Volumes of Revolution

Find the exact volume made when a region turns 360° about the x-axis or the y-axis, including regions between a curve and a line.

In the exam for Edexcel Further Pure Maths 4PM1 · 19 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Spin a region

A potter shapes a vase on a spinning wheel. Each side of the vase is a curve, and the spin turns it into a solid. In this lesson you find the volume of solids made this way.

Take the region under a curve and rotate it 360° about the x-axis. Cut the solid into very thin slices. Each slice is a disc: a short cylinder with radius $y$ and a tiny thickness.

Example

A disc has volume $π r^{2} \times thickness$ , and here $r = y$ . Adding up all the discs is what the integral does.

Formula

V = π \int_{a}^{b} y^{2} d x

$y$ — the curve. It is the radius of each disc, so it is squared.
$a$ — the x-value where the region starts.
$b$ — the x-value where the region ends.
$π$ — from the area of a circle. Keep it in the answer.

Rule

Learn this formula. It is not on the 4PM1 formula sheet.

Worked example

1
Check the formula with a cone. Rotate the line $y = \frac{x}{2}$ , from $x = 0$ to $x = 6$ , about the x-axis.
2
$V = π \int_{0}^{6} (\frac{x}{2})^{2} d x = π \int_{0}^{6} \frac{x ^{2}}{4} d x$ — Square y. The 2 on the bottom is squared too.
3
$= π [\frac{x ^{3}}{12}]_{0}^{6} = π \times \frac{216}{12} = 18 π$ — Integrate and put in the limits.
4
$\frac{1}{3} π (3)^{2} (6) = 18 π$ — The solid is a cone with radius 3 (y at x = 6) and height 6. The cone formula agrees.

Key idea

In 1615 Johannes Kepler published a book on measuring the volume of wine barrels. He imagined the barrel cut into many thin slices, the same idea you use here.

Key idea

Every slice is a disc of radius $y$ . That is why the formula has $π y^{2}$ .

Square y first

Most marks are lost in the squaring. Write $y^{2}$ out in full before you integrate. Then integrate term by term.

Worked example

1
The region under $y = x^{2} + 2$ from $x = 0$ to $x = 1$ is rotated 360° about the x-axis. Find the exact volume.
2
$y^{2} = (x^{2} + 2)^{2} = x^{4} + 4 x^{2} + 4$ — Square the whole bracket. Do not forget the middle term.
3
$V = π \int_{0}^{1} (x^{4} + 4 x^{2} + 4) d x$ — Put the squared expression into the formula.
4
$= π [\frac{x ^{5}}{5} + \frac{4 x ^{3}}{3} + 4 x]_{0}^{1}$ — Integrate each term.
5
$= π (\frac{1}{5} + \frac{4}{3} + 4) = \frac{83}{15} π$ — Put in the limits. The common denominator is 15.

Worked example

1
The region under $y = 2 x$ from $x = 0$ to $x = k$ is rotated about the x-axis. The volume is $18 π$ . Find $k$ .
2
$y^{2} = 4 x$ — Squaring a square root removes it: $(2 x)^{2} = 4 x$ .
3
$V = π \int_{0}^{k} 4 x d x = π [2 x^{2}]_{0}^{k} = 2 π k^{2}$ — Integrate and put in the limits.
4
$2 π k^{2} = 18 π, k^{2} = 9, k = 3$ — Set equal to the volume. Divide by $2 π$ . $k$ is positive.

Questions say “use algebraic integration”. A calculator answer with no working scores nothing, so show $y^{2}$ , the integral and the limits.

Common mistake

Squaring term by term. $(x + 3)^{2}$ is $x^{2} + 6 x + 9$ , not $x^{2} + 9$ .

Common mistake

Forgetting π. Write it at the start and keep it to the end. “Exact” means leave the answer in terms of π.

Key idea

Square first, in full. Then integrate. Then put in the limits.

Curves with e and sin

Squaring an exponential doubles its power: $(e^{x})^{2} = e^{2 x}$ . Squaring sin gives $sin^{2} x$ , which needs an identity before you can integrate it.

Worked example

1
The region under $y = e^{x} + e^{- x}$ from $x = 0$ to $x = 1$ is rotated about the x-axis. Find the exact volume.
2
$y^{2} = e^{2 x} + 2 + e^{- 2 x}$ — Square the bracket. The middle term is $2 \times e^{x} \times e^{- x} = 2 e^{0} = 2$ .
3
$V = π [\frac{1}{2} e^{2 x} + 2 x - \frac{1}{2} e^{- 2 x}]_{0}^{1}$ — Integrate: divide by 2 and by −2.
4
$= π (\frac{1}{2} e^{2} + 2 - \frac{1}{2} e^{- 2})$ $- π (\frac{1}{2} + 0 - \frac{1}{2})$ — Top minus bottom. At 0, $e^{0} = 1$ .
5
$= π (\frac{1}{2} e^{2} + 2 - \frac{1}{2} e^{- 2})$ — The bottom value is 0. Leave the answer exact.

