Volumes with a hole

5 minAS Level1.8.4

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can find the volume when a region that doesn't touch the axis is rotated about the $x$ -axis.

What they ask

The region between the curve $y = 5 - x^2$ and the line $y = 1$ is rotated $360°$ about the $x$ -axis. Find the volume.

The region floats above the axis, so the solid has a tunnel through the middle, like a bead or a doughnut.

The method

Find the limits: where the two graphs meet.
Name the radii: $R$ is the graph farther from the axis (the outer edge), $r$ the nearer one (the hole).
Square each radius on its own, never $(R - r)^2$ : $\;V = \pi\displaystyle\int_a^b \big(R^2 - r^2\big)\,dx$ . This is the step people get wrong.
Integrate, then top limit minus bottom limit.

This only works when the whole region is on one side of the axis. Both radii are then distances measured from the axis.

Worked example: The easiest one: a tube

The rectangle between $y = 3$ and $y = 1$ , from $x = 0$ to $x = 2$ , is rotated $360°$ about the $x$ -axis. Find the volume.

Have a go first?

Quick check

CalculateQuestion 1 · Recall

The rectangle between $y = 4$ and $y = 2$ , from $x = 0$ to $x = 3$ , is rotated $360°$ about the $x$ -axis. Find the volume, in terms of $\pi$ . (Type it as, for example, 5π.)

Answer the 2 questions above to continue.

Previous lessonVolumes of revolution Next lessonModulus: distance from zero

Each slice is a washer: a disc of radius

R

with a hole of radius

r

cut out. Its area is

\pi R^2 - \pi r^2

Multiple choiceQuestion 2 · Recall

The region between $y = x + 2$ and $y = 1$ , from $x = 0$ to $x = 2$ , is rotated $360°$ about the $x$ -axis. Which integral gives the volume?

\pi\displaystyle\int_0^2 \big((x + 2)^2 - 1\big)\,dx

\pi\displaystyle\int_0^2 (x + 2 - 1)^2\,dx

\pi\displaystyle\int_0^2 \big((x + 2)^2 + 1\big)\,dx

\pi\displaystyle\int_0^2 (x + 2)^2\,dx