Home / A-Level Further Maths 9231 / Further integration / Arc length and surface area of revolution
Cambridge International AS & A Level · Topic 2.4
A-Level Further Maths 9231 / Topic 2.4 / AS & A Level

Measure a curve by adding up tiny straight pieces.

Zoom in on a smooth curve and a short piece of it is almost straight, so Pythagoras gives its length \(\delta s\). Add the pieces with an integral to find arc lengths for Cartesian, parametric and polar curves, then spin each piece about an axis to find the area of a surface of revolution.

Topic 2.4All further integration lessons →
The key idea

Pythagoras on a small piece of curve gives \((\delta s)^{2} \approx (\delta x)^{2} + (\delta y)^{2}\). So the arc length is \(s = \displaystyle\int_{a}^{b} \sqrt{1 + \left(\tfrac{dy}{dx}\right)^{2}}\,dx\) (Cartesian), \(s = \displaystyle\int \sqrt{\left(\tfrac{dx}{dt}\right)^{2} + \left(\tfrac{dy}{dt}\right)^{2}}\,dt\) (parametric) or \(s = \displaystyle\int \sqrt{r^{2} + \left(\tfrac{dr}{d\theta}\right)^{2}}\,d\theta\) (polar). Rotating the curve through \(2\pi\) about the \(x\)-axis turns each piece into a band of area \(2\pi y\,\delta s\), so \(S = \displaystyle\int 2\pi y\,ds\); about the \(y\)-axis, \(S = \displaystyle\int 2\pi x\,ds\).

x O y ds y = f(x) 2π radians about Ox arc element ds band it sweeps out radius y, width ds δS ≈ 2πy δs S = ∫ 2πy ds
Fig. 1: the arc y = f(x) rotated through 2π about the x-axis. A short arc element of length δs, at height y, sweeps out a thin band of radius y and slant width δs, so its area is about 2πy δs. Adding the bands gives S = ∫ 2πy ds
Section 01

Pythagoras on a tiny piece.

Take two close points on the curve \(y = f(x)\), separated by \(\delta x\) across and \(\delta y\) up. The arc between them is almost a straight line, so its length is \(\delta s \approx \sqrt{(\delta x)^{2} + (\delta y)^{2}}\). Take \(\delta x\) out of the root, add the pieces from \(x = a\) to \(x = b\) and let \(\delta x \to 0\).

\[\delta s \approx \sqrt{1 + \left(\frac{\delta y}{\delta x}\right)^{2}}\;\delta x\]

so the length of the arc from \(x = a\) to \(x = b\) is

\[s = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}}\,dx.\]

Most curves give an integrand with no simple antiderivative. The ones set in exams are chosen so that \(1 + \left(\frac{dy}{dx}\right)^{2}\) is a perfect square, or a single power of a linear expression. So always expand and simplify \(1 + \left(\frac{dy}{dx}\right)^{2}\) BEFORE you take the root.

Worked example The catenary y = cosh x
Find the exact length of the arc of \(y = \cosh x\) from \(x = 0\) to \(x = \ln 3\).
  1. \(\dfrac{dy}{dx} = \sinh x\), so \(1 + \left(\dfrac{dy}{dx}\right)^{2} = 1 + \sinh^{2} x = \cosh^{2} x\).
  2. \(\cosh x \gt 0\), so the root is \(\cosh x\) and
  3. \[s = \int_{0}^{\ln 3} \cosh x\,dx = \Big[\sinh x\Big]_{0}^{\ln 3} = \sinh(\ln 3)\]
  4. \[\sinh(\ln 3) = \frac{e^{\ln 3} - e^{-\ln 3}}{2} = \frac{3 - \frac{1}{3}}{2}\]
  5. \(= \dfrac{4}{3}\).
Answer: \(s = \dfrac{4}{3}\)
Worked example A power curve
Find the exact length of the curve \(y = x^{3/2}\) from \(x = 0\) to \(x = 4\).
  1. \(\dfrac{dy}{dx} = \tfrac{3}{2}x^{1/2}\), so \(1 + \left(\dfrac{dy}{dx}\right)^{2} = 1 + \tfrac{9}{4}x\).
  2. \[s = \int_{0}^{4} \left(1 + \tfrac{9}{4}x\right)^{1/2} dx\]
  3. Reverse the chain rule: an antiderivative of \(\left(1 + \tfrac{9}{4}x\right)^{1/2}\) is \(\dfrac{\left(1 + \tfrac{9}{4}x\right)^{3/2}}{\tfrac{3}{2} \times \tfrac{9}{4}} = \tfrac{8}{27}\left(1 + \tfrac{9}{4}x\right)^{3/2}\).
  4. \[s = \tfrac{8}{27}\Big[10^{3/2} - 1^{3/2}\Big]\]
  5. \(= \tfrac{8}{27}\big(10\sqrt{10} - 1\big) \approx 9.07\).
Answer: \(s = \tfrac{8}{27}\big(10\sqrt{10} - 1\big)\)
Your turn
Find the length of the arc of \(y = \cosh x\) between \(x = -\ln 2\) and \(x = \ln 2\).
Examiner trap

Do not drop the 1, and do not forget the square root: \(\int \big(1 + (y')^{2}\big)\,dx\) and \(\int y'\,dx\) are both wrong. A quick check: the arc is never shorter than the straight chord between its ends, and never shorter than \(b - a\).

