Washer Method About Other Lines — AP® Calculus AB

Part 1 · Hook

Why this matters

Spin the same flat region about the

x

-axis, then about a line one unit below it, and you get two different hollow solids: the second has a wider hole and more material. The washer method handles both. The whole difference is in measuring each radius from the right line.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The region between $y = \sqrt{x}$ and $y = 1$ for $1 \le x \le 4$ is revolved about the $x$ -axis. Which integral gives the volume?

$\pi\int_1^4 (x - 1)\,dx$
$\pi\int_1^4 (\sqrt{x} - 1)^2\,dx$
$\pi\int_1^4 x\,dx$
$\pi\int_1^4 (1 - x)\,dx$

Show the answer

Washers with $R = \sqrt{x}$ and $r = 1$ : $\pi\int_1^4 \left[(\sqrt{x})^2 - 1^2\right]\,dx$ .

Correct: $\pi\int_1^4 (x - 1)\,dx$ :
$\pi\int_1^4 (\sqrt{x} - 1)^2\,dx$ :
$\pi\int_1^4 x\,dx$ :
$\pi\int_1^4 (1 - x)\,dx$ :

2. The region between $y = x^2$ and $y = 1$ is revolved about the line $y = 1$ . What is the radius of the disc at $x$ ?

$1 - x^2$
$x^2$
$x^2 - 1$
$1 + x^2$

Show the answer

The axis is above the curve: $r = 1 - x^2$ .

Correct: $1 - x^2$ :
$x^2$ :
$x^2 - 1$ :
$1 + x^2$ :

Part 4 · See it

See it first

Left: the region between y = √x and y = 1 from x = 1 to 4, with the line y = −1 dashed below it as the axis of rotation, and arrows from the axis to the thin rectangle: the outer radius R up to the curve and the inner radius r up to the line y = 1. Right: the solid made by revolving the region about y = −1, a flared horn with a hole of radius 2 around the axis, with one washer highlighted in blue: R = √x + 1 and r = 2. Below: V = the integral from 1 to 4 of π[(√x + 1)² − 2²] dx = 47π/6. — The region between $y = \sqrt{x}$ and $y = 1$ revolved about $y = -1$ . Both radii grow by 1: $R = \sqrt{x} + 1$ and $r = 2$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The axis is the line $y = k$ and the region does not touch iteach slice is a washer centered on that line
Draw a rectangle perpendicular to the axisits far end gives $R$ and its near end gives $r$ , both measured from $y = k$
The axis is below the region $R = (\text{top}) - k$ and $r = (\text{bottom}) - k$
The axis is above the regionthe bottom curve is now the far edge: $R = k - (\text{bottom})$ and $r = k - (\text{top})$
Square each radius and subtract $V$ $= \pi\int_a^b \left(R^2 - r^2\right)\,dx$

Part 6 · Key ideas

Key ideas

Washers about a line: $V = \pi\int \left(R^2 - r^2\right)$ with both radii measured from the line $y = k$ or $x = h$ .
Axis below the region ( $y = k$ , $k$ small): $R = f(x) - k$ , $r = g(x) - k$ . With $k = -1$ , add 1 to each height.
Axis above the region: the far edge is the bottom curve, so $R = k - g(x)$ and $r = k - f(x)$ .
Vertical axis $x = h$ : horizontal washers in $dy$ ; the far edge may be the left curve if the axis is on the right.
Square each radius separately; $(R - r)^2$ is the region's height squared, not a washer's area.

Part 7 · Misconception

A common mistake

The wrong idea: To revolve about $y = -1$ , subtract 1 from each radius, since $-1$ is one unit away.

What actually happens: A radius is (larger height) minus (smaller height). From $y = -1$ up to the curve is $f(x) - (-1) = f(x) + 1$ : both radii grow by 1. Subtracting 1 instead measures toward the wrong side of the $x$ -axis.

Part 8 · Check yourself

Check yourself

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

Graph

The region $R$ between $y = x^2$ and $y = 4$

1. $R$ is revolved about the line $y = -1$ . What are the outer and inner radii of the washer at $x$ ?

$R(x) = 5$ and $r(x) = x^2 + 1$
$R(x) = 4$ and $r(x) = x^2$
$R(x) = 3$ and $r(x) = x^2 - 1$
$R(x) = 5$ and $r(x) = x^2 - 1$

Show the answer

Measure from $y = -1$ : to the far edge, $4 - (-1) = 5$ ; to the near edge, $x^2 - (-1) = x^2 + 1$ .

