Volume of Revolution, Including from Parametric Equations

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

What a volume of revolution is, and the one formula the spec actually requires

Take the region trapped between a curve $y=f(x)$ , the x-axis, and two vertical lines $x=a$ and $x=b$ , and spin it a full turn — 360°, i.e. $2\pi$ radians — about the x-axis. It sweeps out a genuine three-dimensional solid, and spec 6.1 asks for its volume. The spec's own guidance is unusually specific about which version of the formula is examinable: 'π∫y²dx is required, but not π∫x²dy' (WMA14 spec, section 6.1 guidance). Rotation about the y-axis exists as a mathematical idea, using the mirror-image formula $\pi\int x^2\,dy$ — but it is explicitly not part of this unit, so every volume-of-revolution question on this paper rotates about the x-axis.

Before trusting the formula on anything unfamiliar, check it against something you already know the answer to. Rotating the horizontal line $y=3$ about the x-axis between $x=0$ and $x=2$ sweeps out a cylinder of radius 3 and length 2 — and $\pi\int_0^2 3^2\,dx = \pi\int_0^2 9\,dx = 18\pi$ , exactly $\pi r^2 h$ with $r=3,h=2$ . Rotating the line $y=2x$ between $x=0$ and $x=3$ sweeps out a cone of base radius 6 and height 3 — and $\pi\int_0^3 (2x)^2\,dx = \pi\int_0^3 4x^2\,dx = \pi\left[\frac{4x^3}{3}\right]_0^3 = 36\pi$ , matching the mensuration formula $\frac13\pi r^2h$ with $r=6,h=3$ exactly. Neither check is new content — it's a way of confirming the formula does what it claims, using a shape whose volume you could already state before integrating anything.

That check also exposes what the formula is silently assuming, which matters the moment a curve dips below the x-axis: it squares $y$ before doing anything else, so a negative value of $y$ contributes exactly the same volume as the positive value at the same $x$ . A solid of revolution can't have negative volume, and $\pi\int y^2\,dx$ never produces one — reassuring, but it also means the formula alone can't tell you whether the curve crossed the axis partway through; that has to come from your own inspection of $f(x)$ , the same way a sketch would.

One more thing worth being honest about: nowhere in the spec's own 'formulae you must remember' list (section P4.2.3 — narrowly the scalar-product-in-components formula from the vectors content) or in the exam formula booklet's P4 section does $\pi\int y^2\,dx$ actually appear. It exists only as guidance text describing what the assessed skill is. No examiner report reviewed for this course ever records a candidate penalised for not knowing the formula itself — only for what happens after they write it down correctly — so the safest reading is that it's expected to be known regardless (it's standard content well beyond this one qualification), not that it's something to look up mid-exam. Either way, it isn't worth spending exam-room time hunting for on the booklet's pages: it isn't there.

Mechanism

Where π∫y²dx actually comes from — thin disks, not a rule to memorise

Cut the solid into thin vertical slices, each of thickness $\delta x$ , taken at some x-value between $a$ and $b$ . Provided $\delta x$ is small, one slice is almost exactly a cylinder: its two flat faces are (almost) parallel circles, both of radius $y=f(x)$ — because every point on the boundary curve traces out a circle of that radius once it's rotated — and its length is $\delta x$ . A cylinder's volume is $\pi r^2 h$ , so this one slice has volume approximately $\pi y^2 \delta x$ . Add up every slice from $x=a$ to $x=b$ : $V \approx \sum \pi y^2 \delta x$ . That sum is only an approximation, because $y$ changes slightly across each slice's own thickness — but the approximation gets better the thinner the slices are, and in the limit as $\delta x\to0$ the sum becomes exactly the $V=\pi\int_a^b y^2\,dx$ . Nobody chose this formula for its own sake; it's what 'add up infinitely many infinitesimally thin cylinders' turns into once the adding is written as an integral instead of a sum — the same limiting argument that turns a sum of rectangle areas into $\int f(x)\,dx$ in the first place, just with a rectangle's area $f(x)\,\delta x$ replaced by a disk's volume $\pi y^2\,\delta x$ . This is also the entire reason the formula squares $y$ rather than using it directly: a disk's area depends on the square of its radius, not the radius itself.

Diagram — The disk method, drawn: one representative slice of the solid formed by y = √x

x-axis: x · y-axis: y

y = √x: The boundary curve. Rotating the region between this curve, the x-axis, x=0 and x=4 by 360° about the x-axis generates the solid — every point on the curve becomes the outer edge of a circular cross-section once the rotation happens, which is exactly why the curve is the RADIUS function of the solid, not just its outline.

Representative disk at x = 2: A thin slice of the solid, thickness δx, taken at x = 2. Its radius is y(2) = √2 (read straight off the curve), so its volume is approximately π(√2)²δx = 2πδx — an ordinary cylinder volume, πr²h, with r = y and h = δx.
Axis of rotation, y = 0: The x-axis. Every disk's flat faces are perpendicular to it and its centre sits on it — which is exactly why the disk method only ever needs the single dimension y as a radius, never a full 2D cross-section.
Right boundary x = 4, y = 2: The largest disk in the solid, radius 2. As x grows the curve rises, so the disks get wider — a useful sanity check when a computed volume looks implausibly small or large: does it grow the direction the picture suggests it should?

