| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Session | Specimen |
| Marks | 5 |
| Topic | Volumes of Revolution |
| Type | Surface area of revolution: Cartesian curve |
| Difficulty | Challenging +1.2 This is a standard surface area of revolution question requiring application of memorized formulas for arc length and surface area. Part (i) is routine differentiation and substitution into the arc length formula. Part (ii) requires the surface area formula S = 2π∫y√(1+(dy/dx)²)dx and algebraic manipulation to express the result in terms of s, but the steps are mechanical once the formulas are recalled. It's above average difficulty due to the algebraic manipulation required and being a Further Maths topic, but doesn't require novel insight. |
| Spec | 4.08f Integrate using partial fractions |
**(i)** $\frac{dy}{dx} = x^2$ and range of $x$ is $[0, 1]$ $\Rightarrow s = \int_0^1 \sqrt{1+x^4}\, dx$ (AG) M1A1 **[2]**
**(ii)** $S = 2\pi\int_0^1 \left(\frac{1}{3}x^3 + 1\right)\left(1+x^4\right)^{\frac{1}{2}} dx$ (AEF) M1A1
$S = 2\pi s + \frac{2}{3}\pi\int_0^1 x^3\left(1+x^4\right)^{\frac{1}{2}} dx$ A1
$\int_0^1 x^3\left(1+x^4\right)^{\frac{1}{2}} dx = \frac{1}{6}\left(1+x^4\right)^{\frac{3}{2}}$ B1
$S = \frac{1}{9}\pi(18s + 2\sqrt{2} - 1)$ (AG; CWO) A1 **[5]**3 A curve has equation
$$y = \frac { 1 } { 3 } x ^ { 3 } + 1$$
The length of the arc of the curve joining the point where $x = 0$ to the point where $x = 1$ is denoted by $s$.\\
(i) Show that
$$s = \int _ { 0 } ^ { 1 } \sqrt { 1 + x ^ { 4 } } \mathrm {~d} x$$
The surface area generated when this arc is rotated through one complete revolution about the $x$-axis is denoted by $S$.\\
(ii) Show that
$$S = \frac { 1 } { 9 } \pi ( 18 s + 2 \sqrt { 2 } - 1 )$$
[Do not attempt to evaluate $s$ or $S$.]
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 Q3 [5]}}