Spin a curve about an axis and the solid's volume is π∫y2dx (or π∫x2dy). The mean value of f on [a,b] is b−a1∫abfdx. An improper integral has an infinite limit or an infinite integrand: replace it by a limit and see whether that limit exists.
Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2AQA 7367 Paper 1AQA 7367 Paper 2≈ 45 minRevision 1 · updated 2026-10-27
Find volumes of revolution about the x- and y-axes.
Find the mean value of a function over an interval.
Evaluate improper integrals as limits and decide whether they converge.
Pause and think
Can an infinitely long region have a finite area?
Section 2 of 14
Core idea
Start herePaint that fills but can't cover
Spin y=x1 (for x≥1) around the x-axis and you get Gabriel's horn: infinitely long, yet its volume is exactly π. Integrals over infinite ranges can still have finite values, if the integrand fades fast enough.
The big idea
Vx=π∫y2dx, Vy=π∫x2dy. Mean value b−a1∫abfdx. Improper integrals are limits: ∫a∞f=limb→∞∫abf.
1. Volumes of revolution
V=π∫y2dx
δV≈πy2δx
A thin slice is a disc of radius y, thickness δx.
V=lim∑πy2δx=π∫aby2dx
Add the discs.
About the y-axis use discs of radius x: π∫x2dy.
Worked exampley=x2 about the y-axis, 0≤y≤2
Volume.
Goal
V.
Strategy
Express x2 in terms of y.
1
x2=y
From the curve.
2
V=π∫02ydy=π[2y2]02
Discs.
3
=2π
Evaluate.
Answer2π.
✓ CheckHalf the cylinder of radius 2, height 2 (4π): a paraboloid fills exactly half ✓.
2. Mean value
fˉ=b−a1∫abf(x)dx
where
fˉ
the height of the rectangle with the same area and width.
3. Improper integrals
∫1∞x−pdx
∫1bx−pdx=1−pb1−p−1
For p=1.
p>1:b1−p→0⇒p−11
Converges.
p<1:b1−p→∞
Diverges.
p=1:lnb→∞
Diverges.
Worked example∫01xdx
Evaluate.
Goal
Value.
Strategy
Integrand infinite at 0: limit from the right.
1
a→0+lim∫a1x−1/2dx
Write as a limit.
2
=a→0+lim(2−2a)=2
a→0.
Answer2.
✓ CheckHere p=21<1, so the integral near 0 converges ✓.
⚠️ Common misconception"Plugging in ∞"Write the limit explicitly and say what happens to the term in b.
Summary
Disc method about either axis.
Mean value: area ÷ width.
Improper integrals as limits.
Key words
Converges
The limit exists and is finite.
Diverges
No finite limit.
📜 HistoryGabriel's horn.Evangelista Torricelli (1643) showed that rotating y=x1 for x≥1 gives a solid of finite volume π but infinite surface area: you could fill it with paint but not paint it.
🤯 Fun factAverage temperature.Weather services report the mean value of the temperature function over a day: literally 241∫024T(t)dt.
Section 3 of 14
Key formulae
VolumesVx=π∫aby2dx,Vy=π∫cdx2dy
Mean valuefˉ=b−a1∫abf(x)dx
Improper integral∫a∞f(x)dx=b→∞lim∫abf(x)dx
🪄 TrickPowers of 1/x.∫1∞xpdx converges exactly when p>1; ∫01xpdx converges exactly when p<1.
Section 4 of 14
Worked examples
Example 1 · Improper integral0 / 2 steps
Evaluate ∫1∞x2dx.
Example 2 · About the y-axis0 / 2 steps
y=x2, 0≤y≤2, rotated about the y-axis. Find the volume.
Section 5 of 14
Explore the graph
📈Area under 1/x² out to b
Loading the interactive graph…
Try this
The horizontal line shows the area 1−b1 from 1 to b.
As b grows, it creeps up to 1 but never passes it.
∫1bx2dx=1−b1→1.
🌍 In the real worldEscape velocity.The work to escape a planet is ∫R∞r2GMmdr=RGMm: an improper integral with a finite answer.
Section 6 of 14
Common mistakes
IncorrectV=π∫ydx
Correctπ∫y2dx
Each slice is a disc of area πy2.
IncorrectSubstituting ∞ directly
CorrectUse a limit
∞ is not a number; show the limit.
Incorrect∫1∞xdx converges
CorrectIt diverges
lnb→∞.
Infinitely long region
🤯
Area: 1. Exactly 1.
Section 7 of 14
Exam tip
🎯 Exam tipWrite the limit.Mark schemes give a mark for writing limb→∞ and showing what happens to the term in b.
Section 8 of 14
Guided practice
SA-FP-VI-G012 marksL1 FoundationCalculatorVolumes
The line y=x for 0≤x≤3 is rotated 2π about the x-axis. Find the volume (to 3 s.f.).
Salam Academy Original Practice · Proposed mark scheme