Surface Area of Revolution

When a curve is spun about an axis it sweeps out a surface of revolution — a sphere from a semicircle, a cone from a line, a torus from an off-axis circle. To find its area, slice the curve into tiny arcs $\mathrm{d}s$ds; each sweeps a thin band whose area is its circumference times its width. Summing these bands gives $S = \int 2\pi y\,\mathrm{d}s$S=∫2πyds. This lesson builds the formula from the arc-length element of the previous lesson, derives the sphere, cone and torus from scratch, and connects everything through Pappus's elegant centroid theorem.

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1 Spec mapping (AQA 7367)

Surface area of revolution is compulsory Pure content for Paper 1 and Paper 2 (each 2 h, 100 marks, 33⅓%, with per-paper weighting AO1 55% / AO2 25% / AO3 20%), in the "Further calculus" strand and built directly on arc length. It is among the most algebraically demanding topics in the course, so it is rich in AO2: examiners almost always scaffold with "Show that the integrand simplifies to …", rewarding the surd simplification and clear communication, before the AO1 integration. Deriving a known area (sphere $=4\pi r^2$=4πr2, cone $=\pi r\ell$=πrℓ) tests whether you can prove, not merely quote, a result. Because the working is long, disciplined layout — formula first, simplify the root, substitute carefully — is what protects the marks; a single dropped factor of $\tfrac12$21​ or a confused $y$y-vs-$x$x circumference can lose several. Surface area also sits at the intersection of three earlier skills — differentiation (for $y'$y′), the arc-length element $\mathrm{d}s$ds, and integration by substitution — so it is a genuinely synoptic topic that rewards fluency across the whole Further-calculus strand rather than a single isolated technique. Treat it as the natural sequel to arc length, with the added twist of weighting each element by its circumference.

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2 Core theory — the band element $2\pi y\,\mathrm{d}s$2πyds

Take a small arc of $y=f(x)$y=f(x) of length $\mathrm{d}s$ds at height $y$y above the $x$x-axis. Rotating it a full turn about the $x$x-axis sweeps a thin band — essentially the curved surface of a tiny frustum (cone slice). Its circumference is $2\pi y$2πy (the circle traced by the point at height $y$y) and its width is the slant $\mathrm{d}s$ds, so its area is

$\mathrm{d}S = 2\pi y\,\mathrm{d}s.$dS=2πyds.

Since $\mathrm{d}s = \sqrt{1+\left(\tfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\,\mathrm{d}x$ds=1+(dxdy​)2​dx (arc-length lesson), integrating gives the surface area about the $x$x-axis:

$\boxed{\;S = 2\pi\int_a^{b} y\,\sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}}\;\mathrm{d}x\;}.$S=2π∫ab​y1+(dxdy​)2​dx​.

The key contrast with volume of revolution: a volume sums discs $\pi y^2\,\mathrm{d}x$πy2dx (flat, area times thickness), whereas a surface sums bands $2\pi y\,\mathrm{d}s$2πyds (curved, circumference times slant). The slant $\mathrm{d}s$ds — not the flat $\mathrm{d}x$dx — is what makes surface area carry the square root. The parametric and $y$y-axis versions just swap which quantity plays the radius and which the element:

$S_x = 2\pi\int y\,\sqrt{\Big(\tfrac{\mathrm{d}x}{\mathrm{d}t}\Big)^2 + \Big(\tfrac{\mathrm{d}y}{\mathrm{d}t}\Big)^2}\;\mathrm{d}t, \qquad S_y = 2\pi\int x\,\mathrm{d}s.$Sx​=2π∫y(dtdx​)2+(dtdy​)2​dt,Sy​=2π∫xds.

For rotation about the $y$y-axis, the radius of each band is the horizontal distance $x$x, so the circumference factor becomes $2\pi x$2πx.

