Volumes of Revolution about the x-axis

When a region in the $xy$xy-plane is rotated through $2\pi$2π radians (a full turn) about the $x$x-axis, it sweeps out a three-dimensional solid of revolution. Integration lets us compute its exact volume by slicing the solid into infinitely thin discs — the disc method. This lesson derives the formula from first principles, works a graded set of examples (including a washer between two curves and a parametric solid), recovers the classical cone and sphere formulae as checks, and drills the exam technique that secures method marks.

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1 Spec mapping (AQA 7367)

Volumes of revolution are compulsory Pure content for Papers 1 and 2 (each 2 h, 100 marks, 33⅓%, with per-paper weighting AO1 55% / AO2 25% / AO3 20%), in the "Further calculus" strand. The topic is mostly AO1 (set up and evaluate the integral) but carries strong AO2 value whenever you are asked to derive the formula, justify which curve is "outer", or confirm a standard solid's volume; an applied "modelling a vase / a lens" version reaches into AO3. It is deeply synoptic: it reuses A-Level integration (powers, $\sin^2$sin2 via double angle, $e^{kx}$ekx), connects to parametric calculus, and previews the surface-area-of-revolution and Pappus results later in this course. Mastering the disc and washer methods here also makes the $y$y-axis case (next lesson) and surface-area formulae feel like small variations rather than new topics.

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2 Core theory — the disc method

Take a curve $y=f(x)$y=f(x) on $[a,b]$[a,b] with $f(x)\ge 0$f(x)≥0. Slice $[a,b]$[a,b] into $n$n thin strips each of width $\delta x$δx. Rotating one strip about the $x$x-axis produces a thin disc (a short cylinder) of radius $y=f(x)$y=f(x) and thickness $\delta x$δx, so its volume is approximately

$\delta V \approx \pi y^2\,\delta x = \pi \big[f(x)\big]^2\,\delta x.$δV≈πy2δx=π[f(x)]2δx.

Summing the discs and letting $\delta x \to 0$δx→0 turns the sum into an integral:

$V = \lim_{\delta x \to 0}\sum \pi y^2\,\delta x = \pi \int_a^{b} y^2\,dx = \pi \int_a^{b} \big[f(x)\big]^2\,dx.$V=limδx→0​∑πy2δx=π∫ab​y2dx=π∫ab​[f(x)]2dx.

This is the disc method (sometimes called the method of discs, or — when two curves are involved — the washer or annulus method). The logic is identical to the way the area under a curve is built from thin rectangles of height $y$y and width $\delta x$δx; here we simply rotate each rectangle into a disc and add up volumes instead of areas. The radius of a typical disc is the height of the curve at that $x$x, namely $y=f(x)$y=f(x), and the disc's circular face has area $\pi r^2 = \pi y^2$πr2=πy2. Multiplying by the thickness $\delta x$δx and summing gives the formula above.

Critical warning. You must square $y$y before integrating. The single commonest error is to write $\pi\int y\,dx$π∫ydx instead of $\pi\int y^2\,dx$π∫y2dx. The $y^2$y2 comes from the disc area $\pi r^2$πr2 — there is no way to get a volume from $\int y\,dx$∫ydx (which has the units of an area). A quick dimensional check catches this slip every time: a volume must come from integrating something with units of (length)$^2$2, and $y^2$y2 supplies exactly that.

A small inline diagram makes the slicing concrete:

flowchart LR
  A["curve y = f(x)"] -->|"rotate strip about x-axis"| B["thin disc, radius y, thickness δx"]
  B -->|"volume πy²·δx"| C["sum and let δx → 0"]
  C -->|"∫ from a to b"| D["V = π ∫ y² dx"]

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3 Worked examples with M1/A1 mark scheme

Worked Example 1 — parabola

Find the volume when the region bounded by $y=x^2$y=x2, the $x$x-axis and the lines $x=0,\ x=3$x=0, x=3 is rotated fully about the $x$x-axis.

$V = \pi \int_0^{3} (x^2)^2\,dx = \pi \int_0^{3} x^4\,dx \qquad \text{(M1 quote } \pi\int y^2\,dx \text{ and square)}$V=π∫03​(x2)2dx=π∫03​x4dx(M1 quote π∫y2dx and square)

$= \pi \left[ \frac{x^5}{5} \right]_0^{3} = \pi\left( \frac{243}{5} \right) = \frac{243\pi}{5} \quad\text{(A1).}$=π[5x5​]03​=π(5243​)=5243π​(A1).