Common error: Writing the volume as 2π∫y²dx, or leaving the π off entirely: V = ∫y²dx.

Correct: V = π∫y²dx, exactly once. The π comes from a single disk's own area formula, πr² — not from the 2π of a full rotation, which is already accounted for by every point on the boundary sweeping out one full circle, not by an extra factor stacked on top.

examiner-report · Jan 2021 · Q9

The same formula, run through a parameter

Nothing about the disk argument above cared how $y$ was written down — only that, at each x-value, there is a single radius to square. When a curve is given parametrically as $x=x(\theta)$ and $y=y(\theta)$ instead of directly as $y=f(x)$ , the disks are still there; only the bookkeeping for $dx$ changes. Since $x$ is itself a function of $\theta$ , $dx = \frac{dx}{d\theta}\,d\theta$ — an ordinary substitution. Putting that into $V=\pi\int y^2\,dx$ gives $V = \pi\int_\alpha^\beta y^2 \frac{dx}{d\theta}\,d\theta$ , where $\alpha$ and $\beta$ are the θ-values corresponding to the two x-limits of the region — not the x-limits themselves, which is the detail this step most often loses.

The spec's own guidance names this directly: 'Students should be able to find a volume of revolution, given parametric equations' (WMA14 spec, section 6.1 guidance) — it is explicitly inside the syllabus, not an extension beyond it. Nothing here is a new formula to learn; it is $\pi\int y^2\,dx$ from before, with $dx$ replaced by what it actually equals once $x$ is written in terms of $\theta$ .

Two things go wrong at this step specifically, both mechanical rather than conceptual. First, forgetting the $\frac{dx}{d\theta}$ factor altogether and integrating $y^2$ with respect to $\theta$ as if it were already $x$ — which silently changes the answer by whatever $\frac{dx}{d\theta}$ was, with no warning that anything has gone wrong, because the integral still evaluates to *something*. Second, keeping the original x-limits instead of converting them to the matching θ-values — an easy slip when $x(\theta)$ isn't the identity function, and one a quick check (does $x(\alpha)$ actually equal the lower x-bound?) catches in seconds.

Two separate skills — and which one the marks are actually testing

Every real WMA14 volume-of-revolution question reviewed for this lesson agrees on where the difficulty sits, and it is not where a first read of the spec point would suggest. The January 2021 general note on the topic states plainly: 'almost all candidates were able to state that the volume of the solid of revolution about the x axis would be π∫y²dx' (Jan 2021, Q9) — writing the formula down, choosing the right axis, setting the right limits: candidates get this right almost universally. Call that Skill 1: setting up the integral. Nothing above this paragraph is where marks are actually lost.

Skill 2 is executing the integral once it's set up — and every one of the three examiner reports reviewed for this topic documents a different specific way that goes wrong, from a different series. Getting the right when $y$ is a sum of sines and cosines (Jan 2021 Q9 — the worked example below). Recognising which integration technique the resulting fraction actually needs, rather than defaulting to one that doesn't fit (Oct 2022 Q5 — the method comparison below). Correctly squaring a compound algebraic expression for $y$ in the first place, before any integration technique is even relevant (Jan 2024 Q7(b) — the marked solution below). Three different series, three different specific breakdowns, and none of them is 'didn't know the formula.'

The practical consequence: revising this topic by re-memorising π∫y²dx one more time spends effort on the one part of the question that essentially nobody gets wrong. The return on effort is on the algebra and integration technique that comes immediately after it — which is also, not coincidentally, ordinary algebra and integration skill from earlier in the Pure strand, applied under the specific pressure of a compound expression that has just been squared.

Worked, in full

The volume when $x=\theta$ , $y=\sin\theta+\cos\theta$ is rotated about the x-axis between $\theta=0$ and $\theta=\frac{\pi}{4}$

