| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| 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 F3 question requiring the surface area of revolution formula for parametric equations, followed by a substitution to evaluate a non-standard integral. Part (a) tests formula application with parametric derivatives; part (b) requires careful algebraic manipulation after substitution, including expressing t² in terms of u. While systematic, it demands technical proficiency beyond standard A-level and multiple precise steps, making it moderately challenging even for Further Maths students. |
| Spec | 1.08h Integration by substitution4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = 12t^3\), \(\frac{dy}{dt} = 12t^2\) | B1 | Correct derivatives |
| \(S = (2\pi)\int y\left(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right)^{\frac{1}{2}} dt = (2\pi)\int 4t^3\sqrt{(12t^3)^2 + (12t^2)^2}\, dt\) | M1 | Substitutes their derivatives into a correct formula (\(2\pi\) not required) |
| \(S = (2\pi)\int 4t^3\left(144t^4\right)^{\frac{1}{2}}\left(t^2+1\right)^{\frac{1}{2}} dt\) | M1 | Attempt to factor out at least \(t^4\) — numerical factor may be left |
| \(S = 96\pi\int_0^1 t^5\left(t^2+1\right)^{\frac{1}{2}} dt\) | A1 | Correct completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u^2 = t^2 + 1 \Rightarrow 2u\frac{du}{dt} = 2t\) or \(2u = 2t\frac{dt}{du}\) | B1 | Correct differentiation |
| \(t=0 \Rightarrow u=1\), \(t=1 \Rightarrow u=\sqrt{2}\) | B1 | Correct limits. ALT: reverse the substitution later (treat as M1, award later when work seen) |
| \(S = (96\pi)\int t^5 \times u \times \frac{u}{t}\, du\) | — | — |
| \(S = (96\pi)\int \left(u^2-1\right)^2 \times u^2\, du\) | M1A1 | M1: Complete substitution. A1: Correct integral in terms of \(u\), ignore limits, need not be simplified |
| \(S = (96\pi)\int\left(u^6 - 2u^4 + u^2\right)du = (96\pi)\left[\frac{u^7}{7} - \frac{2u^5}{5} + \frac{u^3}{3}\right]\) | dM1 | Expands and attempts to integrate |
| \(S = 96\pi\left[\frac{u^7}{7} - \frac{2u^5}{5} + \frac{u^3}{3}\right]_1^{\sqrt{2}} = 96\pi\left\{\left(\frac{\sqrt{2}^7}{7} - \frac{2\sqrt{2}^5}{5} + \frac{\sqrt{2}^3}{3}\right) - \left(\frac{1}{7} - \frac{2}{5} + \frac{1}{3}\right)\right\}\) | ddM1 | Correct use of their changed limits (both to be changed). ALT: if sub reversed, substitute the original limits |
| \(S = \frac{192\pi}{105}\left(11\sqrt{2}-4\right)\) | A1 | cao e.g. \(\frac{64\pi}{35}\) |
## Question 7:
**Given:** $x = 3t^4$, $y = 4t^3$
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 12t^3$, $\frac{dy}{dt} = 12t^2$ | B1 | Correct derivatives |
| $S = (2\pi)\int y\left(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right)^{\frac{1}{2}} dt = (2\pi)\int 4t^3\sqrt{(12t^3)^2 + (12t^2)^2}\, dt$ | M1 | Substitutes their derivatives into a correct formula ($2\pi$ not required) |
| $S = (2\pi)\int 4t^3\left(144t^4\right)^{\frac{1}{2}}\left(t^2+1\right)^{\frac{1}{2}} dt$ | M1 | Attempt to factor out at least $t^4$ — numerical factor may be left |
| $S = 96\pi\int_0^1 t^5\left(t^2+1\right)^{\frac{1}{2}} dt$ | A1 | Correct completion |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u^2 = t^2 + 1 \Rightarrow 2u\frac{du}{dt} = 2t$ or $2u = 2t\frac{dt}{du}$ | B1 | Correct differentiation |
| $t=0 \Rightarrow u=1$, $t=1 \Rightarrow u=\sqrt{2}$ | B1 | Correct limits. ALT: reverse the substitution later (treat as M1, award later when work seen) |
| $S = (96\pi)\int t^5 \times u \times \frac{u}{t}\, du$ | — | — |
| $S = (96\pi)\int \left(u^2-1\right)^2 \times u^2\, du$ | M1A1 | M1: Complete substitution. A1: Correct integral in terms of $u$, ignore limits, need not be simplified |
| $S = (96\pi)\int\left(u^6 - 2u^4 + u^2\right)du = (96\pi)\left[\frac{u^7}{7} - \frac{2u^5}{5} + \frac{u^3}{3}\right]$ | dM1 | Expands and attempts to integrate |
| $S = 96\pi\left[\frac{u^7}{7} - \frac{2u^5}{5} + \frac{u^3}{3}\right]_1^{\sqrt{2}} = 96\pi\left\{\left(\frac{\sqrt{2}^7}{7} - \frac{2\sqrt{2}^5}{5} + \frac{\sqrt{2}^3}{3}\right) - \left(\frac{1}{7} - \frac{2}{5} + \frac{1}{3}\right)\right\}$ | ddM1 | Correct use of their changed limits (both to be changed). ALT: if sub reversed, substitute the original limits |
| $S = \frac{192\pi}{105}\left(11\sqrt{2}-4\right)$ | A1 | cao e.g. $\frac{64\pi}{35}$ |
---7. The curve $C$ has parametric equations
$$x = 3 t ^ { 4 } , \quad y = 4 t ^ { 3 } , \quad 0 \leqslant t \leqslant 1$$
The curve $C$ is rotated through $2 \pi$ radians about the $x$-axis. The area of the curved surface generated is $S$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$S = k \pi \int _ { 0 } ^ { 1 } t ^ { 5 } \left( t ^ { 2 } + 1 \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} t$$
where $k$ is a constant to be found.
\item Use the substitution $u ^ { 2 } = t ^ { 2 } + 1$ to find the value of $S$, giving your answer in the form $p \pi ( 11 \sqrt { 2 } - 4 )$ where $p$ is a rational number to be found.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2016 Q7 [11]}}