| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume using cone or cylinder formula |
| Difficulty | Standard +0.3 This question involves algebraic manipulation to show given results and verification across different cases. While it requires understanding of cross-sectional areas and coordinate relationships, the question explicitly tells students what to show (no discovery needed) and the algebra is straightforward substitution and rearrangement. It's slightly above average due to the multi-part nature and need to work with general formulas, but the scaffolding makes it accessible. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \([\text{Area} =]\ 2xl + \pi x^2\) | M1 | 1.1a |
| \(l = \pi(r-x)\) so Area \(= 2x\pi(r-x) + \pi x^2 = 2\pi rx - \pi x^2 = \pi x(2r-x)\) | A1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(ry + hx = hr\) so \(x = \dfrac{hr - ry}{h}\) and \(\pi x(2r-x) = \pi\left(\dfrac{hr-ry}{h}\right)\left(\dfrac{2rh - hr + ry}{h}\right)\) | M1 | 3.1a |
| \(\pi\left(\dfrac{hr-ry}{h}\right)\left(\dfrac{hr+ry}{h}\right) = \dfrac{\pi}{h^2}(h^2r^2 - r^2y^2) = \dfrac{\pi r^2}{h^2}(h^2 - y^2)\) | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 0\) so area \(= \dfrac{\pi r^2}{h^2}(h^2) = \pi r^2\) | B1 | 3.5a |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = h\) so area \(= \dfrac{\pi r^2}{h^2}(h^2 - h^2) = 0\) | B1 | 3.5a |
## Question 13:
### Part (a)(i):
$[\text{Area} =]\ 2xl + \pi x^2$ | M1 | 1.1a |
$l = \pi(r-x)$ so Area $= 2x\pi(r-x) + \pi x^2 = 2\pi rx - \pi x^2 = \pi x(2r-x)$ | A1 | 2.4 | Convincing completion (answer given); Use of $l = \pi(r-x)$ to get formula in terms of $x$
**[2 marks]**
### Part (a)(ii):
$ry + hx = hr$ so $x = \dfrac{hr - ry}{h}$ and $\pi x(2r-x) = \pi\left(\dfrac{hr-ry}{h}\right)\left(\dfrac{2rh - hr + ry}{h}\right)$ | M1 | 3.1a | Finding $x$ in terms of $y$ correctly and using it |
$\pi\left(\dfrac{hr-ry}{h}\right)\left(\dfrac{hr+ry}{h}\right) = \dfrac{\pi}{h^2}(h^2r^2 - r^2y^2) = \dfrac{\pi r^2}{h^2}(h^2 - y^2)$ | A1 | 2.2a | Convincing completion (answer given) |
**[2 marks]**
### Part (b)(i):
$y = 0$ so area $= \dfrac{\pi r^2}{h^2}(h^2) = \pi r^2$ | B1 | 3.5a |
**[1 mark]**
### Part (b)(ii):
$y = h$ so area $= \dfrac{\pi r^2}{h^2}(h^2 - h^2) = 0$ | B1 | 3.5a |
**[1 mark]**
---13
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the cross-sectional area in Fig. C3.2 is $\pi x ( 2 r - x )$.
\item Hence show that the cross-sectional area is $\frac { \pi r ^ { 2 } } { h ^ { 2 } } \left( h ^ { 2 } - y ^ { 2 } \right)$, as given in line 37 .
\end{enumerate}\item Verify that the formula $\frac { \pi r ^ { 2 } } { h ^ { 2 } } \left( h ^ { 2 } - y ^ { 2 } \right)$ for the cross-sectional area is also valid for
\begin{enumerate}[label=(\roman*)]
\item Fig. C3.1,
\item Fig. C3.3.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2019 Q13 [6]}}