Correct: $R(x) = 5$ and $r(x) = x^2 + 1$ : Both radii measured from the axis $y = -1$ .
$R(x) = 4$ and $r(x) = x^2$ : Those are measured from the $x$ -axis, not from $y = -1$ .
$R(x) = 3$ and $r(x) = x^2 - 1$ : Subtracting $-1$ adds 1: both radii grow by 1, they do not shrink.
$R(x) = 5$ and $r(x) = x^2 - 1$ : The inner radius needs the same shift as the outer one: $x^2 - (-1) = x^2 + 1$ .

2. Find the volume of the solid formed when $R$ is revolved about the line $y = -1$ .

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

Show the answer

$V$ $= \pi\int_{-2}^{2} \left[5^2 - (x^2 + 1)^2\right]\,dx$ $= \pi\int_{-2}^{2} \left(24 - 2x^2 - x^4\right)\,dx$ $= 2\pi\left(48 - \frac{16}{3} - \frac{32}{5}\right)$ $= \frac{1088\pi}{15}$ .

Answer: $\frac{1088\,\pi}{15}$ ≈ 227.870

3. $R$ is revolved about the line $y = 5$ . The volume is $\pi\int_{-2}^{2} g(x)\,dx$ . Write $g(x) = R(x)^2 - r(x)^2$ .

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

Show the answer

From $y = 5$ , the far edge of $R$ is the parabola: $R = 5 - x^2$ . The near edge is $y = 4$ : $r = 5 - 4 = 1$ . So $g(x) = (5 - x^2)^2 - 1^2$ .

Answer: $g(x) =$ $(5 - x^{2})^{2} - 1$

4. Find the volume of the solid formed when $R$ is revolved about the line $y = 5$ .

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

Show the answer

$V$ $= \pi\int_{-2}^{2} \left[(5 - x^2)^2 - 1\right]\,dx$ $= \pi\int_{-2}^{2} \left(24 - 10x^2 + x^4\right)\,dx$ $= 2\pi\left(48 - \frac{80}{3} + \frac{32}{5}\right)$ $= \frac{832\pi}{15}$ .

Answer: $\frac{832\,\pi}{15}$ ≈ 174.254

5. The region between $y = x$ and $y = x^2$ is revolved about the line $y = 1$ . Find the volume.

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

Show the answer

$V$ $= \pi\int_0^1 \left[(1 - x^2)^2 - (1 - x)^2\right]\,dx$ $= \pi\int_0^1 \left(x^4 - 3x^2 + 2x\right)\,dx$ $= \pi\left(\frac{1}{5} - 1 + 1\right)$ $= \frac{\pi}{5}$ .

Answer: $\frac{\pi}{5}$ ≈ 0.628

6. A region between $y = f(x)$ (top) and $y = g(x)$ (bottom), both positive, is revolved about the line $y = -2$ . What is the outer radius?

$f(x) + 2$
$f(x) - 2$
$f(x)$
$g(x) + 2$

Show the answer

The axis is below the region, so the far edge is the top curve: $f(x) - (-2) = f(x) + 2$ .

Correct: $f(x) + 2$ : Top curve minus the axis height $-2$ .
$f(x) - 2$ : Subtracting $-2$ adds 2.
$f(x)$ : That is measured from the $x$ -axis.
$g(x) + 2$ : $g(x) + 2$ is the inner radius, to the nearer curve.

7. A region between $x = g(y)$ (left) and $x = h(y)$ (right) is revolved about the line $x = 5$ , which lies to the right of the region. What is the outer radius?

$5 - g(y)$
$5 - h(y)$
$h(y) - 5$
$h(y) - g(y)$

Show the answer

The axis is on the right, so the left curve is farther from it: $R = 5 - g(y)$ .

Correct: $5 - g(y)$ : From an axis on the right, the left edge is the far edge.
$5 - h(y)$ : $5 - h(y)$ is the inner radius, to the nearer (right) curve.
$h(y) - 5$ : That is negative: $h(y) < 5$ .
$h(y) - g(y)$ : That is the width of the region, not a distance from the axis.

Part 9 · Summary

Summary

Revolving about a line other than an axis uses the same washer integral,

V = \pi\int (R^2 - r^2)

. Draw one rectangle perpendicular to the axis;

R

is the distance from the axis to its far end and

r

to its near end. When the axis is above (or to the right of) the region, the far end is the bottom (or left) curve.

Part 10 · Up next

What comes next

This is the last topic in the course.

Part 11 · Connections

Connections

Back: washers about the axes (8.11)

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=x2y = x^2 and y=4y = 4

Summary

What comes next

Connections

The region $R$ between $y = x^2$ and $y = 4$