AS & A Level · Practice questions · Arc length and surface area of revolution

Lengths and areas, exactly.

Six original Cambridge-style questions on arc length and surface area of revolution: perfect-square integrands, a parametric curve, a polar spiral, the catenary spun about the \(x\)-axis and the surface of a cycloid arch.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Differentiate, simplify, root, integrate.

Pick the formula for the form of the curve: \(1 + \left(\tfrac{dy}{dx}\right)^{2}\), \(\left(\tfrac{dx}{dt}\right)^{2} + \left(\tfrac{dy}{dt}\right)^{2}\) or \(r^{2} + \left(\tfrac{dr}{d\theta}\right)^{2}\).
Simplify before the root: look for a perfect square, \(\sin^{2} + \cos^{2} = 1\), \(\cosh^{2} - \sinh^{2} = 1\) or a half-angle identity.
Root with care: \(\sqrt{u^{2}} = |u|\). Check the sign on the interval; use symmetry or split the range.
Surface area: multiply \(ds\) by \(2\pi y\) for the \(x\)-axis or \(2\pi x\) for the \(y\)-axis, never by \(\pi y^{2}\).
Sense check: a length is at least the chord between the end points, and an area is positive.

Recall

[4 marks]

Show that the length of the arc of \(y = \tfrac{1}{3}(x^{2} + 2)^{3/2}\) from \(x = 0\) to \(x = 3\) is \(12\).

\(\dfrac{dy}{dx} = \tfrac{1}{3} \cdot \tfrac{3}{2}(x^{2} + 2)^{1/2} \cdot 2x = x(x^{2} + 2)^{1/2}\). ✓
\(1 + \left(\dfrac{dy}{dx}\right)^{2} = 1 + x^{2}(x^{2} + 2) = x^{4} + 2x^{2} + 1 = (x^{2} + 1)^{2}\). ✓
\(x^{2} + 1 \gt 0\), so \(s = \displaystyle\int_{0}^{3} (x^{2} + 1)\,dx = \Big[\tfrac{1}{3}x^{3} + x\Big]_{0}^{3}\). ✓
\(= 9 + 3 = 12\). ✓

Calculation

[5 marks]

Find the exact length of the curve \(y = \tfrac{1}{8}x^{2} - \ln x\) from \(x = 1\) to \(x = 4\).

\(\dfrac{dy}{dx} = \tfrac{1}{4}x - \dfrac{1}{x}\), so \(\left(\dfrac{dy}{dx}\right)^{2} = \tfrac{1}{16}x^{2} - \tfrac{1}{2} + \dfrac{1}{x^{2}}\). ✓
\(1 + \left(\dfrac{dy}{dx}\right)^{2} = \tfrac{1}{16}x^{2} + \tfrac{1}{2} + \dfrac{1}{x^{2}} = \left(\tfrac{1}{4}x + \dfrac{1}{x}\right)^{2}\). ✓
\(\tfrac{1}{4}x + \tfrac{1}{x} \gt 0\) for \(x \gt 0\), so ✓
\[s = \int_{1}^{4} \left(\tfrac{1}{4}x + \frac{1}{x}\right)dx = \Big[\tfrac{1}{8}x^{2} + \ln x\Big]_{1}^{4}\] ✓
\(= \left(2 + \ln 4\right) - \tfrac{1}{8} = \tfrac{15}{8} + 2\ln 2\). ✓

Calculation

[5 marks]

A curve has parametric equations \(x = t^{2},\ y = \tfrac{2}{3}t^{3}\), for \(t \ge 0\). Find the length of the arc from \(t = 0\) to \(t = 2\sqrt{2}\).

\(\dfrac{dx}{dt} = 2t\) and \(\dfrac{dy}{dt} = 2t^{2}\). ✓
\(\left(\dfrac{dx}{dt}\right)^{2} + \left(\dfrac{dy}{dt}\right)^{2} = 4t^{2} + 4t^{4} = 4t^{2}(1 + t^{2})\), with root \(2t\sqrt{1 + t^{2}}\) since \(t \ge 0\). ✓
\[s = \int_{0}^{2\sqrt{2}} 2t(1 + t^{2})^{1/2}\,dt\] ✓
\[= \Big[\tfrac{2}{3}(1 + t^{2})^{3/2}\Big]_{0}^{2\sqrt{2}}\] ✓
\(= \tfrac{2}{3}\big(9^{3/2} - 1\big) = \tfrac{2}{3}(27 - 1) = \tfrac{52}{3}\). ✓

Calculation

[4 marks]

A spiral has polar equation \(r = \theta^{2}\). Find the length of the spiral from \(\theta = 0\) to \(\theta = \sqrt{5}\).

