Volumes of Revolution

This lesson covers volumes of revolution — finding the volume of a solid formed by rotating a curve around an axis — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to set up and evaluate integrals for rotation about both the x-axis and the y-axis.

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The Concept

When a region of a graph is rotated 360° (a full turn) about the x-axis or y-axis, it sweeps out a three-dimensional solid called a solid of revolution.

Think of it like a potter's wheel: the curve is the profile, and spinning it creates the 3D shape.

Examples:

• Rotating y = r (a horizontal line) about the x-axis creates a cylinder.
• Rotating y = x about the x-axis creates a cone.
• Rotating y = √(r² - x²) about the x-axis creates a sphere of radius r.

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Volume of Revolution About the x-axis

When the curve y = f(x) is rotated 360° about the x-axis between x = a and x = b:

V = π ∫ from a to b of y² dx

Derivation

Consider a thin disc of thickness δx at position x. The disc has radius y, so its volume is approximately π y² δx. Summing all such discs and taking the limit as δx → 0 gives:

V = π ∫ from a to b of y² dx

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Worked Examples — Rotation About the x-axis

Example 1: Volume of a Cone

Find the volume generated when the line y = 2x is rotated 360° about the x-axis from x = 0 to x = 3.

V = π ∫ from 0 to 3 of (2x)² dx = π ∫ from 0 to 3 of 4x² dx = π [4x³/3] from 0 to 3 = π (4 × 27/3 - 0) = π × 36 = 36π

Example 2: Curve y = √x

Find the volume generated when y = √x is rotated 360° about the x-axis from x = 0 to x = 4.

V = π ∫ from 0 to 4 of (√x)² dx = π ∫ from 0 to 4 of x dx = π [x²/2] from 0 to 4 = π (16/2 - 0) = 8π

Example 3: Curve y = 1/x

Find the volume generated when y = 1/x is rotated 360° about the x-axis from x = 1 to x = 4.

V = π ∫ from 1 to 4 of (1/x)² dx = π ∫ from 1 to 4 of x⁻² dx = π [-x⁻¹] from 1 to 4 = π (-1/4 - (-1)) = π (-1/4 + 1) = 3π/4

Example 4: Trigonometric Curve

Find the volume generated when y = sin x is rotated 360° about the x-axis from x = 0 to x = π.

V = π ∫ from 0 to π of sin²x dx

We need the identity: sin²x = (1 - cos 2x)/2

V = π ∫ from 0 to π of (1 - cos 2x)/2 dx = (π/2) ∫ from 0 to π of (1 - cos 2x) dx = (π/2)[x - sin(2x)/2] from 0 to π = (π/2)[(π - 0) - (0 - 0)] = π²/2

Exam Tip: When finding volumes involving sin²x or cos²x, you will almost always need the double angle identities: sin²x = (1 - cos 2x)/2 and cos²x = (1 + cos 2x)/2.

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Volume of Revolution About the y-axis

When the curve x = g(y) is rotated 360° about the y-axis between y = c and y = d:

V = π ∫ from c to d of x² dy

Key Difference

You need to express x² in terms of y and integrate with respect to y.

Example 1: Curve y = x² About the y-axis

Find the volume generated when the curve y = x² is rotated 360° about the y-axis from y = 0 to y = 4.

First, express x² in terms of y: since y = x², we have x² = y.

V = π ∫ from 0 to 4 of y dy = π [y²/2] from 0 to 4 = π (16/2) = 8π

Example 2: Curve y = eˣ About the y-axis

Find the volume generated when y = eˣ is rotated 360° about the y-axis from y = 1 to y = e.

Express x in terms of y: y = eˣ → x = ln y → x² = (ln y)²

V = π ∫ from 1 to e of (ln y)² dy

This requires integration by parts (twice). Let I = ∫ (ln y)² dy.

First application: u = (ln y)², dv = dy du = 2(ln y)(1/y) dy, v = y

I = y(ln y)² - ∫ 2 ln y dy

Second application (for ∫ ln y dy): u = ln y, dv = dy ∫ ln y dy = y ln y - y

So: I = y(ln y)² - 2(y ln y - y) = y(ln y)² - 2y ln y + 2y

V = π[y(ln y)² - 2y ln y + 2y] from 1 to e = π[(e(1)² - 2e(1) + 2e) - (0 - 0 + 2)] = π[(e - 2e + 2e) - 2] = π[e - 2] = π(e - 2)

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Volume Between Two Curves

If the region between two curves y = f(x) and y = g(x) (where f(x) ≥ g(x) ≥ 0) is rotated about the x-axis, the volume is:

V = π ∫ from a to b of [f(x)]² dx - π ∫ from a to b of [g(x)]² dx

= π ∫ from a to b of {[f(x)]² - [g(x)]²} dx

This is the volume of the outer solid minus the volume of the inner solid (like a ring or washer).