Why a band, not a flat ring? A common temptation is to model the swept element as a flat washer of area $2\pi y\,\mathrm{d}x$2πydx (circumference times horizontal thickness). This under-counts whenever the curve is sloped, because the true element is the slanted curved surface of a tiny frustum, whose width is the arc $\mathrm{d}s$ds, not the projection $\mathrm{d}x$dx. The exact surface area of a frustum of slant height $\ell$ℓ with end-radii $r_1,r_2$r1​,r2​ is $\pi(r_1+r_2)\ell$π(r1​+r2​)ℓ; for a thin band $r_1\approx r_2\approx y$r1​≈r2​≈y and $\ell = \mathrm{d}s$ℓ=ds, recovering $\mathrm{d}S = 2\pi y\,\mathrm{d}s$dS=2πyds exactly. So the $\mathrm{d}s$ds is not a refinement — it is structurally required, and using $\mathrm{d}x$dx instead gives a genuinely wrong (too small) area for any non-horizontal curve.

The presence of $\mathrm{d}s$ds is also why surface area inherits all the surd-simplification machinery of arc length: every surface-area integral contains a $\sqrt{1+(y')^2}$1+(y′)2​ factor that must be tamed, usually multiplied by the radius $y$y (or $x$x). Frequently the radius and the surd interact to cancel — as in the sphere, where $\sqrt{r^2-x^2}$r2−x2​ in the radius meets $\dfrac{1}{\sqrt{r^2-x^2}}$r2−x2​1​ in the slant — which is precisely the engineered simplification examiners look for.

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3 Worked examples with M1/A1 mark scheme

Worked Example 1 — a cone

Find the surface generated by rotating $y = 2x$y=2x (for $0\le x\le 3$0≤x≤3) about the $x$x-axis. Here $\dfrac{\mathrm{d}y}{\mathrm{d}x}=2$dxdy​=2, so $\sqrt{1+4} = \sqrt5$1+4​=5​:

$S = 2\pi\int_0^{3} 2x\,\sqrt5\;\mathrm{d}x = 2\pi\sqrt5\Big[x^2\Big]_0^{3} = 2\pi\sqrt5\cdot 9 = 18\pi\sqrt5.$S=2π∫03​2x5​dx=2π5​[x2]03​=2π5​⋅9=18π5​.

(M1 for $2\pi\int y\sqrt{1+(y')^2}\,dx$2π∫y1+(y′)2​dx with $\sqrt5$5​; A1 for $18\pi\sqrt5$18π5​.) Check against the cone formula $S=\pi r\ell$S=πrℓ: the cone has radius $r = 2\cdot3 = 6$r=2⋅3=6 and slant height $\ell = \sqrt{3^2+6^2} = 3\sqrt5$ℓ=32+62​=35​, so $\pi r\ell = \pi\cdot6\cdot3\sqrt5 = 18\pi\sqrt5$πrℓ=π⋅6⋅35​=18π5​. Agreement. Note that because $\sqrt{1+(y')^2}$1+(y′)2​ is constant for a straight line (the slope never changes), the surd comes out of the integral as a single factor — a feature unique to conical surfaces and a quick way to spot that the answer will involve $\pi r\ell$πrℓ. For any genuinely curved generator the surd varies with $x$x and must stay inside the integral.

Worked Example 2 — the sphere

Derive the surface area of a sphere of radius $r$r by rotating $y=\sqrt{r^2-x^2}$y=r2−x2​ about the $x$x-axis, $-r\le x\le r$−r≤x≤r.

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-x}{\sqrt{r^2-x^2}} \;\implies\; 1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \frac{r^2-x^2+x^2}{r^2-x^2} = \frac{r^2}{r^2-x^2}.$dxdy​=r2−x2​−x​⟹1+(dxdy​)2=r2−x2r2−x2+x2​=r2−x2r2​.

