| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Parametric surface area of revolution |
| Difficulty | Challenging +1.8 This is a Further Maths parametric surface area question requiring the formula S = 2π∫y√((dx/dθ)² + (dy/dθ)²)dθ, followed by a non-trivial trigonometric simplification using half-angle identities (1-cos θ = 2sin²(θ/2)) to evaluate the integral. Part (a) tests careful algebraic manipulation; part (b) requires recognizing the substitution pattern. Significantly harder than standard C3/C4 parametric work but routine for F3 students who know the techniques. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.08d Evaluate definite integrals: between limits8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dx}{d\theta}=1-\cos\theta,\quad \frac{dy}{d\theta}=\sin\theta\) | B1 | Both |
| \((S=2\pi)\int(1-\cos\theta)\sqrt{1-2\cos\theta+\cos^2\theta+\sin^2\theta}\,d\theta\) or \(\int(1-\cos\theta)\sqrt{(1-\cos\theta)^2+\sin^2\theta}\,d\theta\) | M1A1 | M1: Uses correct formula with their derivatives; integrand must be function of \(\theta\); A1: Correct integrand |
| \((S=2\pi)\int(1-\cos\theta)\sqrt{2(1-\cos\theta)}\,d\theta\) | ||
| \(S=2\pi\sqrt{2}\int_0^{2\pi}(1-\cos\theta)^{\frac{3}{2}}\,d\theta^*\) | A1*cso | cso with at least one intermediate step shown; Total: (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S=2\pi\sqrt{2}\int\left(1-\left(1-2\sin^2\frac{\theta}{2}\right)\right)^{\frac{3}{2}}d\theta\) | M1 | Uses \(\cos\theta = \pm1\pm2\sin^2\frac{\theta}{2}\) |
| \(=8\pi\int\sin^3\frac{\theta}{2}\,d\theta\) | A1 | Correct expression |
| \(=8\pi\int\sin\frac{\theta}{2}\left(1-\cos^2\frac{\theta}{2}\right)d\theta\) | dM1 | Uses Pythagoras |
| \(=8\pi\left[-2\cos\frac{\theta}{2}+\frac{2}{3}\cos^3\frac{\theta}{2}\right]_0^{2\pi}\) | ddM1 | Integrates to obtain \(p\cos\frac{\theta}{2}+q\cos^3\frac{\theta}{2},\; p,q\neq 0\) |
| \(=16\pi\left[\left(-\frac{1}{3}+1\right)-\left(\frac{1}{3}-1\right)\right]\) | dddM1 | Include \(16\pi\) and use limits correctly |
| \(=\frac{64\pi}{3}\) | A1 | Total: (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S=2\pi\sqrt{2}\int\left(1-\left(1-2\sin^2\frac{\theta}{2}\right)\right)^{\frac{3}{2}}d\theta\) | M1 | Uses \(\cos\theta=\pm1\pm2\sin^2\frac{\theta}{2}\) |
| \(=8\pi\int\sin^3\frac{\theta}{2}\,d\theta\) | A1 | Correct expression |
| \(\sin^3\frac{\theta}{2}=\frac{3}{4}\sin\frac{\theta}{2}-\frac{1}{4}\sin\frac{3\theta}{2}\) | ||
| \(8\pi\int\left(\frac{3}{4}\sin\frac{\theta}{2}-\frac{1}{4}\sin\frac{3\theta}{2}\right)d\theta\) | dM1 | Uses the above identity (sign errors allowed) |
| \(=8\pi\left(-\frac{3}{4}\times2\cos\frac{\theta}{2}+\frac{1}{4}\times\frac{2}{3}\cos\frac{3\theta}{2}\right)\) | ddM1 | Integrates to obtain \(p\cos\frac{\theta}{2}+q\cos\frac{3\theta}{2}\) |
| \(8\pi\left[\left(\frac{3}{2}-\frac{1}{6}\right)-\left(\frac{3}{2}+\frac{1}{6}\right)\right]=\frac{64\pi}{3}\) | dddM1 A1 | Correct use of limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u=1-\cos\theta;\; du=\sin\theta\,d\theta\) | ||
