| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Session | Specimen |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Surface area of revolution: parametric curve |
| Difficulty | Challenging +1.8 This is a Further Maths surface area of revolution question requiring parametric form of the formula, which is less commonly practiced than volume questions. Students must recall the correct formula (2π∫y√((dx/dt)²+(dy/dt)²)dt), compute derivatives of t²-2ln(t) and 4t, simplify the radical expression (2t-2/t)²+16 to 2(t+1/t), then integrate 8πt(t+1/t) over [1,4]. While systematic, it requires confident handling of parametric formulas and algebraic manipulation beyond standard A-level. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks |
|---|---|
| 1 | 2 2 2 |
| Answer | Marks |
|---|---|
| (cid:32)384(cid:83) so k = 384 | M1 |
| Answer | Marks |
|---|---|
| [4] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | Correct unsimplified |
| Answer | Marks |
|---|---|
| BC | May be seen in their expression for |
Question 1:
1 | 2 2 2
(cid:167)dx(cid:183) (cid:167)dy(cid:183) (cid:167) 2(cid:183)
(cid:14) (cid:32) 2t(cid:16) (cid:14)16
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
(cid:169)dt (cid:185) (cid:169)dt (cid:185) (cid:169) t (cid:185)
4
(cid:180) (cid:167) 2(cid:183) 2
A(cid:32)2(cid:83)(cid:181) 4t 2t(cid:16) (cid:14)16dt
(cid:181) (cid:168) (cid:184)
(cid:169) t (cid:185)
(cid:182)
1
(cid:32)384(cid:83) so k = 384 | M1
A1
M1
A1
[4] | 1.1a
1.1
1.2
1.1 | Correct unsimplified
Use of Surface Area formula
BC | May be seen in their expression for
A.
Must be seen1 A curve is given by $x = t ^ { 2 } - 2 \ln t , y = 4 t$ for $t > 0$. When the arc of the curve between the points where $t = 1$ and $t = 4$ is rotated through $2 \pi$ radians about the $x$-axis, a surface of revolution is formed with surface area $A$.\\
Given that $A = k \pi$, where $k$ is an integer,
\begin{itemize}
\item write down an integral which gives $A$ and
\item find the value of $k$.
\end{itemize}
\hfill \mbox{\textit{OCR Further Additional Pure Q1 [4]}}