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Washer Method
Revolving Around \(x\)/\(y\) Axes

Master the technique for computing volumes of solids of revolution when the cross-section is an annular ring: the washer method.

AP Calculus BC

What Is a Solid of Revolution?

When a planar region is rotated around a line (an axis of revolution), it sweeps out a three-dimensional solid. Computing the volume of that solid is a core skill in integral calculus.

x y
\(f(x)\)
 rotate

Disk vs Washer: Cross-Sections

Disk Cross-Section
A solid circle: one boundary curve, no hole in the middle. Area: \(\pi R^2\).
Washer Cross-Section
An annular ring: outer radius minus inner radius creates a hole. Area: \(\pi(R^2 - r^2)\).
\(R\)
 Disk
\(R\)
\(r\)
 Washer

A transparent lathe reveals the hole inside a washer

A hollow wooden bowl on a stationary lathe accompanies a calibrated generating region and a separate annular cross-section.Rotate the region between y = √x and y = x² on [0,1] about the x-axis. A washer has area π(x − x⁴), and the volume is 3π/10.

A washer subtracts two disk areas

Rotate the region between y = √x and y = x² on [0,1] about the x-axis. A washer has area π(x − x⁴), and the volume is 3π/10.Rotate the region between y = √x and y = x² on [0,1] about the x-axis. A washer has area π(x − x⁴), and the volume is 3π/10.

Why Do We Need the Washer Method?

When the region being revolved is bounded by two curves (an outer curve and an inner curve), each cross-section is not a full disk: it has a hole. We subtract the inner area from the outer area at every slice.

This is a direct extension of the disk method: if \(r = 0\), the washer formula reduces to the disk formula.

The Washer Formula: Revolving Around \(x\)-Axis

\(R(x)\)
\(r(x)\)
 x y
\(a\)
\(b\)
 outer inner

The Washer Formula: Revolving Around \(y\)-Axis

\(R(y)\)
\(r(y)\)
 x y
\(c\)
\(d\)

Disk vs Washer: Key Distinction

Disk Method Washer Method
Bounded by One curve and the axis Two curves (outer and inner)
Cross-section Solid circle Annular ring (donut)
Formula (x-axis) \(\pi\!\int_a^b [f(x)]^2\,dx\) \(\pi\!\int_a^b [R^2 - r^2]\,dx\)
Inner radius \(r = 0\) \(r \neq 0\) (gap exists)
When to use Region touches the axis Region has a gap from the axis

Step-by-Step Washer Method

1
Sketch the region bounded by the given curves and identify the axis of revolution.
2
Identify outer and inner curves: which function is farther from the axis (\(R\)) and which is closer (\(r\))?
3
Express radii as functions of the integration variable. Revolving around the \(x\)-axis: radii are in terms of \(x\). Around the \(y\)-axis: in terms of \(y\).
4
Find integration bounds by solving for intersection points of the curves.
5
Integrate: \(V = \pi\!\int [R^2 - r^2]\,d(\text{variable})\). Simplify and evaluate.

Example 1: Revolving Around the \(x\)-Axis

Find the volume of the solid formed by revolving the region between \(y = x^2\) and \(y = x\) about the \(x\)-axis.

\(y = x\)
\(y = x^2\)
 x y (0,0) (1,1)

Example 1: Solution

1
Expand: \(R^2 - r^2 = x^2 - x^4\).
2
Integrate: \(\displaystyle\pi\!\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1\).
3
Evaluate: \(\displaystyle\pi\!\left(\frac{1}{3} - \frac{1}{5}\right) = \pi\cdot\frac{2}{15}\).

Example 2: Revolving Around the \(y\)-Axis

Find the volume when the region between \(x = y^2\) and \(x = y\) (for \(0 \leq y \leq 1\)) is revolved about the \(y\)-axis.

\(x = y\)
\(x = y^2\)
 x y (0,0) (1,1)
\(R = y\)
\(r = y^2\)

Example 2: Solution

1
Antiderivative: \(\displaystyle\frac{y^3}{3} - \frac{y^5}{5}\).
2
Evaluate at bounds: \(\displaystyle\pi\!\left(\frac{1}{3} - \frac{1}{5} - 0\right) = \frac{2\pi}{15}\).

Example 3: A Harder Problem

Find the volume when the region between \(y = \sqrt{x}\) and \(y = x^2\) is revolved about the \(x\)-axis.

Identifying \(R\) and \(r\) Correctly

Revolve Around \(x\)-Axis
Radii are vertical distances from the \(x\)-axis. \(R\) = the curve with the larger \(y\)-value (farther from axis). \(r\) = the curve with the smaller \(y\)-value.
Revolve Around \(y\)-Axis
Radii are horizontal distances from the \(y\)-axis. \(R\) = the curve with the larger \(x\)-value. \(r\) = the curve with the smaller \(x\)-value. Express everything in terms of \(y\).

Common Mistakes to Avoid

Subtracting Radii First
Writing \(\pi\!\int(R - r)^2\,dx\) instead of \(\pi\!\int(R^2 - r^2)\,dx\). You must square each radius separately, then subtract.
Swapping \(R\) and \(r\)
Mixing up which curve is outer vs inner. Always test a point in the interval to confirm which function is farther from the axis.
Wrong Variable
Using \(dx\) when revolving around the \(y\)-axis (or vice versa). The integration variable must match the axis of revolution's perpendicular direction.

Question 1 of 5

The region between \(y = x\) and \(y = x^3\) on \([0,1]\) is revolved about the \(x\)-axis. Which integral gives the volume?

Question 2 of 5

When revolving a region about the \(y\)-axis using the washer method, the radii are:

Question 3 of 5

Compute: \(\displaystyle\pi\!\int_0^1\bigl[(\sqrt{x})^2 - (x^2)^2\bigr]\,dx\).

Question 4 of 5

The region between \(y = 4\) and \(y = x^2\) is revolved about the \(x\)-axis. What are the outer and inner radii?

Question 5 of 5

The region between \(x = y\) and \(x = y^2\) for \(0 \leq y \leq 1\) is revolved about the \(y\)-axis. What is the volume?