Worked Example 2 — square-root function

Find the volume when the region between $y=\sqrt{x}$y=x​, the $x$x-axis, $x=0$x=0 and $x=4$x=4 is rotated about the $x$x-axis.

$V = \pi \int_0^{4} \left(\sqrt{x}\right)^2 dx = \pi \int_0^{4} x\,dx = \pi \left[ \frac{x^2}{2} \right]_0^{4} = \pi(8) = 8\pi.$V=π∫04​(x​)2dx=π∫04​xdx=π[2x2​]04​=π(8)=8π.

Worked Example 3 — straight line (a cone, as a check)

Find the volume when $y=2x$y=2x (for $0\le x\le 3$0≤x≤3) is rotated about the $x$x-axis.

$V = \pi \int_0^{3} (2x)^2\,dx = \pi \int_0^{3} 4x^2\,dx = \pi \left[ \frac{4x^3}{3} \right]_0^{3} = \pi(36) = 36\pi.$V=π∫03​(2x)2dx=π∫03​4x2dx=π[34x3​]03​=π(36)=36π.

This is a cone of base radius $6$6 (the value of $y=2x$y=2x at $x=3$x=3) and height $3$3. The standard formula agrees: $V = \tfrac13 \pi r^2 h = \tfrac13 \pi (6^2)(3) = 36\pi$V=31​πr2h=31​π(62)(3)=36π. ✓ Such cross-checks against known solids are an excellent way to confirm a disc-method answer.

Worked Example 4 — a trigonometric solid

Find the volume when $y=\sin x$y=sinx (for $0\le x\le \pi$0≤x≤π) is rotated about the $x$x-axis. Use $\sin^2 x = \tfrac12(1-\cos 2x)$sin2x=21​(1−cos2x):

$V = \pi \int_0^{\pi} \sin^2 x\,dx = \frac{\pi}{2}\int_0^{\pi} (1 - \cos 2x)\,dx = \frac{\pi}{2}\left[ x - \frac{\sin 2x}{2} \right]_0^{\pi} = \frac{\pi}{2}\,\pi = \frac{\pi^2}{2}.$V=π∫0π​sin2xdx=2π​∫0π​(1−cos2x)dx=2π​[x−2sin2x​]0π​=2π​π=2π2​.

(M1 use of the double-angle identity; M1 integrate; A1 $\tfrac{\pi^2}{2}$2π2​.)

The double-angle identity is essential: you cannot integrate $\sin^2 x$sin2x directly. Rewriting $\sin^2 x = \tfrac12(1-\cos 2x)$sin2x=21​(1−cos2x) turns the problem into integrating a constant and a cosine, both routine. The same trick ($\cos^2 x = \tfrac12(1+\cos 2x)$cos2x=21​(1+cos2x)) handles any volume generated by a sine or cosine curve — a pattern worth recognising instantly, since trigonometric solids are a recurring exam favourite.

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4 Specimen-style exam question

(Specimen-style — not a real past paper.)

The finite region $R$R is enclosed by the curves $y=x$y=x and $y=x^2$y=x2. (a) Find the coordinates of the points of intersection. (b) Find the exact volume generated when $R$R is rotated through $2\pi$2π about the $x$x-axis.

(a) $x = x^2 \Rightarrow x(1-x)=0 \Rightarrow x=0$x=x2⇒x(1−x)=0⇒x=0 or $x=1$x=1; the curves meet at $(0,0)$(0,0) and $(1,1)$(1,1).

(b) On $(0,1)$(0,1), $x > x^2$x>x2, so $y=x$y=x is the outer curve and $y=x^2$y=x2 the inner — use the washer method $V = \pi\int_a^b\big(y_{\text{out}}^2 - y_{\text{in}}^2\big)\,dx$V=π∫ab​(yout2​−yin2​)dx:

$V = \pi \int_0^{1} \big(x^2 - x^4\big)\,dx = \pi \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_0^{1} = \pi\left( \frac13 - \frac15 \right) = \frac{2\pi}{15}.$V=π∫01​(x2−x4)dx=π[3x3​−5x5​]01​=π(31​−51​)=152π​.