(M1 differentiate; A1 simplify to $\tfrac{r^2}{r^2-x^2}$r2−x2r2​.) So $\sqrt{1+(y')^2} = \dfrac{r}{\sqrt{r^2-x^2}}$1+(y′)2​=r2−x2​r​, and the $\sqrt{r^2-x^2}$r2−x2​ factors cancel beautifully:

$S = 2\pi\int_{-r}^{r}\sqrt{r^2-x^2}\cdot\frac{r}{\sqrt{r^2-x^2}}\;\mathrm{d}x = 2\pi\int_{-r}^{r} r\;\mathrm{d}x = 2\pi r\big[x\big]_{-r}^{r} = 2\pi r\cdot 2r = 4\pi r^2.$S=2π∫−rr​r2−x2​⋅r2−x2​r​dx=2π∫−rr​rdx=2πr[x]−rr​=2πr⋅2r=4πr2.

(M1 for the cancellation; A1 for $4\pi r^2$4πr2.) A clean derivation of the most famous surface area in mathematics. (Archimedes' insight, of which he was so proud he had it carved on his tomb: a sphere's zones have the same area as the corresponding zones of the enclosing cylinder.)

Worked Example 3 — surface about the $y$y-axis

Find the surface when $y = x^2$y=x2 (for $0\le x\le 1$0≤x≤1) is rotated about the $y$y-axis. The radius is $x$x, $\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x$dxdy​=2x:

$S = 2\pi\int_0^{1} x\,\sqrt{1+4x^2}\;\mathrm{d}x.$S=2π∫01​x1+4x2​dx.

Substitute $u = 1+4x^2,\ \mathrm{d}u = 8x\,\mathrm{d}x$u=1+4x2, du=8xdx; limits $x=0\to u=1,\ x=1\to u=5$x=0→u=1, x=1→u=5:

$S = 2\pi\cdot\frac18\int_1^{5}\sqrt u\;\mathrm{d}u = \frac{\pi}{4}\left[\frac23 u^{3/2}\right]_1^{5} = \frac{\pi}{6}\Big(5^{3/2}-1\Big) = \frac{\pi}{6}\big(5\sqrt5 - 1\big).$S=2π⋅81​∫15​u​du=4π​[32​u3/2]15​=6π​(53/2−1)=6π​(55​−1).

(M1 for the $2\pi x$2πx circumference and substitution; A1 for the exact value $\tfrac{\pi}{6}(5\sqrt5-1)\approx 5.33$6π​(55​−1)≈5.33.) Note the circumference factor is $x$x, not $y$y — the single most common error on this topic.

Worked Example 4 — the catenary

Find the surface when $y=\cosh x$y=coshx (for $0\le x\le a$0≤x≤a) is rotated about the $x$x-axis. From the arc-length lesson $\sqrt{1+\sinh^2 x} = \cosh x$1+sinh2x​=coshx, so

$S = 2\pi\int_0^{a}\cosh x\cdot\cosh x\;\mathrm{d}x = 2\pi\int_0^{a}\cosh^2 x\;\mathrm{d}x.$S=2π∫0a​coshx⋅coshxdx=2π∫0a​cosh2xdx.

Using $\cosh^2 x = \tfrac12(1+\cosh 2x)$cosh2x=21​(1+cosh2x):

$S = 2\pi\int_0^{a}\tfrac12(1+\cosh 2x)\;\mathrm{d}x = \pi\left[x + \tfrac12\sinh 2x\right]_0^{a} = \pi\Big(a + \tfrac12\sinh 2a\Big) = \pi\big(a + \sinh a\cosh a\big),$S=2π∫0a​21​(1+cosh2x)dx=π[x+21​sinh2x]0a​=π(a+21​sinh2a)=π(a+sinhacosha),

the last step using $\sinh 2a = 2\sinh a\cosh a$sinh2a=2sinhacosha. This surface — rotating a catenary — is the catenoid, the minimal surface spanning two coaxial rings.

