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

Master the technique of finding volumes of solids of revolution when a region between two curves is rotated around the \(x\)-axis or \(y\)-axis.

AP Calculus AB

What Is a Solid of Revolution?

When you take a 2D region in the coordinate plane and spin it around an axis, the resulting 3D shape is called a solid of revolution. Think of a potter's wheel: the clay profile sweeps around to form a vase.

\(f(x)\)
\(g(x)\)
 Rotate
\(x\)

From Discs to Washers

Disc Method
One curve and the axis. Cross-sections are solid circles: \(A = \pi R^2\).
Washer Method
Two curves. Cross-sections are rings (a big circle minus a small circle): \(A = \pi(R^2 - r^2)\).
Disc
\(A = \pi R^2\)
 Washer
\(A = \pi(R^2 - r^2)\)
 Solid cross-section Hollow cross-section
\(R\)
\(R\)
\(r\)

A ceramic vessel reveals wall material and water capacity

Potter holds a teal ceramic cutaway vessel with an open top, thick wall and water inside a closed-bottom cavity.A washer separates wall material from the cavity: its area is π(R² − r²). Integrating the cavity area instead measures the volume available to hold water.

A ceramic vessel reveals wall material and water capacity

A washer separates wall material from the cavity: its area is π(R² − r²). Integrating the cavity area instead measures the volume available to hold water.A washer separates wall material from the cavity: its area is π(R² − r²). Integrating the cavity area instead measures the volume available to hold water.

Why Does the Washer Method Matter?

Most real-world solids of revolution are not solid all the way through: pipes, bowls, doughnuts, turbine housings. When the region being rotated is bounded by two curves rather than just one curve and the axis, the disc method alone is not enough.

AP Exam
The washer method appears on nearly every AP Calculus AB/BC free-response exam. It is a core integration application.
Engineering
Designing containers, nozzles, and hollow shafts all require computing volume with an inner and outer boundary.

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

\(x\)
\(R(x)\)
\(r(x)\)
\(a\)
\(b\)
\(f(x)\)
\(g(x)\)
\(dx\) slice

Revolving Around the \(y\)-Axis

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

Disc vs Washer vs Shell

Disc Washer Shell
Curves 1 curve + axis 2 curves 1 or 2 curves
Cross-section Solid circle Ring (annulus) Thin cylinder
Slices Perpendicular to axis Perpendicular to axis Parallel to axis
Formula \(\pi\!\int R^2\,dx\) \(\pi\!\int(R^2\!-\!r^2)\,dx\) \(2\pi\!\int r\cdot h\,dx\)

How to Apply the Washer Method

1
Sketch the region. Graph both curves and shade the enclosed area.
2
Identify the axis of rotation. Determine whether you integrate with respect to \(x\) or \(y\).
3
Find \(R\) and \(r\). The outer radius \(R\) is the farther curve; the inner radius \(r\) is the closer curve, measured from the axis.
4
Determine limits of integration. Find intersection points of the two curves.
5
Set up and evaluate. Plug into \(V = \pi\!\int_a^b [R^2 - r^2]\,dx\) (or \(dy\)) and integrate.

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

Find the volume when the region between \(y = x\) and \(y = x^2\) on \([0,1]\) is revolved around the \(x\)-axis.

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

Example 1: Solution

1
Set up the integral: \(\displaystyle V = \pi\int_0^1\!\left[(x)^2 - (x^2)^2\right]dx = \pi\int_0^1\!\left[x^2 - x^4\right]dx\)
2
Integrate: \(\displaystyle \pi\!\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1\)
3
Evaluate: \(\displaystyle \pi\!\left(\frac{1}{3} - \frac{1}{5}\right) = \pi\!\left(\frac{5-3}{15}\right) = \frac{2\pi}{15}\)

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

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

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

Example 2: Solution

1
Set up: \(\displaystyle V = \pi\int_0^1\!\left[(y)^2 - (y^2)^2\right]dy = \pi\int_0^1\!\left[y^2 - y^4\right]dy\)
2
Integrate: \(\displaystyle \pi\!\left[\frac{y^3}{3} - \frac{y^5}{5}\right]_0^1\)
3
Evaluate: \(\displaystyle \pi\!\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}\)

Example 3: Harder Setup

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

\(x\)
\(y\)
 0 1
\(y = \sqrt{x}\)
\(y = x^2\)

Example 3: Solution

1
Simplify radii squared: \(R^2 = (\sqrt{x})^2 = x\), and \(r^2 = (x^2)^2 = x^4\).
2
Integrate: \(\displaystyle \pi\!\int_0^1(x - x^4)\,dx = \pi\!\left[\frac{x^2}{2} - \frac{x^5}{5}\right]_0^1\)
3
Evaluate: \(\displaystyle \pi\!\left(\frac{1}{2} - \frac{1}{5}\right) = \pi\!\cdot\frac{3}{10} = \frac{3\pi}{10}\)

Common Mistakes to Avoid

✕ Subtracting before squaring
Writing \(\pi\!\int(R - r)^2\,dx\) instead of \(\pi\!\int(R^2 - r^2)\,dx\). These are not the same: \((R-r)^2 = R^2 - 2Rr + r^2\).
✕ Swapping R and r
Mixing up which curve is farther from the axis. Always check: the outer radius is the curve with the greater distance from the axis of rotation.
✕ Wrong variable
Integrating with respect to \(x\) when revolving around the \(y\)-axis (or vice versa). Match the variable to the axis: \(dx\) for \(x\)-axis, \(dy\) for \(y\)-axis.

Summary: Washer Method Quick Reference

Around \(x\)-axis Around \(y\)-axis
Variable Integrate with respect to \(x\) Integrate with respect to \(y\)
Radii Vertical distances: \(R(x)\), \(r(x)\) Horizontal distances: \(R(y)\), \(r(y)\)
Formula \(\pi\!\int_a^b\!(R^2 - r^2)\,dx\) \(\pi\!\int_c^d\!(R^2 - r^2)\,dy\)
Limits \(x\)-values of intersections \(y\)-values of intersections
Solve for \(y = f(x)\) \(x = f(y)\)

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 around the \(y\)-axis, the radii \(R(y)\) and \(r(y)\) represent:

Question 3 of 5

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

Question 4 of 5

A student writes \(\displaystyle V = \pi\!\int_0^1\!(\sqrt{x} - x^2)^2\,dx\) for the previous problem. What error did they make?

Question 5 of 5

The region between \(x = 4 - y^2\) and \(x = 0\) (the \(y\)-axis) for \(0 \leq y \leq 2\) is revolved about the \(y\)-axis. Which integral gives the volume?