Volumes of Revolution about the y-axis

In the previous lesson we rotated regions about the $x$x-axis. Rotation about the $y$y-axis is equally important and needs a different setup: the disc method now integrates with respect to $y$y (so you must express $x$x in terms of $y$y), and a powerful alternative — the cylindrical-shell method — lets you keep $y=f(x)$y=f(x) as it is. This lesson derives both, shows how to choose between them, and works examples (including verifying that the two methods always agree).

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1 Spec mapping (AQA 7367)

Volumes of revolution about the $y$y-axis 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%), within "Further calculus". The disc-about-$y$y form is core AQA; the shell method, while not always named on the AQA list, is a legitimate and often faster route that earns full marks provided the method is stated and justified — and it is explicitly examinable on several boards. The work is mainly AO1 (set up and evaluate), with AO2 in the choice and justification of method and in cross-checking by an alternative method, and AO3 in applied modelling. It is synoptic with inverse functions (rearranging $y=f(x)$y=f(x) into $x=g(y)$x=g(y)) and integration by parts (the $(\ln y)^2$(lny)2 example), and it is the direct counterpart of the $x$x-axis lesson — internalising the symmetry between the two cases halves the amount you must remember.

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2 Core theory — disc method about the $y$y-axis

Rotating a curve $x=g(y)$x=g(y) about the $y$y-axis, slice horizontally: a slice at height $y$y of thickness $\delta y$δy sweeps out a disc of radius $x$x, volume $\pi x^2\,\delta y$πx2δy. Summing and letting $\delta y\to 0$δy→0:

$V = \pi \int_c^{d} x^2\,dy,$V=π∫cd​x2dy,

where $c,d$c,d are the $y$y-limits. The structure is exactly the mirror image of the $x$x-axis case: there we sliced vertically and integrated $\pi y^2\,dx$πy2dx; now we slice horizontally and integrate $\pi x^2\,dy$πx2dy. The radius of a horizontal disc is the horizontal distance from the $y$y-axis to the curve, namely $x$x, so the disc face has area $\pi x^2$πx2. The crucial first step is therefore to express $x$x (hence $x^2$x2) in terms of $y$y — rearrange $y=f(x)$y=f(x) into $x=g(y)$x=g(y), and read off the $y$y-values that bound the region.

Key step. Everything inside the integral, and both limits, must be in $y$y. Mixing an $x$x-integrand with $y$y-limits is the classic error, and it produces a meaningless quantity. If you find yourself with $\,dy$dy but an integrand still written in $x$x, you have not finished the rearrangement.

flowchart LR
  A["curve x = g(y)"] -->|"horizontal slice at height y"| B["disc, radius x, thickness δy"]
  B -->|"volume πx²·δy"| C["sum, let δy → 0"]
  C -->|"∫ from c to d in y"| D["V = π ∫ x² dy"]

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3 Worked examples with M1/A1 mark scheme

Worked Example 1 — simple rearrangement

Find the volume when the region bounded by $y=x^2$y=x2, the $y$y-axis and $y=4$y=4 is rotated about the $y$y-axis.

Rearrange $y=x^2 \Rightarrow x^2 = y$y=x2⇒x2=y (first quadrant, $x\ge 0$x≥0). The $y$y-limits are $0$0 to $4$4, and conveniently $x^2$x2 is already in terms of $y$y here, so no further work is needed:

$V = \pi \int_0^{4} x^2\,dy = \pi \int_0^{4} y\,dy = \pi\left[\frac{y^2}{2}\right]_0^{4} = \pi(8) = 8\pi.$V=π∫04​x2dy=π∫04​ydy=π[2y2​]04​=π(8)=8π.

(M1 rearrange to $x^2=y$x2=y with $y$y-limits; M1 integrate; A1 $8\pi$8π.)