Worked Example 4b — a paraboloid (the radius–slant cancellation)

Find the surface when $y = \sqrt{2x}$y=2x​ (for $0\le x\le 4$0≤x≤4) is rotated about the $x$x-axis. Here $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{2x}}$dxdy​=2x​1​, so $1+\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 1 + \dfrac{1}{2x} = \dfrac{2x+1}{2x}$1+(dxdy​)2=1+2x1​=2x2x+1​. The radius $y=\sqrt{2x}$y=2x​ cancels the $\sqrt{2x}$2x​ in the denominator:

$y\sqrt{1+(y')^2} = \sqrt{2x}\cdot\frac{\sqrt{2x+1}}{\sqrt{2x}} = \sqrt{2x+1}.$y1+(y′)2​=2x​⋅2x​2x+1​​=2x+1​.

(M1 for the cancellation — the engineered simplification.) Hence

$S = 2\pi\int_0^{4}\sqrt{2x+1}\;\mathrm{d}x = 2\pi\left[\frac13(2x+1)^{3/2}\right]_0^{4} = \frac{2\pi}{3}\Big(9^{3/2} - 1\Big) = \frac{2\pi}{3}(27-1) = \frac{52\pi}{3}.$S=2π∫04​2x+1​dx=2π[31​(2x+1)3/2]04​=32π​(93/2−1)=32π​(27−1)=352π​.

(M1 integrate; A1 for $\tfrac{52\pi}{3}\approx 54.5$352π​≈54.5.) This paraboloid surface is a textbook example of the radius-meets-slant cancellation that makes surface-area integrals tractable, and it is well worth recognising the pattern: whenever $y$y contains a $\sqrt{\;}$​ that reappears in the slant denominator, expect a clean cancellation that turns a fearsome-looking integral into a one-line power integral.

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4 Specimen-style exam question

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

The curve $C$C has equation $y = \sqrt{x}$y=x​ for $0\le x\le 4$0≤x≤4. The region between $C$C and the $x$x-axis is rotated through $2\pi$2π about the $x$x-axis. (a) Show that the surface area generated is $\displaystyle S = 2\pi\int_0^{4}\sqrt{x + \tfrac14}\;\mathrm{d}x$S=2π∫04​x+41​​dx. (b) Hence show that $S = \dfrac{\pi}{6}\big(17\sqrt{17} - 1\big)$S=6π​(1717​−1).

(a) $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2\sqrt x}$dxdy​=2x​1​, so $1+\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 1 + \dfrac{1}{4x} = \dfrac{4x+1}{4x}$1+(dxdy​)2=1+4x1​=4x4x+1​. Then

$y\sqrt{1+(y')^2} = \sqrt{x}\cdot\sqrt{\frac{4x+1}{4x}} = \sqrt{x}\cdot\frac{\sqrt{4x+1}}{2\sqrt x} = \frac12\sqrt{4x+1} = \sqrt{x + \tfrac14},$y1+(y′)2​=x​⋅4x4x+1​​=x​⋅2x​4x+1​​=21​4x+1​=x+41​​,

so $S = 2\pi\displaystyle\int_0^4 \sqrt{x+\tfrac14}\;\mathrm{d}x$S=2π∫04​x+41​​dx, as required. $\;\square$□ (The $\sqrt x$x​ cancels — the engineered simplification.)

(b) $\displaystyle\int_0^4 \big(x+\tfrac14\big)^{1/2}\,\mathrm{d}x = \left[\frac23\big(x+\tfrac14\big)^{3/2}\right]_0^{4} = \frac23\left[\left(\tfrac{17}{4}\right)^{3/2} - \left(\tfrac14\right)^{3/2}\right] = \frac23\cdot\frac{17\sqrt{17} - 1}{8} = \frac{17\sqrt{17}-1}{12}$∫04​(x+41​)1/2dx=[32​(x+41​)3/2]04​=32​[(417​)3/2−(41​)3/2]=32​⋅81717​−1​=121717​−1​.