01
Square $y$ before touching the integral at all — this is the step the marks actually sit on. $(\sin\theta+\cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$ .
Earns: M1 — attempts $y^2$ with all three terms of the expansion present, including the cross term. Dropping it is the same class of error the January 2024 report names on a different series' question — see the marked solution further down.
02
Simplify using $\sin^2\theta+\cos^2\theta=1$ and $2\sin\theta\cos\theta=\sin2\theta$ : $y^2 = 1+\sin2\theta$ . Since $x=\theta$ , $\frac{dx}{d\theta}=1$ , so $dx=d\theta$ and the θ-limits are already the x-limits, 0 to $\frac{\pi}{4}$ .
Earns: A1 — correct simplified $y^2$ , correct answer only. This is exactly the identity the January 2021 report names candidates failing to apply — see the embedded evidence below.
03
Set up the integral: $V=\pi\int_0^{\pi/4} y^2\,dx = \pi\int_0^{\pi/4}(1+\sin2\theta)\,d\theta$ .
Earns: M1 — correct integral, with the π retained and the correct limits. (Writing down π∫y²dx at all is the step almost every candidate manages — see the teach block above.)
04
Integrate: $\int(1+\sin2\theta)\,d\theta = \theta - \frac12\cos2\theta$ .
Earns: M1 — integrates term by term, with at least one term correctly handled ( $\int\sin2\theta\,d\theta=-\frac12\cos2\theta$ , a spec 5.1 standard result applied with $k=2$ ).
05
Evaluate: $\left[\theta-\frac12\cos2\theta\right]_0^{\pi/4} = \left(\frac{\pi}{4}-\frac12\cos\frac{\pi}{2}\right)-\left(0-\frac12\cos0\right) = \frac{\pi}{4}-\left(-\frac12\right)=\frac{\pi}{4}+\frac12$ . So $V=\pi\left(\frac{\pi}{4}+\frac12\right)=\frac{\pi^2}{4}+\frac{\pi}{2}$ .
Earns: A1 — correct exact final value, both terms present.

Source — Examiner report, Jan 2021

"There were a lot of slips in working with y² with many candidates failing to show that they had used sin2θ = 2sinθcosθ."

Complete it yourself

Complete the chain — the volume when $x=2\theta$ , $y=\cos\theta-\sin\theta$ is rotated about the x-axis between $\theta=0$ and $\theta=\frac{\pi}{6}$

01
Square $y$ first, before touching the integral: $y^2 = (\cos\theta-\sin\theta)^2 = \cos^2\theta - 2\sin\theta\cos\theta + \sin^2\theta$ . Using $\sin^2\theta+\cos^2\theta=1$ and $2\sin\theta\cos\theta=\sin2\theta$ : $y^2 = 1-\sin2\theta$ . (Note the sign — this curve has a MINUS between $\cos\theta$ and $\sin\theta$ , so the cross term comes out negative; the worked example above, with a PLUS, gave $1+\sin2\theta$ instead. Reading which sign belongs to which expression is worth the extra care.)
02
Convert $dx$ to the θ variable: $x=2\theta \Rightarrow \frac{dx}{d\theta}=2$ , so $dx=2\,d\theta$ . The volume is $V=\pi\int_0^{\pi/6} y^2 \frac{dx}{d\theta}\,d\theta = \pi\int_0^{\pi/6} 2(1-\sin2\theta)\,d\theta$ .

Complete it yourself

Complete the chain — the volume when $x=\theta$ , $y=2\sin\theta$ is rotated about the x-axis between $\theta=0$ and $\theta=\frac{\pi}{2}$

01
Square $y$ — and this time there is no compound sum to pull a cross term from. $y=2\sin\theta$ is a single term, so $y^2=4\sin^2\theta$ is already fully expanded; the trouble is a BARE $\sin^2\theta$ still has no elementary antiderivative in that form, and the cross-term identity used earlier in this lesson does not apply here — there is no second term to cross-multiply against. This calls for the OTHER double-angle identity, power-reduction: $\sin^2\theta=\frac{1-\cos2\theta}{2}$ , so $y^2=4\times\frac{1-\cos2\theta}{2}=2-2\cos2\theta$ . Confirmed as its own separate mark on the anchor question cited throughout this lesson: "Attempts to use sin²θ=(1−cos2θ)/2 ... and obtains Volume=∫(P±Qcos2θ)dθ. Depends on the first M." (Jan 2021, Q9).
02
Convert to the integration variable: $x=\theta \Rightarrow \frac{dx}{d\theta}=1$ , so $dx=d\theta$ and the θ-limits are already the x-limits, 0 to $\frac{\pi}{2}$ . $V=\pi\int_0^{\pi/2}(2-2\cos2\theta)\,d\theta$ .

Worked, in full

The volume when $x=\tan\theta$ , $y=\sin2\theta$ ( $\theta\ge0$ ) is rotated about the x-axis, bounded by the curve, the x-axis and $x=1$