Example: Volume Between y = x² and y = x

Find the volume of revolution when the region between y = x and y = x² (from x = 0 to x = 1) is rotated about the x-axis.

Note y = x ≥ y = x² for 0 ≤ x ≤ 1.

V = π ∫ from 0 to 1 of (x² - x⁴) dx = π [x³/3 - x⁵/5] from 0 to 1 = π (1/3 - 1/5) = π (5/15 - 3/15) = 2π/15

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Key Formulae Derived from Volumes of Revolution

Shape	Curve	Limits	Volume
Cone (radius r, height h)	y = (r/h)x	0 to h	(1/3)πr²h
Sphere (radius r)	y = √(r² - x²)	-r to r	(4/3)πr³
Cylinder (radius r, height h)	y = r	0 to h	πr²h

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Common Mistakes

1. Forgetting the π — The formula is V = π ∫ y² dx, not just ∫ y² dx.
2. Not squaring y correctly — If y = x + 1, then y² = (x + 1)² = x² + 2x + 1, NOT x² + 1.
3. Mixing up x-axis and y-axis rotation — For x-axis, integrate y² with respect to x. For y-axis, integrate x² with respect to y.
4. Not converting variables — For y-axis rotation, you must express x² in terms of y.
5. Forgetting double angle identities — For sin²x and cos²x, use the identities before integrating.

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Summary

• About x-axis: V = π ∫ from a to b of y² dx.
• About y-axis: V = π ∫ from c to d of x² dy (express x in terms of y first).
• Between two curves: V = π ∫ [f(x)² - g(x)²] dx.
• Always include the factor of π.
• Square the function carefully — expand brackets if necessary.
• For sin²x or cos²x, use double angle identities.
• Volumes of revolution produce standard shapes (cone, sphere, cylinder) from simple curves.

A-Level Deep Dive: Volumes of Revolution

Spec mapping

Edexcel 9MA0-02 specification section 10 — Integration, sub-strand 10.5 covers integration to find the volume of revolution of a solid formed by rotating a region under a curve about the x-axis or the y-axis (refer to the official specification document for exact wording). This is Year 2 Pure content, examined on 9MA0-02 (Paper 2 — Pure Mathematics) and connected synoptically to section 8 (definite integration), section 5 (trigonometric identities, used to integrate $\sin^2 x$sin2x and $\cos^2 x$cos2x via the double-angle formulas) and section 7 (parametric equations, where the volume integral becomes $\pi \int y^2 \frac{dx}{dt}\,dt$π∫y2dtdx​dt). The Edexcel formula booklet does list the disc-method formulas $V = \pi \int_a^b y^2\,dx$V=π∫ab​y2dx and $V = \pi \int_c^d x^2\,dy$V=π∫cd​x2dy, but the algebraic manipulation that follows must be done from scratch.

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Worked example with full mark scheme

Question (8 marks): The region $R$R is bounded by the curve $y = \sqrt{x}$y=x​, the x-axis and the lines $x = 0$x=0 and $x = 4$x=4. The region $R$R is rotated through $2\pi$2π radians about the x-axis to form a solid of revolution $S$S.

(a) Find the exact volume of $S$S. (5)

(b) The same region $R$R is now rotated through $2\pi$2π radians about the y-axis instead. Find the exact volume of the new solid. (3)

Solution with mark scheme:

(a) Step 1 — quote the formula and substitute.

For rotation about the x-axis: $V = \pi \int_a^b y^2\,dx$V=π∫ab​y2dx.

Here $y = \sqrt{x}$y=x​, so $y^2 = x$y2=x. The limits are $a = 0$a=0, $b = 4$b=4.

$V = \pi \int_0^4 x\,dx$V=π∫04​xdx

M1 — correct formula quoted with $y^2$y2 (not $y$y) and correct limits substituted. A common slip is to integrate $y$y instead of $y^2$y2, which loses both M1s.

A1 — correct simplified integrand $x$x (recognising $(\sqrt{x})^2 = x$(x​)2=x for $x \geq 0$x≥0).