| \(\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-(1-u)^2}=u^{\frac{1}{2}}(2-u)^{\frac{1}{2}}\) | ||
| \(\int(1-\cos\theta)^{\frac{3}{2}}d\theta = \int u(2-u)^{-\frac{1}{2}}du\) | M1A1 | M1: Attempt the substitution, integral in terms of \(u\) only; A1: Correct integral in terms of \(u\) |
| \(=-2u(2-u)^{\frac{1}{2}}-\int -2(2-u)^{\frac{1}{2}}\times 1\,du\) | ||
| \(=-2u(2-u)^{\frac{1}{2}}-2\times\frac{2}{3}(2-u)^{\frac{3}{2}}\) | dM1 | Integrate by parts |
| \(2\times2\pi\sqrt{2}\int_0^{\pi}(1-\cos\theta)^{\frac{3}{2}}d\theta\) | ||
| \(2\times2\pi\sqrt{2}\int_0^2 u(2-u)^{-\frac{1}{2}}du = 2\times2\pi\sqrt{2}\times2\times\frac{2}{3}\times2^{\frac{3}{2}}\) | ddM1 ddM1 | Include constant multiplier; EITHER change limits for \(u\) OR reverse substitution and substitute limits for \(\theta\) |
| \(=\frac{64\pi}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S=8\pi\int\sin^3\frac{\theta}{2}\,d\theta\) | M1A1 | See main scheme |
| \(u=\sin^2\frac{\theta}{2},\quad \frac{dv}{d\theta}=\sin\frac{\theta}{2}\) | ||
| \(\frac{du}{d\theta}=\sin\frac{\theta}{2}\cos\frac{\theta}{2},\quad v=-2\cos\frac{\theta}{2}\) | ||
| \(=-2\sin^2\frac{\theta}{2}\cos\frac{\theta}{2}+\int 2\sin\frac{\theta}{2}\cos^2\frac{\theta}{2}\,d\theta\) | dM1 | Integrate by parts |
| \(\int 2\sin\frac{\theta}{2}\cos^2\frac{\theta}{2}\,d\theta \to k\cos^3\frac{\theta}{2}\) | ddM1 | |
| \(=\frac{64\pi}{3}\) | dddM1A1 | M1: Include constant multiplier and use limits (0 and \(2\pi\)) correctly; A1: Correct answer |
## Question 8:
**Given:** $x=\theta-\sin\theta,\quad y=1-\cos\theta,\quad 0\leq\theta\leq 2\pi$
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dx}{d\theta}=1-\cos\theta,\quad \frac{dy}{d\theta}=\sin\theta$ | B1 | Both |
| $(S=2\pi)\int(1-\cos\theta)\sqrt{1-2\cos\theta+\cos^2\theta+\sin^2\theta}\,d\theta$ or $\int(1-\cos\theta)\sqrt{(1-\cos\theta)^2+\sin^2\theta}\,d\theta$ | M1A1 | M1: Uses correct formula with their derivatives; integrand must be function of $\theta$; A1: Correct integrand |
| $(S=2\pi)\int(1-\cos\theta)\sqrt{2(1-\cos\theta)}\,d\theta$ | | |
| $S=2\pi\sqrt{2}\int_0^{2\pi}(1-\cos\theta)^{\frac{3}{2}}\,d\theta^*$ | A1*cso | cso with at least one intermediate step shown; Total: (4) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S=2\pi\sqrt{2}\int\left(1-\left(1-2\sin^2\frac{\theta}{2}\right)\right)^{\frac{3}{2}}d\theta$ | M1 | Uses $\cos\theta = \pm1\pm2\sin^2\frac{\theta}{2}$ |
| $=8\pi\int\sin^3\frac{\theta}{2}\,d\theta$ | A1 | Correct expression |
| $=8\pi\int\sin\frac{\theta}{2}\left(1-\cos^2\frac{\theta}{2}\right)d\theta$ | dM1 | Uses Pythagoras |
| $=8\pi\left[-2\cos\frac{\theta}{2}+\frac{2}{3}\cos^3\frac{\theta}{2}\right]_0^{2\pi}$ | ddM1 | Integrates to obtain $p\cos\frac{\theta}{2}+q\cos^3\frac{\theta}{2},\; p,q\neq 0$ |
| $=16\pi\left[\left(-\frac{1}{3}+1\right)-\left(\frac{1}{3}-1\right)\right]$ | dddM1 | Include $16\pi$ and use limits correctly |
| $=\frac{64\pi}{3}$ | A1 | Total: (6) |