01
This region is given by an x-limit, not a θ-limit — solve for the matching θ-value before anything else. $x=\tan\theta=1 \Rightarrow \theta=\frac{\pi}{4}$ (the value in $\left[0,\frac{\pi}{2}\right)$ , matching $\theta\ge0$ ). The lower limit needs the same check: $x=0 \Rightarrow \tan\theta=0 \Rightarrow \theta=0$ .
Earns: B1 — solves the parametric relation $x(\theta)$ for the θ-value matching a GIVEN x-limit, rather than being handed the θ-limit directly. Independent of every step below it — nothing about squaring y or converting dx depends on this being correct first — so, matching the equivalent mark in the real mark scheme, it can be earned wherever it is shown, including used silently as the upper limit further down.
02
Convert $dx$ to the θ variable: $x=\tan\theta \Rightarrow \frac{dx}{d\theta}=\sec^2\theta$ . So $V=\pi\int_0^{\pi/4} y^2\sec^2\theta\,d\theta$ .
Earns: M1 — sets up the integral entirely in θ, with the correct $\frac{dx}{d\theta}$ factor and, from stage 1, the correct limits.
03
Square $y$ — and this time EXPAND the double angle inside the square, the opposite direction from every other example in this lesson. $y=\sin2\theta \Rightarrow y^2=\sin^22\theta=(2\sin\theta\cos\theta)^2=4\sin^2\theta\cos^2\theta$ . Multiplying by $\sec^2\theta=\frac{1}{\cos^2\theta}$ from stage 2 cancels the $\cos^2\theta$ : $y^2\sec^2\theta = 4\sin^2\theta\cos^2\theta\times\frac{1}{\cos^2\theta} = 4\sin^2\theta$ .
Earns: dM1 — depends on the M1 above. Uses sin2θ = 2sinθcosθ and proceeds to the cos²θ/sec²θ cancellation, exactly as the real mark scheme requires: "Uses sin2θ = 2sinθcosθ and proceeds to Volume = B∫sin²θ dθ ... Note that if (2sin2θ)² becomes (2sinθcosθ)² with no evidence of a correct identity then score dM0." Squaring out $(2\sin\theta\cos\theta)^2$ without then carrying the cancellation through — left as $4\sin^2\theta\cos^2\theta$ with no further progress — is not enough on its own.
04
The bare $\sin^2\theta$ left over is still not directly integrable — apply power-reduction, a SEPARATE identity from stage 3: $\sin^2\theta=\frac{1-\cos2\theta}{2}$ , so $4\sin^2\theta = 4\times\frac{1-\cos2\theta}{2} = 2-2\cos2\theta$ .
Earns: dM1 — a second, independent double-angle mark, also depending on the first M rather than on stage 3 specifically. Stage 3 EXPANDED sin2θ to enable a cancellation; this one applies POWER-REDUCTION to the resulting bare sin²θ, because a bare even power of sinθ has no elementary antiderivative on its own. Treating this as the same double-angle step already done in stage 3, and skipping it, is exactly how this mark is lost — they are two different identities doing two different jobs.
05
Integrate: $\int(2-2\cos2\theta)\,d\theta = 2\theta-\sin2\theta$ .
Earns: M1 — integrates term by term, with the trig term correctly handled.
06
Evaluate: $\left[2\theta-\sin2\theta\right]_0^{\pi/4} = \left(\frac{\pi}{2}-\sin\frac{\pi}{2}\right)-(0-\sin0) = \left(\frac{\pi}{2}-1\right)-0=\frac{\pi}{2}-1$ . So $V=\pi\left(\frac{\pi}{2}-1\right)=\frac{\pi^2}{2}-\pi$ .
Earns: A1 — correct exact final value, both terms present.

Source — Mark scheme, Jan 2021

"dM1: Uses sin2θ = 2sinθcosθ and proceeds to Volume = B∫sin²θ dθ with or without "dθ" ... Note that if (2sin2θ)² becomes (2sinθcosθ)² with no evidence of a correct identity then score dM0. Depends on the first M."

Same question, every valid method

Find the exact volume when the curve $y=\frac{\sqrt{2x}}{x^2+1}$ between $x=0$ and $x=1$ is rotated 360° about the x-axis. (VERIDIAN-original question, built specifically to need the same choice October 2022's Q5 examiner report names — recognising a reverse-chain-rule integrand, or substituting for it — rather than reaching for integration by parts, a real technique that does not fit this shape of integrand efficiently.)

2 valid methods · every one reaches $\frac{\pi}{2}$ · 3 marks available

01
$y^2 = \left(\frac{\sqrt{2x}}{x^2+1}\right)^2 = \frac{2x}{(x^2+1)^2}$ , so $V=\pi\int_0^1 \frac{2x}{(x^2+1)^2}\,dx$ .
Method mark for squaring the compound expression correctly and writing down the integral with correct limits — Skill 1, the step almost every candidate reaches successfully.
M1
02
Differentiating $-(x^2+1)^{-1}$ by the chain rule gives $2x(x^2+1)^{-2}$ — exactly the integrand. So $\int \frac{2x}{(x^2+1)^2}\,dx = -\frac{1}{x^2+1}$ .
Method mark for recognising the integrand as $f'(x)[f(x)]^n$ in disguise (here $f(x)=x^2+1$ , $n=-2$ ) and writing down the antiderivative that recognition implies.
M1
03
$V=\pi\left[-\frac{1}{x^2+1}\right]_0^1 = \pi\left(-\frac12-(-1)\right)=\frac{\pi}{2}$ .
Accuracy mark for correct evaluation at both limits and the correct final exact value.
A1

Fastest when the chain-rule shape is visible — no extra variable to introduce, no limits to convert. The real cost, confirmed directly by the October 2022 report, is that failing to spot the shape leads candidates to waste 'much time pursuing incorrect methods,' including integration by parts, which doesn't resolve an integrand like this one efficiently. If the shape isn't obvious within a few seconds, switch to the substitution method rather than force a technique that isn't fitting.

