Washer method around the x- or y-axis

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

1. Core Concept of the Washer Method★★☆☆☆⏱ 2 min

The washer method is an integration technique to find the volume of a solid of revolution formed when a region between two curves is rotated around a horizontal or vertical axis. When the rotated region does not touch the axis of rotation, a cross-section cut perpendicular to the axis is a washer: a circular disk with a smaller concentric hole, giving the method its name.

It differs from the simpler disk method, which applies when the region touches the axis (leaving no hole), by accounting for the empty volume between the region's inner boundary and the rotation axis. This topic falls within Unit 8, which counts for 10-15% of your total AP Calculus BC exam score.

📘 Definition

Washer Method

Integration technique for volume of revolution that integrates cross-sectional area of annular washers perpendicular to the axis of rotation.

Example:

Used for regions between two curves rotated around an axis, creating a hollow solid.

2. Washer Method for Rotation Around the x-Axis★★☆☆☆⏱ 4 min

When rotating around the horizontal x-axis, we cut cross-sections perpendicular to the x-axis, so we integrate with respect to $x$ . For a region bounded above by $y = R (x)$ (farther from the x-axis) and below by $y = r (x)$ (closer to the x-axis) between $x = a$ and $x = b$ , the area of a single washer at position $x$ is the area of the outer circle minus the area of the inner circle:

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

Integrating the area across the full interval of $x$ gives the total volume formula:

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

📐 Worked Example

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

1
Find intersection points to get integral bounds. Set $2 x = x^{2}$ , which simplifies to $x (x - 2) = 0$ . Intersections are at $x = 0$ and $x = 2$ , so $a = 0$ , $b = 2$ .
2
Identify outer and inner radii. Test $x = 1$ (inside the interval): $2 (1) = 2 > 1^{2} = 1$ , so $y = 2 x$ is farther from the x-axis. Thus $R (x) = 2 x$ , $r (x) = x^{2}$ .
3
Substitute into the volume formula:
4
$V = \pi \int_0^2 \left( (2x)^2 - (x^2)^2 \right) dx = \pi \int_0^2 (4x^2 - x^4) dx$
5
Evaluate the definite integral to get the final volume:
6
$\int_0^2 (4x^2 - x^4) dx = \left[ \frac{4x^3}{3} - \frac{x^5}{5} \right]_0^2 = \frac{32}{3} - \frac{32}{5} = \frac{64}{15}, \quad V = \frac{64\pi}{15}$

Exam tip:

Always confirm which function is farther from the axis of rotation by testing a value in the interval between intersections; do not assume higher-degree functions are always inner or outer.

3. Washer Method for Rotation Around the y-Axis★★★☆☆⏱ 4 min

When rotating around the vertical y-axis, we cut cross-sections perpendicular to the y-axis, so we must express all boundaries as $x$ in terms of $y$ , and integrate over the interval of $y$ values. For a region bounded right by $x = R (y)$ (farther from the y-axis) and left by $x = r (y)$ (closer to the y-axis) between $y = c$ and $y = d$ , the volume formula is:

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

The only change from rotation around the x-axis is that we integrate with respect to $y$ . If the AP exam explicitly asks for the washer method, you must use this approach even if the shells method is computationally easier.

📐 Worked Example

Find the volume of the solid formed when the region bounded by $y = x^{3}$ , $y = 8$ , and $x = 0$ is rotated around the y-axis.

1
Rewrite $y = x^{3}$ as $x$ in terms of $y$ : $x = y^{1/3}$ , using the positive root since $x \geq 0$ in this region.
2
Find bounds for $y$ : the region spans from $y = 0$ to $y = 8$ , so $c = 0$ , $d = 8$ .
3
Identify outer and inner radii: the rightmost boundary is $x = y^{1/3}$ , which is farther from the y-axis than the left boundary $x = 0$ , so $R (y) = y^{1/3}$ , $r (y) = 0$ .
4
Substitute into the volume formula and evaluate:
5
$V = \pi \int_0^8 y^{2/3} dy = \pi \left[ \frac{3 y^{5/3}}{5} \right]_0^8 = \pi \left(\frac{3}{5} \cdot 32\right) = \frac{96\pi}{5}$

Exam tip:

If your original functions are given as $y = f (x)$ , do not forget to solve explicitly for $x$ in terms of $y$ before setting up the integral; integrating with respect to $x$ for the washer method around a vertical axis is always incorrect.

4. Washer Method for Shifted Axes of Rotation★★★★☆⏱ 5 min

The washer method generalizes to any horizontal or vertical axis of rotation, not just the x-axis ( $y = 0$ ) or y-axis ( $x = 0$ ). The core rule for radii stays the same: the radius of any boundary curve is the absolute distance between the curve and the axis of rotation, regardless of where the axis is located.

For a horizontal axis at $y = k$ , we still integrate with respect to $x$ , with radii $R (x) = ∣ outer boundary - k ∣$ and $r (x) = ∣ inner boundary - k ∣$ . For a vertical axis at $x = h$ , we integrate with respect to $y$ , with radii $R (y) = ∣ outer boundary - h ∣$ and $r (y) = ∣ inner boundary - h ∣$ . The volume formula remains the same after substituting these radii.