Worked Example 1b — a square-root curve

Find the volume when the region bounded by $y = \sqrt{x}$y=x​, the $y$y-axis and $y = 2$y=2 is rotated about the $y$y-axis. Rearrange $y = \sqrt{x} \Rightarrow x = y^2 \Rightarrow x^2 = y^4$y=x​⇒x=y2⇒x2=y4; the $y$y-limits are $0$0 to $2$2:

$V = \pi \int_0^{2} x^2\,dy = \pi \int_0^{2} y^4\,dy = \pi\left[ \frac{y^5}{5} \right]_0^{2} = \pi\cdot\frac{32}{5} = \frac{32\pi}{5}.$V=π∫02​x2dy=π∫02​y4dy=π[5y5​]02​=π⋅532​=532π​.

Note carefully that $x = y^2$x=y2 gives $x^2 = y^4$x2=y4, not $y^2$y2 — squaring the already-squared inverse is the step where marks are most often dropped on this kind of question.

Worked Example 2 — cubic curve

Find the volume when $y=x^3$y=x3 (for $0\le x\le 2$0≤x≤2, i.e. $0\le y\le 8$0≤y≤8) is rotated about the $y$y-axis.

$y=x^3 \Rightarrow x = y^{1/3} \Rightarrow x^2 = y^{2/3}$y=x3⇒x=y1/3⇒x2=y2/3:

$V = \pi \int_0^{8} y^{2/3}\,dy = \pi \left[ \frac{y^{5/3}}{5/3} \right]_0^{8} = \frac{3\pi}{5}\,8^{5/3}.$V=π∫08​y2/3dy=π[5/3y5/3​]08​=53π​85/3.

Since $8^{5/3} = \big(8^{1/3}\big)^5 = 2^5 = 32$85/3=(81/3)5=25=32,

$V = \frac{3\pi}{5}\times 32 = \frac{96\pi}{5}.$V=53π​×32=596π​.

The fractional-index arithmetic is where this example tests you: $x = y^{1/3}$x=y1/3 so $x^2 = y^{2/3}$x2=y2/3, the integral of $y^{2/3}$y2/3 is $\dfrac{y^{5/3}}{5/3} = \dfrac{3}{5}y^{5/3}$5/3y5/3​=53​y5/3, and $8^{5/3}$85/3 is best computed by taking the cube root first ($8^{1/3}=2$81/3=2) then raising to the fifth power. Doing the root before the power keeps the numbers small and avoids errors.

Worked Example 3 — exponential (needs parts)

The region bounded by $y=e^{x}$y=ex, the $y$y-axis, $y=1$y=1 and $y=e^2$y=e2 is rotated about the $y$y-axis.

$y=e^x \Rightarrow x=\ln y \Rightarrow x^2 = (\ln y)^2$y=ex⇒x=lny⇒x2=(lny)2, so $V = \pi\displaystyle\int_1^{e^2}(\ln y)^2\,dy$V=π∫1e2​(lny)2dy. Integrate by parts twice:

$\int (\ln y)^2\,dy = y(\ln y)^2 - \int 2\ln y\,dy = y(\ln y)^2 - 2\big(y\ln y - y\big) = y(\ln y)^2 - 2y\ln y + 2y.$∫(lny)2dy=y(lny)2−∫2lnydy=y(lny)2−2(ylny−y)=y(lny)2−2ylny+2y.

$\Big[ y(\ln y)^2 - 2y\ln y + 2y \Big]_1^{e^2}: \quad \text{at } e^2:\ 4e^2 - 4e^2 + 2e^2 = 2e^2; \quad \text{at } 1:\ 0 - 0 + 2 = 2.$[y(lny)2−2ylny+2y]1e2​:at e2: 4e2−4e2+2e2=2e2;at 1: 0−0+2=2.

$V = \pi(2e^2 - 2) = 2\pi\big(e^2 - 1\big).$V=π(2e2−2)=2π(e2−1).

