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Washer Method
Revolving Around Other Axes

Master the technique of computing volumes of revolution when the axis of rotation is not a coordinate axis: horizontal lines \( y = k \) and vertical lines \( x = k \).

AP Calculus BC

What Is the Washer Method?

When a region between two curves is revolved around an axis, every cross-section perpendicular to that axis looks like a washer: a disk with a hole. The volume is the integral of each washer's area.

R r axis

Why Other Axes Matter

Most textbooks start with rotation about the \( x \)-axis or \( y \)-axis, where the radii are simply the function values. But when the axis shifts to a line like \( y = 4 \) or \( x = -1 \), every radius must be recalculated as a signed distance from the curve to that axis.

Standard Axis
Rotate about \( y = 0 \) or \( x = 0 \). Radii equal function values directly.
Shifted Axis
Rotate about \( y = k \) or \( x = k \). Radii become distances: \( |f(x) - k| \).

Visualizing the Radius Shift

Consider the region between \( y = x^2 \) and \( y = 2x \) revolved about \( y = 5 \). Each radius is the distance from the curve down to the line \( y = 5 \):

x y y = 5 y = x² y = 2x R r 1 2

The General Washer Formula

Key insight: "far" and "near" are relative to the axis of rotation, not to the coordinate axes.

Determining R and r

Axis Above Region
\( R = k - f_{\text{lower}} \)
\( r = k - f_{\text{upper}} \)
Both positive when region is below axis.
Axis Below Region
\( R = f_{\text{upper}} - k \)
\( r = f_{\text{lower}} - k \)
Both positive when region is above axis.
Axis Left / Right
Same logic for vertical axes. Measure horizontal distances from \( x = k \) to each curve.

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 = x² and y = x on [0,1] about y = 2. The outer radius is 2 − x² and the inner radius is 2 − x, giving volume 8π/15.

Farther from the axis means outer

Rotate the region between y = x² and y = x on [0,1] about y = 2. The outer radius is 2 − x² and the inner radius is 2 − x, giving volume 8π/15.Rotate the region between y = x² and y = x on [0,1] about y = 2. The outer radius is 2 − x² and the inner radius is 2 − x, giving volume 8π/15.

Disk vs Washer vs Shell

Disk Washer Shell
Cross-section Solid circle Ring (annulus) Thin cylinder
When to use One boundary Two boundaries Perpendicular axis
Integrand \( \pi R^2 \) \( \pi(R^2 - r^2) \) \( 2\pi r \cdot h \)
Slices ⊥ to Axis of rotation Axis of rotation Axis of rotation (parallel)

How to Apply the Washer Method

1
Sketch the region and draw the axis of rotation. Mark which curve is farther from the axis.
2
Find intersections of the bounding curves to determine limits of integration \( a \) and \( b \) (or \( c \) and \( d \)).
3
Write \( R \) and \( r \) as distances from each curve to the axis: \( R = |\text{far curve} - k| \), \( r = |\text{near curve} - k| \).
4
Set up the integral: \( V = \pi \displaystyle\int_a^b \bigl[R(x)^2 - r(x)^2\bigr]\,dx \).
5
Expand, integrate, and evaluate. Simplify before integrating when possible.

Example 1: Revolving About \( y = 4 \)

Find the volume when the region bounded by \( y = x^2 \) and \( y = 2x \) is revolved about the line \( y = 4 \).

x y y = 4
\(R = 4 - x^2\)
\(r = 4 - 2x\)
\(y = x^2\)
\(y = 2x\)
 0 2

Example 1: Solution

Example 2: Revolving About \( x = -1 \)

Find the volume when the region bounded by \( y = \sqrt{x} \) and \( y = x^2 \) is revolved about \( x = -1 \).

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

Example 2: Solution

Example 3: Axis Between Curves

Find the volume when the region between \( y = 4 \) and \( y = x^2 \) is revolved about \( y = 1 \).

Example Recap

Example Axis Variable Volume
1 \( y = 4 \) (above) \( dx \) \( \frac{32\pi}{5} \)
2 \( x = -1 \) (left) \( dy \) \( \frac{29\pi}{30} \)
3 \( y = 1 \) (between) \( dx \) \( \frac{448\pi}{15} \)

Common Mistakes

✕ Using Function Value as Radius
Writing \( R = f(x) \) instead of \( R = f(x) - k \). The radius is always the distance from the curve to the axis, not the function value itself.
✕ Swapping R and r
Mixing up which curve is farther from the axis. Always check: the outer radius must be the larger distance, or you get a negative integrand.
✕ Wrong Integration Variable
Revolving about a vertical line like \( x = 3 \) but integrating with \( dx \). For vertical axes, you typically integrate with respect to \( y \).

Question 1 of 5

The region bounded by \( y = x \) and \( y = x^2 \) is revolved about \( y = -1 \). What is the outer radius \( R \)?

Question 2 of 5

When revolving about a vertical line \( x = k \), we should integrate with respect to:

Question 3 of 5

The region between \( y = \sqrt{x} \) and \( y = 0 \) on \( [0, 4] \) is revolved about \( y = 3 \). Which integral gives the volume?

Question 4 of 5

A student revolves the region between \( y = x^2 \) and \( y = 4 \) about \( y = 5 \) and writes \( R = 5 - x^2 \) and \( r = 1 \). Is this correct?

Question 5 of 5

The region between \( x = y^2 \) and \( x = 4 \) is revolved about \( x = 5 \). What is the volume?