Maths › Further calculus › Volumes of revolution
Volumes of revolution
A slice perpendicular to the axis of rotation forms a disc or washer. About the x-axis, its area is π(R² − r²), where R and r are radial distances from the axis. Integrate this cross-sectional area over the stated interval, splitting where the boundary or radius changes.
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.
- Subtract solids for a region between two curves, and handle parametric curves.
- Check answers against known solids such as cones.
COMMON MISCONCEPTION
For a region between two curves, subtract the curves first and square after: V = π∫(youter − yinner)² dx.
Each slice is a washer, so the two radii are squared separately: dV = π(R2 − r2) dx with R ≥ r ≥ 0, and the square of a difference is not the difference of the squares. Between y = x and y = x2 from 0 to 1 the volume is π∫(x2 − x4) dx = 2π/15, where squaring the difference would give π/30.
Slicing into discs
Rotate the region under y = f(x) about the x-axis. Each vertical strip becomes a thin disc whose radius is the distance |y| from the axis and whose thickness is dx, so its volume is πy² dx and:
Check the current formula booklet for your awarding organisation; whether or not a volume formula is supplied, you must know when it applies and how to derive it. Slicing perpendicular to the y-axis instead gives V = π∫x² dy, with limits taken on the y-axis. The quantity squared is the radial distance from the axis of rotation, which matters as soon as two curves are involved.
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.
Squaring removes the square root before integration, as it does for any curve of the form y = √(g(x)).
Checking against known solids, and the y-axis
Rotating y = 2x from x = 0 to 3 produces a cone. V = π∫4x² dx = 36π, which matches πr²h/3 with r = 6 and h = 3. Use a known solid to check the integral whenever the curve is a straight 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.
The radius is now x, and x² = y on the curve.
V = π∫x² dy = π∫y dy from 0 to 4 = π × 8 = 8π.
The value matches the first example because this region is the first one reflected in y = x, so the solid is the same, now with limits on y and radius x.
GUIDED PRACTICE
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 gives πr²h/3 = π × 36 × 2/3 = 24π.
The two values agree.
Washers and parametric curves
A region between two curves generates a solid with a hole. About the x-axis, use radii measured perpendicular to the axis. Where R(x) ≥ r(x) ≥ 0, dV = π(R² − r²) dx. If the region crosses the axis or the outer radius changes, split the integral. When the region between y = x and y = x² from 0 to 1 is rotated about the x-axis, both curves are non-negative there and x ≥ x², so R = x and r = x², giving π∫(x² − x⁴) dx = π(1/3 − 1/5) = 2π/15. Square each radius separately: the square of a difference is not the difference of the squares.
Describe the drawing
The sliver between the line y equals x and the parabola y equals x squared shaded on the left, with its dashed mirror image below the axis, and one slice pulled out on the right as a washer: an amber outer rim of radius x from the line, a coral inner hole of radius x squared from the parabola, and navy showing through the middle. Squaring each radius separately and subtracting gives pi times the integral of x squared minus x to the fourth, which is 2 pi over 15.
WORKED EXAMPLE
A parametric volume
The curve x = t², y = t³ for 0 ≤ t ≤ 1 is rotated about the x-axis. Find the volume generated.
Change the variable in π∫y² dx. Here dx = 2t dt.
V = π∫ t⁶ × 2t dt from t = 0 to 1 = 2π∫ t⁷ dt.
= 2π[t⁸/8] = π/4.
Convert the limits to values of t when you replace dx; an integral in t must carry t-limits.
ASSESSMENT FOCUS
- Square the radius before integrating: integrating y and squaring afterwards gives a different value.
- About the y-axis, express x² in terms of y and use y-limits.
- Give answers as exact multiples of π unless the question asks for a decimal.
- For a region between two curves, use π∫(R² − r²) dx with R and r the radial distances, and split the integral where the region crosses the axis or the outer radius changes.
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
V = π∫4x² dx from 0 to 3 = π[4x³/3] = 36π. The solid is a cone with base radius 6 and height 3, and πr²h/3 gives the same 36π.
About the x-axis, V = π∫y² dx between x-limits, with the radius measured perpendicular to the axis; square first, then integrate.
About the y-axis, V = π∫x² dy between y-limits, with x² rewritten in terms of y.
For a washer, dV = π(R² − r²) dx with R ≥ r ≥ 0; split the integral where the region crosses the axis or the outer radius changes.
Parametric curves: substitute for y² and dx, and change the limits to t.
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.
Or read them with their worked answers on the volumes of revolution questions page.
USED LATER IN
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.
- Subtract solids for a region between two curves, and handle parametric curves.
- Check answers against known solids such as cones.
Open the full revision checklist to see every objective in the course in one place.