In your own words

In one sentence: why can a candidate correctly write down $V=\pi\int y^2\,dx$ and still score close to zero on the rest of a volume-of-revolution question?

Marked, line by line

$y = x - \frac{2}{x}$ . The region between this curve, the x-axis, $x=1$ and $x=2$ is rotated 360° about the x-axis. (a) Write down the volume as a definite integral. (1) (b) Expand and simplify $y^2$ . (2) (c) Hence find the exact volume of the solid formed. (3) — VERIDIAN-original question, built specifically to test the algebra step named in the January 2024 report: candidates who state π∫y²dx correctly and then cannot square a compound expression for y. Not a reproduction of any past-paper question.

6 marks available

(a) — 1 mark

01
$V=\pi\int_1^2 y^2\,dx$
Independent mark for the correct formula with the correct limits — Skill 1, and per the January 2021 general report the step almost every candidate gets right; it does not depend on anything that follows.
B1

(b) — 2 marks

101
$y^2 = \left(x-\frac2x\right)^2 = x^2 - 2\cdot x\cdot\frac2x + \frac4{x^2}$
Method mark for attempting to expand the compound expression, with all three terms of $(p-q)^2=p^2-2pq+q^2$ present, including the cross term.
M1
102
$y^2 = x^2 - 4 + \frac{4}{x^2}$
Accuracy mark, correct answer only. The cross term $2\cdot x\cdot\frac2x$ simplifies to a constant (the x's cancel), which is easy to lose if the middle term is rushed.
A1

(c) — 3 marks

201
$\int\left(x^2-4+\frac4{x^2}\right)dx = \frac{x^3}{3}-4x-\frac4x$
Method mark for integrating term by term, with the power of at least one term correctly increased by 1 (here $x^{-2}\to x^{-1}$ , and note the sign: $\int 4x^{-2}\,dx=-4x^{-1}$ ).
M1
202
$\left[\frac{x^3}{3}-4x-\frac4x\right]_1^2 = \left(\frac83-8-2\right)-\left(\frac13-4-4\right) = -\frac{22}{3}-\left(-\frac{23}{3}\right)=\frac13$
Accuracy mark for correct evaluation at both limits.
A1
203
$V=\pi\times\frac13=\frac{\pi}{3}$
Accuracy mark, correct answer only, exact form. A decimal approximation here would not satisfy an 'exact volume' instruction, under the general marking guidance's own convention that marks are normally lost for resorting to rounded decimals where an exact answer is required.
A1

Named traps

wrong-constant-or-missing-pi: Confirmed directly on a real parametric volume-of-revolution question: "A few candidates mistakenly recalled the volume as 2π∫y²dx... as were those candidates who omitted the π" (Jan 2021, Q9). The mark scheme distinguishes the two: a wrong constant multiplier in front of an otherwise-correct integral is a different, more survivable error than losing the π-formula structure entirely — genuinely useful for partial-credit strategy, not just a warning to be careful. The fix is the same disk-method check the mechanism block above derives from scratch: the π comes from a single disk's own area, $\pi r^2$ , once — not from the $2\pi$ of a full rotation, which is already accounted for by every point on the boundary sweeping out one full circle, not by an extra factor stacked on top.
double-angle-identity-not-applied-when-squaring-y: Confirmed on the same question: "There were a lot of slips in working with y² with many candidates failing to show that they had used sin2θ = 2sinθcosθ" (Jan 2021, Q9). Whenever $y(\theta)$ is a sum or difference of $\sin\theta$ and $\cos\theta$ terms, squaring it produces a cross term of the form $2\sin\theta\cos\theta$ — and that term cannot be integrated in that form. It has to be recognised as $\sin2\theta$ (or, depending on the setup, absorbed via $\sin^2\theta+\cos^2\theta=1$ ) before the integral is even attempted. This is a P3 identity, carried forward in the exam formula booklet under its own cumulative rule, not new P4 content; the P4-specific failure is not recognising WHEN it needs to be reached for. The SAME identity is also needed in the opposite direction on this same cited question: whenever $y(\theta)$ already contains a $\sin2\theta$ (or $\cos2\theta$ ) term in its own right and $\frac{dx}{d\theta}$ is a trig expression rather than a constant (e.g. $\sec^2\theta$ , from $x=\tan\theta$ ), the double angle has to be EXPANDED — not condensed — to expose a $\cos^2\theta$ factor that then cancels against the $\sec^2\theta$ ; the worked chain further down models exactly this. On the real Jan 2021 Q9, expanding sin2θ this way is only the first of two separate dependent marks — the second is the power-reduction identity for a bare $\sin^2\theta$ , a genuinely distinct technique named in its own trap-taxonomy item below, not a repeat of this one.
power-reduction-identity-not-used-for-a-bare-sin-or-cos-squared: A second, separate dM1 on the same cited question, earned independently of the double-angle expansion above: "Attempts to use sin²θ=(1−cos2θ)/2 [or cos²θ=(1+cos2θ)/2] ... and obtains Volume=∫(P±Qcos2θ)dθ. Depends on the first M." (Jan 2021, Q9). A BARE $\sin^2\theta$ or $\cos^2\theta$ — one that is not the leftover of squaring a compound sum, and so has no $\sin^2\theta+\cos^2\theta=1$ shortcut to reach for — has no elementary antiderivative in that form; power-reduction is what makes it integrable, rewriting it as a constant plus a $\cos2\theta$ term. It is easy to mistake this for "the double-angle step already done above" and skip it as redundant — it is not: the cross-term identity CONDENSES two terms into one, while this one converts a single even power into an integrable linear combination. Both marks appear in sequence on the same real question; each is earned or lost independently of the other.
wrong-integration-method-chosen-for-the-resulting-fraction: Confirmed on a different series' question, testing the same spec point via a Cartesian rather than parametric curve: "The major stumbling block for the majority of candidates was a failure to choose a correct approach to the integration which was at the heart of the question... few candidates made this choice and wasted much time pursuing incorrect methods which included... using integration by parts [and] integrating to ln(2x²+3)³" (Oct 2022, Q5). The examiner names the actually-efficient route explicitly: "recognise the integrand as being the result of a chain rule differentiation of (2x²+3)⁻² or else using a substitution." Both are shown, side by side, in the method comparison above — the report's own point is that a candidate reaching for integration by parts on this shape of integrand is reaching for a real, legitimate technique applied to the wrong problem, not making an arithmetic slip.
compound-expression-not-squared-correctly: Confirmed on a third, independent series: "Many wrote the correct formula... but made no further progress, usually due to being unable to see how to square y" (Jan 2024, Q7(b)). Skill 1 — writing $V=\pi\int y^2\,dx$ — was intact; the block was purely algebraic, squaring a $y$ that is itself a sum or difference of two terms rather than a single one. $(p+q)^2=p^2+2pq+q^2$ : the cross term $2pq$ is not optional, and it is the term that most often goes missing, because dropping it still leaves something on the page that looks like a plausible squared expression. The marked solution above models exactly this failure and what it costs.