Hence $S = 2\pi\cdot\dfrac{17\sqrt{17}-1}{12} = \dfrac{\pi}{6}\big(17\sqrt{17} - 1\big) \approx 36.18$S=2π⋅121717​−1​=6π​(1717​−1)≈36.18.

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5 Surface area in parametric form, and the torus

Many surfaces — the torus especially — are described most naturally by a parameter, so the parametric surface-area formula is essential. In parametric form the band radius is still the physical distance to the axis ($y$y for $x$x-axis rotation, $x$x for $y$y-axis rotation), but the element $\mathrm{d}s$ds now uses the parametric speed:

$S = 2\pi\int_{t_1}^{t_2} y\,\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}\;\mathrm{d}t.$S=2π∫t1​t2​​y(dtdx​)2+(dtdy​)2​dt.

Worked Example 5 — the sphere again, parametrically

With $x = r\cos t,\ y = r\sin t,\ 0\le t\le\pi$x=rcost, y=rsint, 0≤t≤π: $\tfrac{\mathrm{d}x}{\mathrm{d}t}=-r\sin t,\ \tfrac{\mathrm{d}y}{\mathrm{d}t}=r\cos t$dtdx​=−rsint, dtdy​=rcost, so $\sqrt{\cdots}=r$⋯​=r. Then

$S = 2\pi\int_0^{\pi} r\sin t\cdot r\;\mathrm{d}t = 2\pi r^2\big[-\cos t\big]_0^{\pi} = 2\pi r^2(1+1) = 4\pi r^2,$S=2π∫0π​rsint⋅rdt=2πr2[−cost]0π​=2πr2(1+1)=4πr2,

matching Worked Example 2 — a satisfying cross-check by a different route.

Worked Example 6 — the torus

A circle of radius $b$b centred at $(a,0)$(a,0) with $a>b$a>b is rotated about the $y$y-axis. Parametrise it as $x = a+b\cos t,\ y = b\sin t,\ 0\le t\le 2\pi$x=a+bcost, y=bsint, 0≤t≤2π. The distance from the surface to the $y$y-axis is $x = a+b\cos t$x=a+bcost, and $\sqrt{(x')^2+(y')^2} = \sqrt{b^2\sin^2 t + b^2\cos^2 t} = b$(x′)2+(y′)2​=b2sin2t+b2cos2t​=b. Thus

$S = 2\pi\int_0^{2\pi}(a+b\cos t)\,b\;\mathrm{d}t = 2\pi b\Big[at + b\sin t\Big]_0^{2\pi} = 2\pi b\cdot 2\pi a = 4\pi^2 ab.$S=2π∫02π​(a+bcost)bdt=2πb[at+bsint]02π​=2πb⋅2πa=4π2ab.

Note $4\pi^2 ab = (2\pi a)(2\pi b)$4π2ab=(2πa)(2πb): the circumference of the central circle times the circumference of the tube — a direct illustration of Pappus's theorem below.

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6 Pappus's centroid theorem (a powerful shortcut)

Pappus (surface). When an arc of length $L$L is rotated about an axis it does not cross, the surface area swept is $S = 2\pi\bar{y}\,L$S=2πyˉ​L, where $\bar y$yˉ​ is the distance from the centroid of the arc to the axis.

This is immediate from $S = \int 2\pi y\,\mathrm{d}s = 2\pi\int y\,\mathrm{d}s = 2\pi\,\bar y\,L$S=∫2πyds=2π∫yds=2πyˉ​L, using the definition $\bar y = \tfrac{1}{L}\int y\,\mathrm{d}s$yˉ​=L1​∫yds. For the torus, the generating circle has $L = 2\pi b$L=2πb and its centroid is at the centre $(a,0)$(a,0), so $\bar y$yˉ​ (distance to the $y$y-axis) $= a$=a, giving $S = 2\pi\cdot a\cdot 2\pi b = 4\pi^2 ab$S=2π⋅a⋅2πb=4π2ab in one line — no integral. Pappus also runs in reverse: if you know $S$S and $L$L, you can locate the centroid.