📐 Worked Example

Find the volume of the solid formed when the region bounded by $y = 2 x$ and $y = x^{2}$ is rotated around the horizontal line $y = - 2$ .

1
Bounds for $x$ are still the intersection points from our earlier example: $a = 0$ , $b = 2$ .
2
Calculate the distance from each boundary to the axis $y = - 2$ . Since both boundaries are above $y = - 2$ , distances are positive: for $y = 2 x$ , distance is $2 x - (- 2) = 2 x + 2$ ; for $y = x^{2}$ , distance is $x^{2} - (- 2) = x^{2} + 2$ .
3
Between $0$ and $2$ , $2 x > x^{2}$ , so $2 x + 2 > x^{2} + 2$ , meaning $R (x) = 2 x + 2$ and $r (x) = x^{2} + 2$ .
4
Substitute, expand, and evaluate:
5
$V = \pi \int_0^2 \left( (2x+2)^2 - (x^2+2)^2 \right) dx = \pi \int_0^2 (8x - x^4) dx$
6
$= \pi \left[ 4x^2 - \frac{x^5}{5} \right]_0^2 = \pi \left(16 - \frac{32}{5}\right) = \frac{48\pi}{5}$

Exam tip:

Shifted axes are one of the most commonly tested variations on AP FRQ. Always calculate the distance from the axis to every boundary before identifying inner and outer radii.

5. Common Pitfalls

Wrong move:

For rotation around the y-axis, leave functions as $y = f (x)$ and integrate with respect to $x$

Why:

You confuse the washer method with the shell method, which integrates with respect to $x$ for vertical rotation axes

Correct move:

Always solve for $x$ as a function of $y$ and integrate with respect to $y$ when using washer method for any vertical axis.

Wrong move:

Write the integrand as $π (R - r)^{2}$ instead of $π (R^{2} - r^{2})$

Why:

You incorrectly factor the square when subtracting the areas of the two circles

Correct move:

Always expand the area subtraction first: $π R^{2} - π r^{2} = π (R^{2} - r^{2})$ , never $π (R - r)^{2}$ .

Wrong move:

When rotating around a horizontal axis $y = k$ , subtract $k$ only from the outer radius, not the inner radius

Why:

You assume the inner radius is still measured from the origin, not the new axis of rotation

Correct move:

Calculate the distance from the axis of rotation to every boundary curve before identifying inner and outer radii.

Wrong move:

Do not split the integral when the inner/outer radius changes at an intermediate point in the interval

Why:

You do not check if the region boundary changes across the integration interval

Correct move:

Test boundaries at multiple points across the interval; split the integral at any point where the identity of inner or outer radius changes.

Wrong move:

Use the washer method when the region is bounded by one curve and touches the axis of rotation

Why:

You confuse the washer and disk methods; there is no hole so you do not need to subtract an inner area

Correct move:

If the region touches the axis, use the disk method, which is just a special case of the washer method with inner radius $r = 0$ .

6. Quick Reference Cheatsheet

Category	Formula	Notes
Washer around x-axis	$V = π \int_{a}^{b} ([R (x)]^{2} - [r (x)]^{2}) d x$	$R (x)$ = outer radius (farther from x-axis), $a, b$ = x-bounds
Washer around y-axis	$V = π \int_{c}^{d} ([R (y)]^{2} - [r (y)]^{2}) d y$	$R (y)$ = outer radius (farther from y-axis), $c, d$ = y-bounds
Washer around horizontal $y = k$	$V = π \int_{a}^{b} ([R (x)]^{2} - [r (x)]^{2}) d x$	$R (x) = ∣ f (x) - k ∣$ , $r (x) = ∣ g (x) - k ∣$ , $R$ = larger distance
Washer around vertical $x = h$	$V = π \int_{c}^{d} ([R (y)]^{2} - [r (y)]^{2}) d y$	$R (y) = ∣ f (y) - h ∣$ , $r (y) = ∣ g (y) - h ∣$ , $R$ = larger distance
Area of one washer	$A = π (R^{2} - r^{2})$	Never use $A = π (R - r)^{2}$ , the most common student error
Special case: disk method	$V = π \int_{a}^{b} R^{2} d x$ (x-axis)	Disk method = washer with inner radius $r = 0$

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 · BC
FRQ volume of revolution problem
2022 · BC
MCQ correct integral setup

What's Next

Mastery of the washer method is required for all further topics involving volumes of revolution, which are a regular component of both multiple-choice and free-response questions on the AP Calculus BC exam. The washer method reinforces the core conceptual idea of integrating cross-sectional area to find total volume, which unifies all applications of integration. Next, you will learn the cylindrical shells method, an alternative approach for volumes of revolution that is often easier for rotation around vertical axes when you cannot easily solve for $x$ as a function of $y$ . You will also apply washer method reasoning to find volumes of solids with arbitrary known cross-sections, another core topic in Unit 8. Without mastering the process of identifying inner and outer radii and setting up the correct definite integral, these subsequent topics will be much harder to master.