Retrieval — with feedback on every choice

Question 1

2 marks

A curve $y=f(x)$ , lying entirely above the x-axis, is rotated 360° about the x-axis between $x=1$ and $x=5$ . Which integral gives the volume of the solid formed?

Question 2

3 marks

A curve is given parametrically by $x=x(\theta)$ , $y=y(\theta)$ . To find the volume when the curve between $\theta=\alpha$ and $\theta=\beta$ is rotated about the x-axis, which integral do you evaluate?

Question 3

2 marks

$y = 2x - \frac{1}{x}$ . What is $y^2$ ?

Question 4

3 marks

A volume-of-revolution question simplifies down to $\int \frac{6x}{(x^2+2)^3}\,dx$ . Which is the most efficient way to evaluate it?

Question 5

2 marks

While finding a volume of revolution from parametric equations, you reach $y^2 = \cos^2\theta - 2\sin\theta\cos\theta + \sin^2\theta$ partway through squaring $y=\cos\theta-\sin\theta$ . What is the correct next simplification?

Reference — not a study method, a lookup

V = π∫ y² dx, about the x-axis ONLY. π∫x²dy is not required on this paper.
Parametric: dx = (dx/dθ)dθ, so V = π∫ y² (dx/dθ) dθ — convert the θ-limits too, not just dx.
Squaring a compound y: (p±q)² = p² ± 2pq + q². Never drop the cross term.
y = sinθ ± cosθ type: y² = 1 ± sin2θ, from sin²θ+cos²θ=1 and 2sinθcosθ=sin2θ.
Bare sin²θ or cos²θ (not from squaring a compound sum): sin²θ = (1 − cos2θ)/2, cos²θ = (1 + cos2θ)/2 — makes an otherwise-unintegrable bare even power integrable.
Stuck on the integral: check for a reverse chain rule shape (f′(x)[f(x)]ⁿ) before reaching for parts.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme, examiner report or spec in this lesson was independently verified against the primary Pearson document (WMA14-verified-facts.md), not carried over from prior course material — and each of the three examiner-report traps used here (Jan 2021 Q9, Oct 2022 Q5, Jan 2024 Q7(b)) is cited from exactly the one series it was found in, never inflated into a claim about multiple series agreeing on the same error. No primary document reviewed for this topic supplied a full worked numeric scenario, so every question in this lesson — prequestion, worked chain, chain drill, method comparison, marked solution and MCQ alike — is VERIDIAN-original wording and VERIDIAN-original numbers, built specifically to exercise one of the three documented failure points and checked symbolically (sympy) before being written down. Because the questions are original, the per-line mark allocations attached to them are modelled on the verified mark-scheme conventions in the facts bank (what M, A and B marks mean, when follow-through applies) rather than transcribed from a real mark scheme, which for an original question does not exist.

