a numerical volume.

Why:

Students remember the core integrand but omit the constant from the circle area formula.

Correct move:

Write $π$ outside the integral when setting up, before any other steps.

Wrong move:

Leaving functions in terms of $x$ when integrating with respect to $y$ for rotation around a vertical axis.

Why:

Students are more comfortable with $y$ as a function of $x$ and forget to invert.

Correct move:

As soon as you confirm you are integrating with respect to $y$ , invert all functions to get $x$ as a function of $y$ before calculating radii.

7. Quick Reference Cheatsheet

Category	Formula	Notes
General Washer Cross-Section Area	$A = π (R^{2} - r^{2})$	$R$ = outer radius (farthest from axis), $r$ = inner radius (closest to axis), both always positive
Volume, Revolution Around Horizontal Axis $y = k$	$V = π \int_{a}^{b} ([R (x)]^{2} - [r (x)]^{2}) d x$	Integrate with respect to $x$ ; $R (x) = ∣ y_{farthest} - k ∣$ , $r (x) = ∣ y_{closest} - k ∣$
Volume, Revolution Around x-axis ( $y = 0$ )	$V = π \int_{a}^{b} ([y_{upper} (x)]^{2} - [y_{lower} (x)]^{2}) d x$	Special case of horizontal axis, $k = 0$ , radii equal to function $y$ -values
Volume, Revolution Around Vertical Axis $x = k$	$V = π \int_{c}^{d} ([R (y)]^{2} - [r (y)]^{2}) d y$	Integrate with respect to $y$ ; rewrite all functions as $x (y)$ first
Volume, Revolution Around y-axis ( $x = 0$ )	$V = π \int_{c}^{d} ([x_{right} (y)]^{2} - [x_{left} (y)]^{2}) d y$	Special case of vertical axis, $k = 0$ , radii equal to function $x$ -values
Radius for Shifted Non-Coordinate Axis	$R a d i u s = ∣ curve - axis ∣$	Absolute value ensures positive radius; farthest curve from axis = outer radius
Disk Method (Special Case of Washer)	$V = π \int R^{2} d (variable)$	When region touches the axis of rotation, inner radius $r = 0$ , washer becomes a solid disk

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Volume around horizontal non-coordinate axis
2022 · FRQ
Set up volume around vertical axis
2021 · MCQ
Identify correct volume integral

What's Next

The washer method builds on your knowledge of area between two curves, extending 2D integration to calculate 3D volumes of revolution, a core application heavily tested on the AP Calculus AB exam. Mastery of this method is a prerequisite for the cylindrical shells method (the other major volume technique for solids of revolution) and shares the same cross-section integration logic used for volumes with known cross-sections, which you will study next. Incorrect radius setup or wrong integration variable is a top cause of lost points on Unit 8 free-response questions, so consistent practice will help you earn full points.

→Washer method around other axes