Section 02

Parametric and polar curves.

For a curve \(x = x(t),\ y = y(t)\), divide the Pythagoras result by \((\delta t)^{2}\) instead of \((\delta x)^{2}\). For a polar curve \(r = f(\theta)\), use \(x = r\cos\theta,\ y = r\sin\theta\) with \(\theta\) as the parameter; the cross terms cancel and leave \(r^{2} + \left(\frac{dr}{d\theta}\right)^{2}\).

CurveArc length
\(y = f(x)\)\(\displaystyle\int_{a}^{b} \sqrt{1 + \left(\tfrac{dy}{dx}\right)^{2}}\,dx\)
\(x = x(t),\ y = y(t)\)\(\displaystyle\int_{t_{1}}^{t_{2}} \sqrt{\left(\tfrac{dx}{dt}\right)^{2} + \left(\tfrac{dy}{dt}\right)^{2}}\,dt\)
\(r = f(\theta)\)\(\displaystyle\int_{\alpha}^{\beta} \sqrt{r^{2} + \left(\tfrac{dr}{d\theta}\right)^{2}}\,d\theta\)

Trigonometric curves usually simplify through \(\sin^{2} + \cos^{2} = 1\) and the half-angle identities \(1 - \cos t = 2\sin^{2}\tfrac{t}{2}\) and \(1 + \cos\theta = 2\cos^{2}\tfrac{\theta}{2}\). The square root of a square is a MODULUS: \(\sqrt{\cos^{2} u} = |\cos u|\). Check its sign on your interval, and split the range or use symmetry where it changes sign.

Worked example One arch of a cycloid
A cycloid has parametric equations \(x = t - \sin t,\ y = 1 - \cos t\). Find the length of one arch, \(0 \le t \le 2\pi\).
  1. \(\dfrac{dx}{dt} = 1 - \cos t\) and \(\dfrac{dy}{dt} = \sin t\).
  2. \[(1 - \cos t)^{2} + \sin^{2} t = 2 - 2\cos t\]
  3. \(2 - 2\cos t = 4\sin^{2}\tfrac{t}{2}\), and \(\sin\tfrac{t}{2} \ge 0\) for \(0 \le t \le 2\pi\), so the integrand is \(2\sin\tfrac{t}{2}\).
  4. \[s = \int_{0}^{2\pi} 2\sin\tfrac{t}{2}\,dt = \Big[-4\cos\tfrac{t}{2}\Big]_{0}^{2\pi}\]
  5. \(= -4(-1) - (-4)(1) = 8\).
Answer: \(s = 8\)
Worked example The cardioid
Find the total length of the cardioid \(r = a(1 + \cos\theta)\), where \(a \gt 0\).
  1. \(\dfrac{dr}{d\theta} = -a\sin\theta\), so \(r^{2} + \left(\dfrac{dr}{d\theta}\right)^{2} = a^{2}\big(1 + 2\cos\theta + \cos^{2}\theta + \sin^{2}\theta\big)\).
  2. \[= 2a^{2}(1 + \cos\theta) = 4a^{2}\cos^{2}\tfrac{\theta}{2}\]
  3. The root is \(2a\left|\cos\tfrac{\theta}{2}\right|\). The curve is symmetric about the initial line, so find the upper half, \(0 \le \theta \le \pi\), where \(\cos\tfrac{\theta}{2} \ge 0\), and double it.
  4. \[s = 2\int_{0}^{\pi} 2a\cos\tfrac{\theta}{2}\,d\theta\]
  5. \[= 2\Big[4a\sin\tfrac{\theta}{2}\Big]_{0}^{\pi}\]
  6. \(= 2 \times 4a = 8a\).
Answer: \(s = 8a\)
Your turn
The astroid has parametric equations \(x = \cos^{3} t,\ y = \sin^{3} t\), \(0 \le t \le 2\pi\). Find its total length.
Examiner trap

Integrating \(2a\cos\tfrac{\theta}{2}\) straight from \(0\) to \(2\pi\) gives \(4a\big[\sin\tfrac{\theta}{2}\big]_{0}^{2\pi} = 0\), an impossible length. The root is \(2a\left|\cos\tfrac{\theta}{2}\right|\), which changes form at \(\theta = \pi\). A length of zero, or a negative length, always means a lost modulus.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Which integral gives the length of the curve \(y = f(x)\) from \(x = a\) to \(x = b\)?