Question 12 marks

A curve $y=f(x)$ , lying entirely above the x-axis, is rotated 360° about the x-axis between $x=1$ and $x=5$ . Which integral gives the volume of the solid formed?

$\pi\int_1^5 y^2\,dx$
Correct — the disk method's formula, with the correct axis and correct limits.
B $2\pi\int_1^5 y^2\,dx$
A real, documented error on this exact topic: "A few candidates mistakenly recalled the volume as 2π∫y²dx" (Jan 2021, Q9). The 2 doesn't belong — the π already accounts for the disk's circular cross-section once per slice.
C $\pi\int_1^5 y\,dx$
This is $\pi$ times the area under the curve, not a volume — $y$ was never squared, so the formula has lost its connection to a disk's area, $\pi r^2$ .
D $\pi\int_1^5 x^2\,dy$
This is the formula for rotation about the y-axis — explicitly excluded by the spec's guidance: 'π∫y²dx is required, but not π∫x²dy.'

Traps tested: Wrong constant or missing pi · Y not squared · Wrong axis formula

Question 23 marks

A curve is given parametrically by $x=x(\theta)$ , $y=y(\theta)$ . To find the volume when the curve between $\theta=\alpha$ and $\theta=\beta$ is rotated about the x-axis, which integral do you evaluate?

$\pi\int_\alpha^\beta y^2\frac{dx}{d\theta}\,d\theta$
Correct. $V=\pi\int y^2\,dx$ with $dx=\frac{dx}{d\theta}\,d\theta$ substituted in, and the limits converted from x-values to the matching θ-values.
B $\pi\int_\alpha^\beta y^2\,d\theta$
The $\frac{dx}{d\theta}$ factor has been dropped entirely — this integrates $y^2$ with respect to $\theta$ as if $\theta$ and $x$ were the same variable, which silently changes the answer whenever $\frac{dx}{d\theta}\ne1$ .
C $\pi\int_{x(\alpha)}^{x(\beta)} y^2\,d\theta$
The limits belong to $x$ but the integral is being taken with respect to $\theta$ — an inconsistency. Limits and the variable of integration have to match: either both in $x$ (with $dx$ left as $dx$ ) or both in $\theta$ (with the limits converted to $\alpha,\beta$ and $dx$ replaced by $\frac{dx}{d\theta}\,d\theta$ ).
D $\pi\int_\alpha^\beta x^2\frac{dy}{d\theta}\,d\theta$
This is the parametric version of rotating about the y-axis — squaring $x$ instead of $y$ . Explicitly excluded by the same spec guidance that rules out $\pi\int x^2\,dy$ in the Cartesian case.

Traps tested: Dx dtheta factor dropped · Limits and variable mismatched · Wrong axis formula

Question 32 marks

$y = 2x - \frac{1}{x}$ . What is $y^2$ ?

$4x^2 - 4 + \frac{1}{x^2}$
Correct. $(2x)^2=4x^2$ , $\left(\frac1x\right)^2=\frac1{x^2}$ , and the cross term is $-2\cdot2x\cdot\frac1x=-4$ .
B $4x^2 + \frac{1}{x^2}$
The cross term is missing — this squares $2x$ and $\frac1x$ separately and just adds the results, the same 'unable to see how to square y' failure a real examiner report names on this topic (Jan 2024, Q7(b)).
C $4x^2 - 2 + \frac{1}{x^2}$
The cross term has lost its factor of 2: it should be $-2\times2x\times\frac1x=-4$ , not $-2x\times\frac1x=-2$ . $(p-q)^2=p^2-2pq+q^2$ — the 2 is part of the term, not optional.
D $4x^2 + 4 + \frac{1}{x^2}$
The sign of the cross term is flipped — this is what $(2x+\frac1x)^2$ would give. The original expression has a minus sign between the terms, which the cross term has to carry through as a negative.

Traps tested: Compound expression not squared correctly · Cross term halved · Cross term sign error

Question 43 marks

A volume-of-revolution question simplifies down to $\int \frac{6x}{(x^2+2)^3}\,dx$ . Which is the most efficient way to evaluate it?

