Washer Method About the Axes — AP® Calculus AB

Part 1 · Hook

Why this matters

A bagel, a pipe, a bundt cake: each is a solid of revolution with a hole down the middle. Slice one and you get a ring, a disc with a smaller disc cut out. That ring, called a washer after the metal ring under a bolt, is the slice you get whenever the spinning region does not reach the axis.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The region under $y = 2x$ for $0 \le x \le 2$ is revolved about the $x$ -axis. What is its volume?

$\frac{32\pi}{3}$
$4\pi$
$\frac{32}{3}$
$8\pi$

Show the answer

Discs of radius $2x$ : $\pi\int_0^2 4x^2\,dx$ $= \frac{32\pi}{3}$ .

Correct: $\frac{32\pi}{3}$ :
$4\pi$ :
$\frac{32}{3}$ :
$8\pi$ :

2. On $(0, 2)$ , which is larger, $2x$ or $x^2$ ?

$2x$
$x^2$
They are equal
It changes at $x = 1$

Show the answer

$2x - x^2 = x(2 - x) > 0$ for $0 < x < 2$ . Test $x = 1$ : $2 > 1$ .

Correct: $2x$ :
$x^2$ :
They are equal:
It changes at $x = 1$ :

Part 4 · See it

See it first

Left: the region between y = √x (top) and y = 1 (bottom) from x = 1 to 4, shaded, with a thin vertical rectangle and arrows from the x-axis to the curve (outer radius R) and to the line (inner radius r). Right: the solid made by revolving the region about the x-axis, a flared horn with a round hole of radius 1 along the axis, with one washer highlighted in blue: R = √x reaches the curve and r = 1 reaches the line, both from the axis. Below: V = the integral from 1 to 4 of π[(√x)² − 1²] dx = 9π/2. — Revolving the region between $y = \sqrt{x}$ and $y = 1$ about the $x$ -axis. Each slice is a washer: outer radius $\sqrt{x}$ , inner radius 1. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The region does not touch the axis along its lengtheach rectangle starts above the axis and sweeps out a ring, a washer
The far edge of the rectangle is on one curve, the near edge on the otherouter radius $R$ = axis to the far curve; inner radius $r$ = axis to the near curve
A washer is a big disc minus a small discits area is $\pi R^2 - \pi r^2 = \pi\left(R^2 - r^2\right)$
Integrate the washer areas along the axis $V$ $= \pi\int_a^b \left(R(x)^2 - r(x)^2\right)\,dx$

Part 6 · Key ideas

Key ideas

Washer method: $V$ $= \pi\int_a^b \left(R^2 - r^2\right)\,dx$ , outer radius squared minus inner radius squared.
Both radii are measured from the axis: $R$ to the far curve, $r$ to the near one. About the $x$ -axis, $R = f(x)$ (top) and $r = g(x)$ (bottom).
Square each radius, then subtract: $R^2 - r^2$ , not $(R - r)^2$ .
About the $y$ -axis, the washers are horizontal: write each curve as $x$ in terms of $y$ , and the farther curve from the $y$ -axis is the outer radius.
Discs are washers with $r = 0$ : use washers whenever the region leaves a gap at the axis.

Part 7 · Misconception

A common mistake

The wrong idea: A washer's area is $\pi(R - r)^2$ , since $R - r$ is the width of the region.

What actually happens: The washer is a big disc with a small disc removed, so its area is $\pi R^2 - \pi r^2$ . With $R = 3$ and $r = 2$ that is $5\pi$ , while $\pi(3 - 2)^2 = \pi$ is far too small: the ring sits far from the axis, where it sweeps out a lot of area.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The region $R$ between $y = 2x$ and $y = x^2$

1. $R$ is revolved about the $x$ -axis. What are the outer radius $R(x)$ and the inner radius $r(x)$ of the washer at $x$ ?

$R(x) = 2x$ and $r(x) = x^2$
$R(x) = x^2$ and $r(x) = 2x$
$R(x) = 2x - x^2$ and $r(x) = 0$
$R(x) = 2x$ and $r(x) = 2x - x^2$

Show the answer

Both radii start at the $x$ -axis. The far edge of the region is the line ( $2x$ ); the near edge is the parabola ( $x^2$ ), which leaves a hole.

Correct: $R(x) = 2x$ and $r(x) = x^2$ : Outer to the far curve, inner to the near curve.
$R(x) = x^2$ and $r(x) = 2x$ : On $(0, 2)$ , $2x > x^2$ : the line is farther from the axis, so it gives the outer radius.
$R(x) = 2x - x^2$ and $r(x) = 0$ : That treats the region's height as a solid disc. The region does not reach the axis, so each slice has a hole of radius $x^2$ .
$R(x) = 2x$ and $r(x) = 2x - x^2$ : The inner radius runs from the axis to the near curve, $x^2$ , not across the region.

