Volumes of revolution: the washer method

Warm-up

Three quick questions from earlier lessons.

Discover

Revolve the region between $y = 2$ and $y = 1$ , $0 \le x \le 3$ , about the $x$ -axis: a thick pipe. Each slice is a washer with radii 2 and 1. What is its area?

$3\pi$
$\pi$

Key idea

A hole in the middle: subtract the inner disc.

Washer method:

V = \pi\displaystyle\int_a^b \big(R(x)^2 - r(x)^2\big)\,dx

, with

R

the outer radius and

r

the inner radius, both measured from the axis of rotation.

Worked example

Region between $y = x$ and $y = x^2$ revolved about the $x$ -axis.

R = x

,

r = x^2

They meet at 0 and 1; y = x is farther from the axis.

V = \pi\displaystyle\int_0^1 (x^2 - x^4)\,dx

R² − r².

= \tfrac{2\pi}{15}

1/3 − 1/5.

Your turn, guided

Your turn: revolve the region under $y = 1$ , $0 \le x \le 2$ , about $y = -1$ .

V = 6\pi

R = 2, r = 1: π(4 − 1) × 2.

Practice ★

The region between $y = 16$ and $y = x^2$ is revolved about the $x$ -axis. Which integral is its volume?
1. $\pi\displaystyle\int_{-4}^{ 4 } \big(256 - x^4\big)\,dx$
2. $\pi\displaystyle\int_{-4}^{ 4 } \big(16 - x^2\big)^2\,dx$
3. $\pi\displaystyle\int_{-4}^{ 4 } \big(16 - x^2\big)\,dx$

The region between $y = 3x$ and $y = x^2$ is revolved about the $x$ -axis. Find the volume.
The region between $y = 1x$ and $y = x^2$ is revolved about the $x$ -axis. Find the volume.

Watch out!

Practice ★★

The region under $y = \sqrt{x}$ , $0 \le x \le 9$ , is revolved about the line $y = -1$ . Find the volume.
The region under $y = \sqrt{x}$ , $0 \le x \le 1$ , is revolved about the line $y = -1$ . Find the volume.

Boss problem. The region $R$ between $y = 2x$ and $y = x^2$ .

Exit ticket

Three questions to check you have got it.

The region between $y = 3x$ and $y = x^2$ is revolved about the $x$ -axis. Find the volume.
The region between $y = 1x$ and $y = x^2$ is revolved about the $x$ -axis. Find the volume.
The region between $y = 2x$ and $y = x^2$ is revolved about the $x$ -axis. Find the volume.

Summary card

Washer method

$V = \pi\displaystyle\int_a^b (R^2 - r^2)\,dx$ , radii from the axis.

Watch out: $R^2 - r^2$ , not $(R - r)^2$ .

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