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5 The washer method (between two curves)

When the region between an outer curve $y=f(x)$y=f(x) and an inner curve $y=g(x)$y=g(x) is rotated about the $x$x-axis, subtract the inner disc from the outer disc at each $x$x:

$V = \pi \int_a^{b} \Big( \big[f(x)\big]^2 - \big[g(x)\big]^2 \Big)\,dx.$V=π∫ab​([f(x)]2−[g(x)]2)dx.

Beware: $\big[f\big]^2 - \big[g\big]^2 \ne (f-g)^2$[f]2−[g]2=(f−g)2. You square each curve separately then subtract — squaring the difference is wrong. Geometrically, $\pi f^2$πf2 is the area of the outer disc and $\pi g^2$πg2 is the area of the inner hole; the cross-section of the solid is the annulus (ring) between them, of area $\pi(f^2 - g^2)$π(f2−g2). The quantity $(f-g)^2$(f−g)2 would instead be the square of the gap between the curves, which has no geometric meaning here.

The washer method only applies when both curves lie on the same side of the axis (so that the inner curve really does carve a hole out of the outer disc). If the region straddles the axis of rotation, or if the "inner" curve crosses to the other side, you must split the region first — a point examiners sometimes test by giving curves that intersect partway along.

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6 More worked examples

Worked Example 5 — exponential

Find the volume when $y=e^{-x}$y=e−x (for $0\le x\le 2$0≤x≤2) is rotated about the $x$x-axis.

$V = \pi \int_0^{2} \big(e^{-x}\big)^2\,dx = \pi \int_0^{2} e^{-2x}\,dx = \pi \left[ -\tfrac12 e^{-2x} \right]_0^{2} = \pi\left( \tfrac12 - \tfrac12 e^{-4} \right) = \frac{\pi}{2}\big(1 - e^{-4}\big).$V=π∫02​(e−x)2dx=π∫02​e−2xdx=π[−21​e−2x]02​=π(21​−21​e−4)=2π​(1−e−4).

Numerically $e^{-4}\approx 0.0183$e−4≈0.0183, so $V \approx \tfrac{\pi}{2}(0.9817) \approx 1.542$V≈2π​(0.9817)≈1.542. The crucial step is squaring $(e^{-x})^2 = e^{-2x}$(e−x)2=e−2x — doubling the exponent, not leaving it as $e^{-x}$e−x; this is exactly the kind of index slip that quietly loses accuracy marks.

Worked Example 6 — hyperbola

Find the volume when $y=\dfrac1x$y=x1​ (for $1\le x\le 3$1≤x≤3) is rotated about the $x$x-axis.

$V = \pi \int_1^{3} \frac{1}{x^2}\,dx = \pi \left[ -\frac1x \right]_1^{3} = \pi\left( -\tfrac13 + 1 \right) = \frac{2\pi}{3}.$V=π∫13​x21​dx=π[−x1​]13​=π(−31​+1)=32π​.

Here squaring gives $(\tfrac1x)^2 = \tfrac{1}{x^2} = x^{-2}$(x1​)2=x21​=x−2, which integrates to $-x^{-1}$−x−1; this is the same integrand whose improper version over $[1,\infty)$[1,∞) gives the finite-volume Torricelli's trumpet of $\S 12$§12.

Worked Example 7 — parametric solid

If $x = x(t),\ y = y(t)$x=x(t), y=y(t) with $t$t from $\alpha$α to $\beta$β, then the substitution $\,dx = \dfrac{dx}{dt}\,dt$dx=dtdx​dt converts the $x$x-integral into a $t$t-integral:

$V = \pi \int y^2\,dx = \pi \int_{\alpha}^{\beta} y^2 \,\frac{dx}{dt}\,dt.$V=π∫y2dx=π∫αβ​y2dtdx​dt.

The disc-method logic is unchanged — radius $y$y, area $\pi y^2$πy2 — but everything is now expressed through the parameter $t$t. Remember to convert the limits to the corresponding $t$t-values (here $t$t already runs over $[\alpha,\beta]$[α,β]).

For $x=t^2,\ y=t,\ 0\le t\le 2$x=t2, y=t, 0≤t≤2: $\dfrac{dx}{dt}=2t$dtdx​=2t, so

$V = \pi \int_0^{2} t^2 \cdot 2t\,dt = \pi \int_0^{2} 2t^3\,dt = \pi \left[ \frac{t^4}{2} \right]_0^{2} = \pi(8) = 8\pi.$V=π∫02​t2⋅2tdt=π∫02​2t3dt=π[2t4​]02​=π(8)=8π.