Recognise it as the result of differentiating $-\frac{3}{2}(x^2+2)^{-2}$ by the chain rule, and write the antiderivative down directly.
Correct — check it: $\frac{d}{dx}\left[-\frac32(x^2+2)^{-2}\right] = -\frac32\times(-2)(x^2+2)^{-3}\times2x = 6x(x^2+2)^{-3}$ , exactly the integrand.
BIntegrate by parts, treating $6x$ and $(x^2+2)^{-3}$ as the two factors.
Integration by parts is a real technique — it's just the wrong one here. A real examiner report on this exact question type records candidates who 'wasted much time pursuing incorrect methods which included... using integration by parts' (Oct 2022, Q5), when the integrand was a chain-rule shape all along.
CRewrite it as $6x\times\ln(x^2+2)^{-3}$ and integrate the log term.
This mistakes a power for a logarithm. $(x^2+2)^{-3}$ is a reciprocal power — its antiderivative involves another power of $(x^2+2)$ , not a natural log. A log antiderivative belongs to a $\frac{1}{x^2+2}$ -shaped integrand, which this isn't. The same report records a genuinely attempted wrong log form on this question type: 'integrating to ln(2x²+3)³' (Oct 2022, Q5).
DExpand $(x^2+2)^{-3}$ using the binomial series for rational n, then integrate term by term.
This produces an infinite series, not an exact closed-form antiderivative — useless for an exact volume, and the general marking guidance's own convention is that marks are normally lost where an exact answer is required and a candidate resorts to an approximation instead. The binomial series solves a different problem (approximating a value near $x=0$ ), not this one.

Traps tested: Wrong integration method chosen for the resulting fraction · Reverse chain rule confused with log form · Binomial series misapplied to an exact integral

Question 52 marks

While finding a volume of revolution from parametric equations, you reach $y^2 = \cos^2\theta - 2\sin\theta\cos\theta + \sin^2\theta$ partway through squaring $y=\cos\theta-\sin\theta$ . What is the correct next simplification?

$y^2 = 1 - \sin2\theta$
Correct. $\sin^2\theta+\cos^2\theta=1$ , and the cross term $-2\sin\theta\cos\theta = -\sin2\theta$ , using $\sin2\theta=2\sin\theta\cos\theta$ .
B $y^2 = 1 + \sin2\theta$
The sign of the cross term is wrong. This expression is $\cos^2\theta-2\sin\theta\cos\theta+\sin^2\theta$ — the cross term is negative — so it simplifies to $1-\sin2\theta$ , not $1+\sin2\theta$ (which is what squaring $\sin\theta+\cos\theta$ instead would give).
C $y^2 = 1 - 2\sin\theta\cos\theta$
Mathematically true but not simplified into an integrable form. This is exactly the gap a real examiner report names on the sister question: candidates 'failing to show that they had used sin2θ = 2sinθcosθ' (Jan 2021, Q9) — the term $2\sin\theta\cos\theta$ can't be integrated as it stands and has to be rewritten as $\sin2\theta$ first.
D $y^2 = \cos2\theta$
This is a different, non-equivalent expression, from a different double-angle identity ( $\cos2\theta=\cos^2\theta-\sin^2\theta$ ) applied to the wrong situation. Check with a value: at $\theta=\frac{\pi}{8}$ , $1-\sin2\theta\approx0.293$ but $\cos2\theta\approx0.707$ — they don't agree.

Traps tested: Cross term sign error · Double angle identity not applied when squaring y · Wrong double angle identity used

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jan 2021 · Q9 — cited directly in this lesson
Mark scheme: Jan 2021 · Q9 — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA14.

Volume of Revolution, Including from Parametric Equations

What a volume of revolution is, and the one formula the spec actually requires

Where π∫y²dx actually comes from — thin disks, not a rule to memorise

The same formula, run through a parameter

Two separate skills — and which one the marks are actually testing

The volume when x=θx=\thetax=θ, y=sin⁡θ+cos⁡θy=\sin\theta+\cos\thetay=sinθ+cosθ is rotated about the x-axis between θ=0\theta=0θ=0 and θ=π4\theta=\frac{\pi}{4}θ=4π​

Complete the chain — the volume when x=2θx=2\thetax=2θ, y=cos⁡θ−sin⁡θy=\cos\theta-\sin\thetay=cosθ−sinθ is rotated about the x-axis between θ=0\theta=0θ=0 and θ=π6\theta=\frac{\pi}{6}θ=6π​

Complete the chain — the volume when x=θx=\thetax=θ, y=2sin⁡θy=2\sin\thetay=2sinθ is rotated about the x-axis between θ=0\theta=0θ=0 and θ=π2\theta=\frac{\pi}{2}θ=2π​

The volume when x=tan⁡θx=\tan\thetax=tanθ, y=sin⁡2θy=\sin2\thetay=sin2θ (θ≥0\theta\ge0θ≥0) is rotated about the x-axis, bounded by the curve, the x-axis and x=1x=1x=1

Deep dive

Practice this for real

The volume when $x=\theta$ , $y=\sin\theta+\cos\theta$ is rotated about the x-axis between $\theta=0$ and $\theta=\frac{\pi}{4}$

Complete the chain — the volume when $x=2\theta$ , $y=\cos\theta-\sin\theta$ is rotated about the x-axis between $\theta=0$ and $\theta=\frac{\pi}{6}$

Complete the chain — the volume when $x=\theta$ , $y=2\sin\theta$ is rotated about the x-axis between $\theta=0$ and $\theta=\frac{\pi}{2}$

The volume when $x=\tan\theta$ , $y=\sin2\theta$ ( $\theta\ge0$ ) is rotated about the x-axis, bounded by the curve, the x-axis and $x=1$