Step 2 — integrate.

$\int_0^4 x\,dx = \left[\dfrac{x^2}{2}\right]_0^4$∫04​xdx=[2x2​]04​

M1 — applying the power rule $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$∫xndx=n+1xn+1​ with $n = 1$n=1.

Step 3 — evaluate and multiply by $\pi$π.

$\left[\dfrac{x^2}{2}\right]_0^4 = \dfrac{16}{2} - 0 = 8 \implies V = 8\pi$[2x2​]04​=216​−0=8⟹V=8π

A1 A1 — correct evaluation and final exact answer with $\pi$π retained (not $25.13...$25.13...).

(b) Step 1 — switch to the y-axis formula.

For rotation about the y-axis: $V = \pi \int_c^d x^2\,dy$V=π∫cd​x2dy.

Rearrange $y = \sqrt{x}$y=x​ to $x = y^2$x=y2, so $x^2 = y^4$x2=y4. When $x = 0$x=0, $y = 0$y=0; when $x = 4$x=4, $y = 2$y=2.

M1 — rearranging $x$x as a function of $y$y and converting limits.

Step 2 — integrate and reconcile geometry.

$V_{\text{curve-to-y-axis}} = \pi \int_0^2 y^4\,dy = \pi \left[\dfrac{y^5}{5}\right]_0^2 = \dfrac{32\pi}{5}$Vcurve-to-y-axis​=π∫02​y4dy=π[5y5​]02​=532π​

The region $R$R is bounded by the x-axis (not the y-axis), so the wanted solid is the cylinder of radius 4 and height 2 minus the inner solid:

$V = \pi (4)^2 (2) - \dfrac{32\pi}{5} = 32\pi - \dfrac{32\pi}{5} = \dfrac{128\pi}{5}$V=π(4)2(2)−532π​=32π−532π​=5128π​

M1 A1 — correct integration and cylinder-minus-inner subtraction giving the exact final answer.

Total: 8 marks (M4 A4, split as shown).

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): The curve $C$C has equation $y = 2\sin x$y=2sinx for $0 \leq x \leq \pi$0≤x≤π. The region under $C$C is rotated through $2\pi$2π radians about the x-axis to form a solid $S$S.

(a) Show that the volume of $S$S is $2\pi^2$2π2. (5)

(b) Hence write down the volume when $y = 2\sin x + 1$y=2sinx+1 is rotated about the x-axis on the same interval. (1)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — quoting $V = \pi \int_0^{\pi} (2\sin x)^2\,dx = 4\pi \int_0^{\pi} \sin^2 x\,dx$V=π∫0π​(2sinx)2dx=4π∫0π​sin2xdx.
• M1 (AO1.1b) — applying the identity $\sin^2 x = \dfrac{1 - \cos 2x}{2}$sin2x=21−cos2x​ to convert into an integrable form.
• M1 (AO1.1b) — integrating: $\int \dfrac{1 - \cos 2x}{2}\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$∫21−cos2x​dx=2x​−4sin2x​+C.
• A1 (AO1.1b) — evaluating between $0$0 and $\pi$π: $\left[\dfrac{x}{2} - \dfrac{\sin 2x}{4}\right]_0^{\pi} = \dfrac{\pi}{2}$[2x​−4sin2x​]0π​=2π​.
• A1 (AO2.1) — combining: $V = 4\pi \cdot \dfrac{\pi}{2} = 2\pi^2$V=4π⋅2π​=2π2, matching the printed answer.

(b)

• B1 (AO3.1a) — recognising the new volume is not $2\pi^2$2π2 plus a constant; it requires expanding $(2\sin x + 1)^2 = 4\sin^2 x + 4\sin x + 1$(2sinx+1)2=4sin2x+4sinx+1 and integrating each term, giving $3\pi^2 + 8\pi$3π2+8π.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. Volumes-of-revolution questions usually weigh AO1 most heavily (procedural integration) but reserve an AO2 mark for trig-identity choice and an AO3 mark when a "hence" extension demands recognition that squaring breaks linearity.

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Synoptic links

Connects to:

1. Section 8 — Definite integration: the volume integral is a definite integral with integrand $\pi y^2$πy2. Every technique that earns marks in section 8 — substitution, integration by parts, partial fractions, trig identities — earns marks here too. A volumes question is fundamentally a definite-integration question dressed in geometric clothing.