(Take care with sign: if $x$x decreases as $t$t increases, $\tfrac{dx}{dt}<0$dtdx​<0 and you adjust to keep $V>0$V>0.)

Worked Example 7b — a quadratic requiring expansion

Find the volume when $y = x + 1$y=x+1 (for $0\le x\le 2$0≤x≤2) is rotated about the $x$x-axis. Square first, then integrate:

$V = \pi \int_0^{2} (x+1)^2\,dx = \pi \int_0^{2} (x^2 + 2x + 1)\,dx = \pi \left[ \frac{x^3}{3} + x^2 + x \right]_0^{2} = \pi\left( \frac{8}{3} + 4 + 2 \right) = \frac{26\pi}{3}.$V=π∫02​(x+1)2dx=π∫02​(x2+2x+1)dx=π[3x3​+x2+x]02​=π(38​+4+2)=326π​.

The lesson is procedural: expand $(x+1)^2$(x+1)2 into $x^2 + 2x + 1$x2+2x+1 before integrating — you cannot integrate $(x+1)^2$(x+1)2 term-by-term without expanding (or using a substitution).

Worked Example 7c — a washer between a line and a curve

Find the volume when the region between $y = 2$y=2 (outer) and $y = x$y=x (inner) for $0\le x\le 2$0≤x≤2 is rotated about the $x$x-axis. Here $2 \ge x$2≥x on the interval, so $y=2$y=2 is outer:

$V = \pi \int_0^{2} \big( 2^2 - x^2 \big)\,dx = \pi \int_0^{2} (4 - x^2)\,dx = \pi \left[ 4x - \frac{x^3}{3} \right]_0^{2} = \pi\left( 8 - \frac{8}{3} \right) = \frac{16\pi}{3}.$V=π∫02​(22−x2)dx=π∫02​(4−x2)dx=π[4x−3x3​]02​=π(8−38​)=316π​.

Note the cross-section here is genuinely a ring: at each $x$x the solid has an outer radius $2$2 and an inner radius $x$x, so the washer area is $\pi(4 - x^2)$π(4−x2).

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7 Deriving the sphere — Worked Example 8

Rotate the semicircle $y=\sqrt{r^2 - x^2}$y=r2−x2​ for $-r\le x\le r$−r≤x≤r about the $x$x-axis:

$V = \pi \int_{-r}^{r} \big(r^2 - x^2\big)\,dx = \pi \left[ r^2 x - \frac{x^3}{3} \right]_{-r}^{r}.$V=π∫−rr​(r2−x2)dx=π[r2x−3x3​]−rr​.

$= \pi \left[ \left( r^3 - \tfrac{r^3}{3} \right) - \left( -r^3 + \tfrac{r^3}{3} \right) \right] = \pi \left[ \tfrac{2r^3}{3} + \tfrac{2r^3}{3} \right] = \frac{4}{3}\pi r^3.$=π[(r3−3r3​)−(−r3+3r3​)]=π[32r3​+32r3​]=34​πr3.

The classical volume of a sphere — recovered cleanly by integration. The key step where students slip is the lower limit: $(-r)^3 = -r^3$(−r)3=−r3, so the term $-\tfrac{x^3}{3}$−3x3​ evaluated at $-r$−r is $-\tfrac{(-r)^3}{3} = +\tfrac{r^3}{3}$−3(−r)3​=+3r3​, and $r^2 x$r2x at $-r$−r is $-r^3$−r3; together the lower-limit contribution is $-r^3 + \tfrac{r^3}{3} = -\tfrac{2r^3}{3}$−r3+3r3​=−32r3​, which is subtracted, giving the second $+\tfrac{2r^3}{3}$+32r3​. Because $y^2 = r^2 - x^2$y2=r2−x2 is even, you could equally integrate from $0$0 to $r$r and double: $V = 2\pi\int_0^r (r^2 - x^2)\,dx = 2\pi\big[r^3 - \tfrac{r^3}{3}\big] = 2\pi\cdot\tfrac{2r^3}{3} = \tfrac43\pi r^3$V=2π∫0r​(r2−x2)dx=2π[r3−3r3​]=2π⋅32r3​=34​πr3 — often the cleaner route, and a useful trick whenever the integrand is even and the limits are symmetric.

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8 Mark-scheme literacy