**Alternative 1 for (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S=2\pi\sqrt{2}\int\left(1-\left(1-2\sin^2\frac{\theta}{2}\right)\right)^{\frac{3}{2}}d\theta$ | M1 | Uses $\cos\theta=\pm1\pm2\sin^2\frac{\theta}{2}$ |
| $=8\pi\int\sin^3\frac{\theta}{2}\,d\theta$ | A1 | Correct expression |
| $\sin^3\frac{\theta}{2}=\frac{3}{4}\sin\frac{\theta}{2}-\frac{1}{4}\sin\frac{3\theta}{2}$ | | |
| $8\pi\int\left(\frac{3}{4}\sin\frac{\theta}{2}-\frac{1}{4}\sin\frac{3\theta}{2}\right)d\theta$ | dM1 | Uses the above identity (sign errors allowed) |
| $=8\pi\left(-\frac{3}{4}\times2\cos\frac{\theta}{2}+\frac{1}{4}\times\frac{2}{3}\cos\frac{3\theta}{2}\right)$ | ddM1 | Integrates to obtain $p\cos\frac{\theta}{2}+q\cos\frac{3\theta}{2}$ |
| $8\pi\left[\left(\frac{3}{2}-\frac{1}{6}\right)-\left(\frac{3}{2}+\frac{1}{6}\right)\right]=\frac{64\pi}{3}$ | dddM1 A1 | Correct use of limits |
**Alternative 2 for (b) — substitution $u=1-\cos\theta$:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u=1-\cos\theta;\; du=\sin\theta\,d\theta$ | | |
| $\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-(1-u)^2}=u^{\frac{1}{2}}(2-u)^{\frac{1}{2}}$ | | |
| $\int(1-\cos\theta)^{\frac{3}{2}}d\theta = \int u(2-u)^{-\frac{1}{2}}du$ | M1A1 | M1: Attempt the substitution, integral in terms of $u$ only; A1: Correct integral in terms of $u$ |
| $=-2u(2-u)^{\frac{1}{2}}-\int -2(2-u)^{\frac{1}{2}}\times 1\,du$ | | |
| $=-2u(2-u)^{\frac{1}{2}}-2\times\frac{2}{3}(2-u)^{\frac{3}{2}}$ | dM1 | Integrate by parts |
| $2\times2\pi\sqrt{2}\int_0^{\pi}(1-\cos\theta)^{\frac{3}{2}}d\theta$ | | |
| $2\times2\pi\sqrt{2}\int_0^2 u(2-u)^{-\frac{1}{2}}du = 2\times2\pi\sqrt{2}\times2\times\frac{2}{3}\times2^{\frac{3}{2}}$ | ddM1 ddM1 | Include constant multiplier; EITHER change limits for $u$ OR reverse substitution and substitute limits for $\theta$ |
| $=\frac{64\pi}{3}$ | A1 | |
**Alternative 3 for (b) — integration by parts:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S=8\pi\int\sin^3\frac{\theta}{2}\,d\theta$ | M1A1 | See main scheme |
| $u=\sin^2\frac{\theta}{2},\quad \frac{dv}{d\theta}=\sin\frac{\theta}{2}$ | | |
| $\frac{du}{d\theta}=\sin\frac{\theta}{2}\cos\frac{\theta}{2},\quad v=-2\cos\frac{\theta}{2}$ | | |
| $=-2\sin^2\frac{\theta}{2}\cos\frac{\theta}{2}+\int 2\sin\frac{\theta}{2}\cos^2\frac{\theta}{2}\,d\theta$ | dM1 | Integrate by parts |
| $\int 2\sin\frac{\theta}{2}\cos^2\frac{\theta}{2}\,d\theta \to k\cos^3\frac{\theta}{2}$ | ddM1 | |
| $=\frac{64\pi}{3}$ | dddM1A1 | M1: Include constant multiplier and use limits (0 and $2\pi$) correctly; A1: Correct answer |
---8. The curve $C$ has parametric equations
$$x = \theta - \sin \theta , \quad y = 1 - \cos \theta , \quad 0 \leqslant \theta \leqslant 2 \pi$$
The curve $C$ is rotated through $2 \pi$ radians about the $x$-axis. The area of the curved surface generated is given by $S$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$S = 2 \pi \sqrt { 2 } \int _ { 0 } ^ { 2 \pi } ( 1 - \cos \theta ) ^ { \frac { 3 } { 2 } } \mathrm {~d} \theta$$
\item Hence find the exact value of $S$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2017 Q8 [10]}}