This example showcases the two skills that make the $y$y-axis disc method demanding: the rearrangement ($y=e^x \Rightarrow x = \ln y$y=ex⇒x=lny, so $x^2 = (\ln y)^2$x2=(lny)2) and then a non-trivial integration of $(\ln y)^2$(lny)2 requiring parts twice. Take the parts carefully: the first application reduces $(\ln y)^2$(lny)2 to $\ln y$lny, and the second reduces $\ln y$lny to a constant. A common slip is to stop after one application; the presence of $\ln y$lny (still not elementary to integrate without parts) is the signal to go again.

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4 Specimen-style exam question

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

The region $R$R is bounded by $y = 4 - x^2$y=4−x2 and the $x$x-axis (with $x\ge 0$x≥0). (a) Find the volume generated when $R$R is rotated through $2\pi$2π about the $y$y-axis using the disc method. (b) Verify your answer using the shell method.

(a) Rearrange: $y = 4 - x^2 \Rightarrow x^2 = 4 - y$y=4−x2⇒x2=4−y; the $y$y-limits run $0$0 to $4$4 (from the $x$x-axis up to the vertex at $y=4$y=4):

$V = \pi \int_0^{4} (4 - y)\,dy = \pi\left[ 4y - \frac{y^2}{2} \right]_0^{4} = \pi(16 - 8) = 8\pi.$V=π∫04​(4−y)dy=π[4y−2y2​]04​=π(16−8)=8π.

(b) The curve meets the $x$x-axis at $x=2$x=2 (solving $4-x^2=0$4−x2=0 with $x\ge 0$x≥0); each shell has radius $x$x and height $4-x^2$4−x2 (the height of the curve above the $x$x-axis at that $x$x):

$V = 2\pi \int_0^{2} x(4 - x^2)\,dx = 2\pi \int_0^{2} (4x - x^3)\,dx = 2\pi\left[ 2x^2 - \frac{x^4}{4} \right]_0^{2} = 2\pi(8 - 4) = 8\pi. \ \checkmark$V=2π∫02​x(4−x2)dx=2π∫02​(4x−x3)dx=2π[2x2−4x4​]02​=2π(8−4)=8π. ✓

Both methods give $8\pi$8π, as they must — a reassuring confirmation that the setup is correct. Part (b) is worth doing in full even though part (a) already produced the answer, because "verify" explicitly asks for the independent second calculation; simply asserting "shells give the same" would earn nothing.

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5 The shell method

Consider a thin vertical strip at $x$x of height $f(x)$f(x) and width $\delta x$δx. Rotating it about the $y$y-axis sweeps out a thin cylindrical shell (like a tin can with no top or bottom) of radius $x$x, height $f(x)$f(x), thickness $\delta x$δx. Imagine slitting the shell down its side and "unrolling" it into a flat thin slab: its length is the circumference $2\pi x$2πx, its height is $f(x)$f(x), and its thickness is $\delta x$δx. So its volume is

$\delta V \approx (\text{circumference})\times(\text{height})\times(\text{thickness}) = 2\pi x \cdot f(x)\cdot \delta x \quad\Longrightarrow\quad V = 2\pi \int_a^{b} x\,f(x)\,dx.$δV≈(circumference)×(height)×(thickness)=2πx⋅f(x)⋅δx⟹V=2π∫ab​xf(x)dx.

The factor $2\pi x$2πx is the circumference of the circle the strip traces; the factor $f(x)$f(x) is its height. The beauty is that you keep $y=f(x)$y=f(x) as given — no rearrangement to $x=g(y)$x=g(y) is needed — so shells shine precisely when inverting $f$f is hard or impossible (for instance $y = e^{-x^2}$y=e−x2, which has no elementary inverse).

flowchart LR
  A["vertical strip at x, height f(x)"] -->|"rotate about y-axis"| B["cylindrical shell, radius x"]
  B -->|"unroll: 2πx · f(x) · δx"| C["sum, let δx → 0"]
  C -->|"∫ from a to b in x"| D["V = 2π ∫ x·f(x) dx"]

Worked Example 4 — shell method, with a disc check

Use shells for $y=x^2$y=x2, $0\le x\le 2$0≤x≤2, about the $y$y-axis:

$V = 2\pi \int_0^{2} x\cdot x^2\,dx = 2\pi \int_0^{2} x^3\,dx = 2\pi\left[ \frac{x^4}{4} \right]_0^{2} = 2\pi(4) = 8\pi.$V=2π∫02​x⋅x2dx=2π∫02​x3dx=2π[4x4​]02​=2π(4)=8π.