\(\dfrac{dr}{d\theta} = 2\theta\), so \(r^{2} + \left(\dfrac{dr}{d\theta}\right)^{2} = \theta^{4} + 4\theta^{2} = \theta^{2}(\theta^{2} + 4)\). ✓
For \(\theta \ge 0\) the root is \(\theta\sqrt{\theta^{2} + 4}\). ✓
\[s = \int_{0}^{\sqrt{5}} \theta(\theta^{2} + 4)^{1/2}\,d\theta\] ✓
\[= \Big[\tfrac{1}{3}(\theta^{2} + 4)^{3/2}\Big]_{0}^{\sqrt{5}}\] ✓
\(= \tfrac{1}{3}(27 - 8) = \tfrac{19}{3}\). ✓

Analysis

[5 marks]

The arc of the catenary \(y = \cosh x\) from \(x = 0\) to \(x = \ln 2\) is rotated through \(2\pi\) radians about the \(x\)-axis.

Show that the area of the surface generated is \(\pi\left(\ln 2 + \dfrac{15}{16}\right)\).

\(1 + \sinh^{2} x = \cosh^{2} x\), so \(ds = \cosh x\,dx\) and \(S = \displaystyle\int_{0}^{\ln 2} 2\pi\cosh^{2} x\,dx\). ✓
Use \(\cosh^{2} x = \tfrac{1}{2}(1 + \cosh 2x)\): ✓
\[S = \pi\int_{0}^{\ln 2} (1 + \cosh 2x)\,dx = \pi\Big[x + \tfrac{1}{2}\sinh 2x\Big]_{0}^{\ln 2}\] ✓
\[\sinh(2\ln 2) = \sinh(\ln 4) = \frac{4 - \frac{1}{4}}{2} = \frac{15}{8}\] ✓
So \(S = \pi\left(\ln 2 + \tfrac{1}{2} \times \tfrac{15}{8}\right) = \pi\left(\ln 2 + \tfrac{15}{16}\right)\). ✓

Analysis

[7 marks]

One arch of the cycloid \(x = t - \sin t,\ y = 1 - \cos t\), \(0 \le t \le 2\pi\), is rotated through \(2\pi\) radians about the \(x\)-axis.

(a) Show that \(\dfrac{ds}{dt} = 2\sin\tfrac{1}{2}t\) for \(0 \le t \le 2\pi\).

(b) Hence find the exact area of the surface generated.

(a) \(\left(\dfrac{dx}{dt}\right)^{2} + \left(\dfrac{dy}{dt}\right)^{2} = (1 - \cos t)^{2} + \sin^{2} t = 2 - 2\cos t = 4\sin^{2}\tfrac{1}{2}t\). ✓
\(\sin\tfrac{1}{2}t \ge 0\) for \(0 \le t \le 2\pi\), so \(\dfrac{ds}{dt} = 2\sin\tfrac{1}{2}t\). ✓
(b) \(y = 1 - \cos t = 2\sin^{2}\tfrac{1}{2}t\), so ✓
\[S = \int_{0}^{2\pi} 2\pi \cdot 2\sin^{2}\tfrac{1}{2}t \cdot 2\sin\tfrac{1}{2}t\,dt\] ✓
\[= 8\pi\int_{0}^{2\pi} \sin^{3}\tfrac{1}{2}t\,dt\] ✓
Let \(u = \tfrac{1}{2}t\), \(dt = 2\,du\): \(S = 16\pi\displaystyle\int_{0}^{\pi} \sin^{3} u\,du\). ✓
\(\sin^{3} u = \sin u\,(1 - \cos^{2} u)\), so ✓
\[\int_{0}^{\pi} \sin^{3} u\,du = \Big[-\cos u + \tfrac{1}{3}\cos^{3} u\Big]_{0}^{\pi}\] ✓
\(= \left(1 - \tfrac{1}{3}\right) - \left(-1 + \tfrac{1}{3}\right) = \tfrac{4}{3}\). ✓
\(S = 16\pi \times \tfrac{4}{3} = \tfrac{64\pi}{3} \approx 67.0\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.