Maths › Further calculus › Volumes of revolution
Volumes of revolution
Spin a curve round an axis and it sweeps out a solid; slice the solid into discs and integration adds them up. One formula turns areas into volumes.
Builds on Definite integrals and areas and Areas and the limit of a sum.
IN THIS TOPIC
- Use V = π∫y² dx for rotation about the x-axis, with the right limits.
- Swap to V = π∫x² dy for rotation about the y-axis.
- Check answers against known solids such as cones.
WHAT YOU PROBABLY THINK
Integration can only ever measure flat area; volume is beyond it.
Slicing into discs
Rotate the region under y = f(x) about the x-axis and every vertical strip becomes a thin disc of radius y and thickness dx. A disc's volume is πy² dx, and integration stacks the discs:
The same slicing sideways handles rotation about the y-axis: V = π∫x² dy, with limits read off the y-axis. Flat areas were never the ceiling, whatever the opening claim says: the squared radius carries the third dimension.
WORKED EXAMPLE
A curve made solid
The region under y = √x from x = 0 to 4 is rotated fully about the x-axis. Find the volume.
V = π∫y² dx = π∫x dx from 0 to 4.
= π[x²/2] = π × 8 = 8π.
Note how squaring tamed the surd before any integration happened; rotation about the x-axis often simplifies exactly this way.
Sanity checks and the y-axis
Rotating y = 2x from x = 0 to 3 should make a cone, and it does: V = π∫4x² dx = 36π, matching the formula πr²h/3 with r = 6 and h = 3. Known solids are free marking checks; use them whenever the curve is a line.
WORKED EXAMPLE
About the y-axis instead
The region between y = x², the y-axis and y = 4 is rotated about the y-axis. Find the volume.
Radius is now x, and x² = y on the curve.
V = π∫x² dy = π∫y dy from 0 to 4 = π × 8 = 8π.
Same number as the first example by coincidence, but the setup is the mirror image: limits on y, radius x.
YOUR TURN
A cone by integration
By rotating y = 3x between x = 0 and x = 2 about the x-axis, verify the cone volume formula for r = 6, h = 2.
Show the working
V = π∫9x² dx from 0 to 2 = π[3x³] = 24π.
The formula: πr²h/3 = π × 36 × 2/3 = 24π.
The two answers agree, which is exactly what the disc method promises.
THE EXAM BIT
- Square before integrating; the classic error is integrating y and squaring after.
- About the y-axis, everything flips: express x² in terms of y and use y-limits.
- Leave answers as exact multiples of π unless the question asks for decimals.
- For a region between two curves, subtract the two solids: π∫(outer² − inner²).
CHECK YOURSELF
The region under y = 2x from x = 0 to 3 is rotated about the x-axis. Find the volume, and name the solid.
Show a hint
π∫4x² dx, then compare with πr²h/3.
Show the answer
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About the x-axis: V = π∫y² dx between x-limits; square first, integrate second.
About the y-axis: V = π∫x² dy between y-limits, with x² rewritten in terms of y.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Use V = π∫y² dx for rotation about the x-axis, with the right limits.
- Swap to V = π∫x² dy for rotation about the y-axis.
- Check answers against known solids such as cones.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.