Quick check+10 XP

For a polar curve \(r = f(\theta)\), the arc length integrand is

Quick check+10 XP

For \(y = \cosh x\), the expression \(1 + \left(\dfrac{dy}{dx}\right)^{2}\) simplifies to

Quick check+10 XP

A curve \(x = x(t),\ y = y(t)\), with \(y \ge 0\), is rotated through \(2\pi\) radians about the \(x\)-axis. The surface area is

Section 03

Spinning the arc: surface area.

Rotate the arc through \(2\pi\) radians about the \(x\)-axis. Each short piece of length \(\delta s\) at height \(y\) sweeps out a thin band, nearly a slice of a cylinder of radius \(y\) and width \(\delta s\) (Fig. 1). Its area is about \(2\pi y\,\delta s\). Adding the bands gives the area of the curved surface.

The radius is the distance from the AXIS of rotation, so it is \(y\) for the \(x\)-axis and \(x\) for the \(y\)-axis. Surface areas for polar curves are not in the 9231 syllabus.

Worked example A paraboloid
The arc of \(y^{2} = 4x\) with \(y \ge 0\), from \(x = 0\) to \(x = 3\), is rotated through \(2\pi\) radians about the \(x\)-axis. Find the exact area of the surface generated.
  1. \(y = 2x^{1/2}\), so \(\dfrac{dy}{dx} = x^{-1/2}\) and \(1 + \left(\dfrac{dy}{dx}\right)^{2} = 1 + \dfrac{1}{x} = \dfrac{x + 1}{x}\).
  2. \[S = \int_{0}^{3} 2\pi \cdot 2\sqrt{x}\,\sqrt{\frac{x + 1}{x}}\,dx\]
  3. The \(\sqrt{x}\) cancels: \(S = 4\pi\displaystyle\int_{0}^{3} (x + 1)^{1/2}\,dx\).
  4. \[S = 4\pi\Big[\tfrac{2}{3}(x + 1)^{3/2}\Big]_{0}^{3}\]
  5. \(= \tfrac{8\pi}{3}(8 - 1) = \tfrac{56\pi}{3}\).
Answer: \(S = \dfrac{56\pi}{3} \approx 58.6\)
Worked example Spinning the astroid
The astroid \(x = \cos^{3} t,\ y = \sin^{3} t\) is rotated through \(2\pi\) radians about the \(x\)-axis. Find the total surface area generated.
  1. The part with \(y \ge 0\), \(0 \le t \le \pi\), generates the whole surface. By symmetry in the \(y\)-axis, find the surface from \(0 \le t \le \tfrac{\pi}{2}\) and double it.
  2. From the Your turn in the previous section, \(\dfrac{ds}{dt} = 3\sin t\cos t\) on \(0 \le t \le \tfrac{\pi}{2}\).
  3. \[S = 2\int_{0}^{\pi/2} 2\pi\sin^{3} t \cdot 3\sin t\cos t\,dt\]
  4. \[= 12\pi\int_{0}^{\pi/2} \sin^{4} t\cos t\,dt\]
  5. \[= 12\pi\Big[\tfrac{1}{5}\sin^{5} t\Big]_{0}^{\pi/2}\]
  6. \(= \tfrac{12\pi}{5}\).
Answer: \(S = \dfrac{12\pi}{5}\)
Your turn
The arc of \(y = \tfrac{1}{3}x^{3}\) from \(x = 0\) to \(x = 1\) is rotated through \(2\pi\) radians about the \(x\)-axis. Find the exact surface area.
Examiner trap

The surface area is \(\int 2\pi y\,ds\), NOT \(\int 2\pi y\,dx\) (that leaves out the slope of the band) and NOT \(\int \pi y^{2}\,dx\) (that is the volume of revolution). For rotation about the \(y\)-axis the radius is \(x\), but \(ds\) is unchanged.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Further Mathematics Paper 1 to 4 standard.

Finish the four checks above to unlock the exam questions
Exam style+20 XP

Find the exact length of the arc of \(y = \tfrac{2}{3}x^{3/2}\) from \(x = 0\) to \(x = 3\).

Exam style+20 XP

Find the total length of the cardioid \(r = 2(1 + \cos\theta)\).

Exam style+20 XP

The arc of \(y = x^{3}\) from \(x = 0\) to \(x = 1\) is rotated through \(2\pi\) radians about the \(x\)-axis. Find the exact surface area.

Exam style+20 XP

The arc of \(y = x^{2}\) from \((0, 0)\) to \((1, 1)\) is rotated through \(2\pi\) radians about the \(y\)-axis. Find the exact surface area.

Skill unlocked

Curves measured and surfaces spun.

This skill is now recorded on your profile. Keep the chain going.

-Rank -Level -Score -Topics
Go deeper · practice
Six original Cambridge-style questions
Arc lengths of Cartesian, parametric and polar curves that simplify to a perfect square, and surface areas of revolution for the catenary and the cycloid.