The Washer Method

When a region doesn't reach the axis of revolution, each slice perpendicular to the axis sweeps out a disc with a hole in it, a washer.

Outer and inner radius

A washer of thickness

\Delta x

has an outer radius

R

, from the axis to the far edge of the region. Its inner radius

r

runs from the axis to the near edge. Its volume is the big disc minus the hole:

\pi R^2\,\Delta x - \pi r^2\,\Delta x = \pi\left(R^2 - r^2\right)\Delta x.

Each radius is squared on its own before subtracting. The area of a washer is a difference of two disc areas.

Example 1

The region between

y = \sec x

and the line

y = 2

, for

-\frac{\pi}{3} \le x \le \frac{\pi}{3}

, is revolved about the

x

-axis. Find the volume of the solid.

On the interval,

1 \le \sec x \le 2

, so the curve is below the line and above the axis. The far edge is the line and the near edge is the curve:

R(x) = 2

and

r(x) = \sec x

The region, its mirror image across the x-axis, and the washer at x = 0.5 seen edge-on. The hole comes from the gap between the axis and the curve.

\begin{aligned} V &= \pi\int_{-\pi/3}^{\pi/3} \left(4 - \sec^2 x\right)dx \\[4pt] &= \pi\left(4x - \tan x\right)\Big|_{-\pi/3}^{\pi/3} \\[4pt] &= \pi\left(\frac{8\pi}{3} - 2\sqrt{3}\right) \approx 15.436. \end{aligned}

Example 2

The region bounded by the curve

x = 2 + \sin y

, the line

x = 1

, and the lines

y = 0

and

y = \pi

is revolved about the

y

-axis. Find the volume of the solid.

The axis is vertical, so slice horizontally. At height

y

the region runs from

x = 1

x = 2 + \sin y

, so

r(y) = 1

and

R(y) = 2 + \sin y

The region right of x = 1, its mirror image, and the washer at y = 1. The solid has a hole of radius 1 all the way up.

\begin{aligned} V &= \pi\int_{0}^{\pi} \Big((2 + \sin y)^2 - 1^2\Big)\,dy \\[4pt] &= \pi\int_{0}^{\pi} \left(3 + 4\sin y + \sin^2 y\right)dy. \end{aligned}

[0, \pi]

the integral of

\sin y

is 2. For the squared sine, use the double-angle identity:

\begin{aligned} &\int_{0}^{\pi} \sin^2 y\,dy \\ &\quad = \frac{1}{2}\int_{0}^{\pi} (1 - \cos 2y)\,dy \\[4pt] &\quad = \frac{1}{2}\left(y - \frac{\sin 2y}{2}\right)\Big|_{0}^{\pi} = \frac{\pi}{2}. \end{aligned}

\begin{aligned} V &= \pi\left(3\pi + 8 + \frac{\pi}{2}\right) \\[4pt] &= \pi\left(8 + \frac{7\pi}{2}\right) \approx 59.676. \end{aligned}

Revolving about other lines

For the axis

y = k

x = h

, measure both radii from that line. For an axis below or to the left of the region, each radius is the edge minus the axis value. For an axis above or to the right, it's the axis value minus the edge.

Example 3

The region bounded by

y = e^x

y = e^{-x}

, and the line

x = 1

is revolved about the line

y = -1

. Find the volume of the solid.

The curves meet at

(0, 1)

, and for

0 < x \le 1

the curve

y = e^x

is on top. The axis is below the region, so each radius is an edge minus

-1

. The far edge is the top curve and the near edge is the bottom one:

\begin{aligned} R(x) &= e^x + 1, \\ r(x) &= e^{-x} + 1. \end{aligned}

The region between the two curves, its mirror image across y = −1, and the washer at x = 0.6. Both radii change with x.

\begin{aligned} V &= \pi\int_{0}^{1} \Big((e^x + 1)^2 \\ &\qquad\qquad - (e^{-x} + 1)^2\Big)\,dx \\[4pt] &= \pi\int_{0}^{1} \big(e^{2x} + 2e^x \\ &\qquad\qquad - e^{-2x} - 2e^{-x}\big)\,dx \\[4pt] &= \pi\Big(\frac{e^{2x}}{2} + 2e^x \\ &\qquad\qquad + \frac{e^{-2x}}{2} + 2e^{-x}\Big)\Big|_{0}^{1}. \end{aligned}

x = 0

the bracket is

\frac{1}{2} + 2 + \frac{1}{2} + 2 = 5

, so

\begin{aligned} V &= \pi\left(\frac{e^2}{2} + 2e + \frac{1}{2e^2} + \frac{2}{e} - 5\right) \\[4pt] &\approx 15.502. \end{aligned}

Example 4

The region bounded by

y = \frac{2}{x}

, the line

x = 2

, and the line

y = 2

is revolved about the line

x = 3

. Find the volume of the solid.

The curve meets

y = 2

(1, 2)

and

x = 2

(2, 1)

, so the region spans

1 \le y \le 2

. At height

y

it runs from the curve,

x = \frac{2}{y}

, to the line

x = 2

. The axis is to the right, so each radius is 3 minus an edge, and the far edge is the curve:

\begin{aligned} R(y) &= 3 - \frac{2}{y}, \\ r(y) &= 3 - 2 = 1. \end{aligned}

The region, its mirror image across x = 3, and the washer at y = 1.6.

\begin{aligned} V &= \pi\int_{1}^{2} \left(\left(3 - \frac{2}{y}\right)^2 - 1\right)dy \\[4pt] &= \pi\int_{1}^{2} \left(8 - \frac{12}{y} + \frac{4}{y^2}\right)dy \\[4pt] &= \pi\left(8y - 12\ln y - \tfrac{4}{y}\right)\Big|_{1}^{2} \\[4pt] &= \pi\big((14 - 12\ln 2) - 4\big) \\[4pt] &= \pi(10 - 12\ln 2). \end{aligned}

The volume is

\pi(10 - 12\ln 2) \approx 5.285