Disc check: $x^2=y$x2=y, $V = \pi\int_0^4 y\,dy = \pi[\tfrac{y^2}{2}]_0^4 = 8\pi$V=π∫04​ydy=π[2y2​]04​=8π. ✓ The two methods slice the same solid in perpendicular directions — vertically for shells, horizontally for discs — so they must give the same volume. Performing both here is quick and demonstrates the equivalence; in an exam, doing so when time permits turns a single answer into a self-verified one.

Worked Example 5 — shell method with parts

Use shells for $y=\sin x$y=sinx, $0\le x\le \pi$0≤x≤π, about the $y$y-axis:

$V = 2\pi \int_0^{\pi} x\sin x\,dx.$V=2π∫0π​xsinxdx.

By parts ($u=x,\ dv=\sin x\,dx,\ v=-\cos x$u=x, dv=sinxdx, v=−cosx):

$\int_0^{\pi} x\sin x\,dx = \big[-x\cos x\big]_0^{\pi} + \int_0^{\pi}\cos x\,dx = \big[-\pi\cos\pi\big] + \big[\sin x\big]_0^{\pi} = \pi + 0 = \pi.$∫0π​xsinxdx=[−xcosx]0π​+∫0π​cosxdx=[−πcosπ]+[sinx]0π​=π+0=π.

$V = 2\pi\cdot \pi = 2\pi^2.$V=2π⋅π=2π2.

This is a perfect illustration of why shells are the natural choice here. Rotating $y=\sin x$y=sinx about the $y$y-axis by the disc method would require inverting $y=\sin x$y=sinx to $x=\arcsin y$x=arcsiny and integrating $(\arcsin y)^2$(arcsiny)2 — a genuinely unpleasant integral. Shells sidestep all of that: keep $y=\sin x$y=sinx, multiply by the radius $x$x, and integrate $x\sin x$xsinx by a single application of parts. Whenever the curve is given as $y=f(x)$y=f(x) and $f$f is awkward to invert, this is the tell-tale sign to use shells.

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6 Washer method about the $y$y-axis — Worked Example 6

Between two curves, subtract inner from outer (radii now measured horizontally):

$V = \pi \int_c^{d} \big( x_{\text{out}}^2 - x_{\text{in}}^2 \big)\,dy.$V=π∫cd​(xout2​−xin2​)dy.

For $x=y$x=y and $x=y^2$x=y2 on $0\le y\le 1$0≤y≤1: here $y \ge y^2$y≥y2, so $x=y$x=y is outer (further from the $y$y-axis), $x=y^2$x=y2 inner. Square each separately: $x_{\text{out}}^2 = y^2$xout2​=y2, $x_{\text{in}}^2 = y^4$xin2​=y4:

$V = \pi \int_0^{1} \big(y^2 - y^4\big)\,dy = \pi\left[ \frac{y^3}{3} - \frac{y^5}{5} \right]_0^{1} = \pi\left( \frac13 - \frac15 \right) = \frac{2\pi}{15}.$V=π∫01​(y2−y4)dy=π[3y3​−5y5​]01​=π(31​−51​)=152π​.

The radii are measured horizontally from the $y$y-axis, so the cross-section at height $y$y is an annulus with outer radius $y$y and inner radius $y^2$y2, of area $\pi(y^2 - y^4)$π(y2−y4). Note this is the exact $y$y-axis counterpart of the $x$x-axis washer between $y=x$y=x and $y=x^2$y=x2, which also gave $\tfrac{2\pi}{15}$152π​ — a pleasing symmetry arising because the region is symmetric under swapping $x$x and $y$y.

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6a Synoptic links