2. Find the volume of the solid formed when $R$ is revolved about the $x$ -axis.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$V$ $= \pi\int_0^2 \left[(2x)^2 - (x^2)^2\right]\,dx$ $= \pi\left[\frac{4x^3}{3} - \frac{x^5}{5}\right]_0^2$ $= \pi\left(\frac{32}{3} - \frac{32}{5}\right)$ $= \frac{64\pi}{15}$ .

Answer: $\frac{64\,\pi}{15}$ ≈ 13.404

3. The volume of the solid formed by revolving $R$ about the $x$ -axis is $\pi\int_0^2 g(x)\,dx$ . Write $g(x)$ .

Type an expression for $g(x) =$ .

Show the answer

$g(x)$ $= R^2 - r^2$ $= (2x)^2 - (x^2)^2$ $= 4x^2 - x^4$ .

Answer: $g(x) =$ $4x^{2} - x^{4}$

4. Instead, $R$ is revolved about the $y$ -axis. Find the volume.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Horizontal washers, $0 \le y \le 4$ . The line is $x = \frac{y}{2}$ and the parabola is $x = \sqrt{y}$ ; at $y = 1$ , $\sqrt{1} > \frac{1}{2}$ . $V$ $= \pi\int_0^4 \left(y - \frac{y^2}{4}\right)\,dy$ $= \pi\left(8 - \frac{16}{3}\right)$ $= \frac{8\pi}{3}$ .

Answer: $\frac{8\,\pi}{3}$ ≈ 8.378

5. When $R$ is revolved about the $y$ -axis, what are the radii of the washer at height $y$ ?

$R = \sqrt{y}$ and $r = \frac{y}{2}$
$R = \frac{y}{2}$ and $r = \sqrt{y}$
$R = 2x$ and $r = x^2$
$R = \sqrt{y} - \frac{y}{2}$ and $r = 0$

Show the answer

Horizontal slices: write each curve as $x$ in terms of $y$ . The parabola $x = \sqrt{y}$ is farther from the $y$ -axis than the line $x = \frac{y}{2}$ .

Correct: $R = \sqrt{y}$ and $r = \frac{y}{2}$ : Outer to the far curve, inner to the near one, both in $y$ .
$R = \frac{y}{2}$ and $r = \sqrt{y}$ : At $y = 1$ , $\frac{y}{2} = 0.5 < 1 = \sqrt{y}$ : the line is the nearer curve.
$R = 2x$ and $r = x^2$ : Those are the radii about the $x$ -axis, in $x$ .
$R = \sqrt{y} - \frac{y}{2}$ and $r = 0$ : That is the width of the region, not a radius from the $y$ -axis.

6. The region bounded by $y = \sqrt{x}$ , $y = 1$ and $x = 4$ is revolved about the $x$ -axis. Find the volume.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

From $x = 1$ (where $\sqrt{x} = 1$ ) to $x = 4$ : $R = \sqrt{x}$ , $r = 1$ , so $V$ $= \pi\int_1^4 (x - 1)\,dx$ $= \pi\left(\frac{15}{2} - 3\right)$ $= \frac{9\pi}{2}$ .

Answer: $\frac{9\,\pi}{2}$ ≈ 14.137

7. A washer has outer radius 3 and inner radius 2. What is the area of its face?

$5\pi$
$\pi$
$25\pi$
$13\pi$

Show the answer

Big disc minus small disc: $\pi(3)^2 - \pi(2)^2$ $= 9\pi - 4\pi$ $= 5\pi$ .

Correct: $5\pi$ : $\pi(R^2 - r^2)$ .
$\pi$ : $\pi(3 - 2)^2$ squares the difference; the ring is wider than that.
$25\pi$ : $\pi(3 + 2)^2$ is a disc of radius 5.
$13\pi$ : $\pi(3^2 + 2^2)$ adds the hole instead of removing it.

Part 9 · Summary

Summary

When the region does not touch the axis, each slice is a washer. Measure both radii from the axis: the outer radius

R

to the far curve and the inner radius

r

to the near one. Then

V

= \pi\int_a^b \left(R^2 - r^2\right)\,dx

: the difference of squares, never the square of the difference.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The region RR between y=2xy = 2x and y=x2y = x^2

Summary

What comes next

Connections

The region $R$ between $y = 2x$ and $y = x^2$