| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Topic | Volumes of Revolution |
| Type | Surface area of revolution: parametric curve |
| Difficulty | Challenging +1.8 This question requires knowledge of the surface area of revolution formula for parametric curves, which is a specialized topic beyond standard A-level. Students must correctly apply S = ∫2πy√((dx/dt)² + (dy/dt)²)dt, differentiate parametric equations, simplify the resulting expression involving t and 1/t, and integrate over the given bounds to find an exact answer. While the integration itself is manageable once set up, the parametric surface area formula is less commonly encountered than volume of revolution, and the multi-step process with exact answer requirement elevates this above typical A-level questions. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
**Question 4**
$\dot{x} = t - \dfrac{1}{t}$ and $\dot{y} = 2$ **B1** at least $\dot{x}$ correct
$(\dot{x})^2 + (\dot{y})^2 = t^2 - 2 + \dfrac{1}{t^2} + 4$ **M1** attempted
$= \left(t + \dfrac{1}{t}\right)^2$ **A1** Here or in the integral for $S$ (2nd fraction of line below)
$S = 2\pi\displaystyle\int_1^4 2t\cdot\left(t+\dfrac{1}{t}\right)\text{d}t$ **M1** Use of formula (Ignore limits until final answer)
$= 4\pi\displaystyle\int_1^4 (t^2+1)\,\text{d}t$ **A1** In a form ready to integrate
$= 4\pi\!\left[\dfrac{t^3}{3}+t\right]_1^4$ **B1** Correct integration (FT provided it is polynomial)
$= 96\pi$ **A1**
**Total: 7 marks**4 The curve $C$ has parametric equations $x = \frac { 1 } { 2 } t ^ { 2 } - \ln t , y = 2 t$, for $1 \leqslant t \leqslant 4$. When $C$ is rotated through $2 \pi$ radians about the $x$-axis, a surface of revolution is formed of surface area $S$. Determine the exact value of $S$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2017 Q4 [7]}}