Volumes of revolution

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 of revolution made by spinning a region about the $x$ -axis or the $y$ -axis.

What they ask

The region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ is rotated through $360°$ about the $x$ -axis. Find the volume of the solid formed, giving your answer in terms of $\pi$ .

Picture the region spinning round the axis like a potter's wheel: it sweeps out a solid.

Definition

Volume of revolution

About the $x$ -axis, between $x = a$ and $x = b$ :

$V = \pi\int_a^b y^2\,dx$

$V$ is the volume.
$y$ is the curve's equation, in terms of $x$ ; it is the radius of each disc.
$a$ and $b$ are the $x$ -limits.

About the $y$ -axis, between $y = c$ and $y = d$ :

$V = \pi\int_c^d x^2\,dy$

The method

Find the limits (about the $x$ -axis these are $x$ -values).
Square $y$ first, then simplify it. Expand any bracket: $(x^2 + 1)^2 = x^4 + 2x^2 + 1$ .
Integrate $y^2$ and work out top limit minus bottom limit.
Multiply by $\pi$ . Leave it as "something $\pi$ " if they say in terms of $\pi$ or exact.

Worked example: The easiest one: a cone

The region under $y = x$ from $x = 0$ to $x = 3$ is rotated $360°$ about the $x$ -axis. Find the volume.

Have a go first?

Quick check

CalculateQuestion 1 · Recall

The region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ 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 lessonArea between two graphs Next lessonVolumes with a hole

Slice the solid into thin discs. Each is a cylinder of radius $y$ , so its volume is about $\pi y^2 \times$ its thickness.

Add them all up and you get the formula.

Multiple choiceQuestion 2 · Recall

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

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

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

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

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