Worked example

1
One arch of $y = sin x$ , from $x = 0$ to $x = π$ , is rotated about the x-axis. Find the exact volume.
2
$sin^{2} x = \frac{1}{2} - \frac{1}{2} cos 2 x$ — Put A = B = x in the formula sheet's $cos (A + B)$ to get $cos 2 x = cos^{2} x - sin^{2} x = 1 - 2 sin^{2} x$ . Then rearrange.
3
$V = π \int_{0}^{π} (\frac{1}{2} - \frac{1}{2} cos 2 x) d x$ $= π [\frac{x}{2} - \frac{1}{4} sin 2 x]_{0}^{π}$ — Cosine gives sine, divided by 2.
4
$= π (\frac{π}{2} - 0) - 0 = \frac{π ^{2}}{2}$ — $sin 2 π = 0$ and $sin 0 = 0$ .

Common mistake

Writing $\int sin^{2} x d x = \frac{s i n ^{3} x}{3}$ . The power rule does not work on sin. Use the identity.

Key idea

Square, simplify with the index laws or an identity, then integrate.

About the y-axis

Now spin the region about the y-axis. The discs lie flat, stacked up the y-axis. Each one has radius $x$ and a tiny thickness in the y-direction.

Formula

V = π \int_{c}^{d} x^{2} d y

$x$ — the radius of each disc. Write $x^{2}$ in terms of $y$ .
$c$ — the lower y-value.
$d$ — the upper y-value.

Rule

Learn this one too. It is not on the formula sheet.

Worked example

1
The region between $y = x^{2} + 1$ , the y-axis and the line $y = 5$ is rotated 360° about the y-axis. Find the exact volume.
2
$x^{2} = y - 1$ — Rearrange for $x^{2}$ . You need $x^{2}$ , not $x$ , so there is no square root.
3
$y from 1 to 5$ — The curve meets the y-axis at $y = 1$ . The top is the line $y = 5$ .
4
$V = π \int_{1}^{5} (y - 1) d y = π [\frac{y ^{2}}{2} - y]_{1}^{5}$ — Integrate with respect to $y$ .
5
$= π (\frac{15}{2} - (- \frac{1}{2})) = 8 π$ — At 5: 12.5 − 5 = 7.5. At 1: 0.5 − 1 = −0.5.

Common mistake

Using x-values as the limits. With $d y$ , the limits are y-values.

Key idea

About the y-axis: swap the roles. $x^{2}$ in terms of $y$ , y-limits, $d y$ .

Between a curve and a line

When the region lies between two graphs, the solid has a hole in it, like a ring. Find the volume of the outer solid, then take away the volume of the hole.

Formula

V = π \int_{a}^{b} (y_{1}^{2} - y_{2}^{2}) d x

$y_{1}$ — the outer graph, further from the x-axis.
$y_{2}$ — the inner graph, nearer the x-axis. It makes the hole.
$a, b$ — the x-values where the two graphs meet.

Worked example

1
The region between $y = 6 - x^{2}$ and the line $y = 2$ is rotated 360° about the x-axis. Find the exact volume.
2
$6 - x^{2} = 2 \Rightarrow x = \pm 2$ — Where the graphs meet: these are the limits.
3
$y_{1}^{2} - y_{2}^{2} = (6 - x^{2})^{2} - 2^{2}$ $= x^{4} - 12 x^{2} + 32$ — Square each one first, then subtract.
4
$V = π [\frac{x ^{5}}{5} - 4 x^{3} + 32 x]_{- 2}^{2}$ — Integrate each term.
5
$= π \times 2 (\frac{32}{5} - 32 + 64) = \frac{384}{5} π$ — The value at −2 is minus the value at 2, so top minus bottom doubles it.

Common mistake

Squaring the difference: $π \int (y_{1} - y_{2})^{2} d x$ . Here that gives $\frac{512}{15} π$ , which is wrong. Square each graph, then subtract.

Example

If the line goes through the origin, its part of the solid is a cone. You may use $\frac{1}{3} π r^{2} h$ for it instead of integrating.

Key idea

Outer volume minus inner volume. Square first, subtract second.

Which method? Read the question

Read which axis the region turns about, and what bounds it. That tells you what to do.

You see	You do
Rotated about the x-axis	$V = π \int y^{2} d x$ with x-limits
Rotated about the y-axis	$V = π \int x^{2} d y$ with y-limits
$y$ is a bracket or a sum	Square it in full before integrating
$sin^{2} x$ or $cos^{2} x$ after squaring	Use the double angle identity first
Region between two graphs	Find where they meet; $π \int (y_{1}^{2} - y_{2}^{2}) d x$
Volume given, find $k$	Integrate with $k$ as a limit, set equal, solve
“Exact” or “in terms of π”	Keep π, e and surds; no decimals

Common mistake

Using the area formula by habit. Volume needs $π$ and $y^{2}$ .

Key idea

Picture the discs. Their radius tells you what to square, and their direction tells you $d x$ or $d y$ .

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