| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Rotation about y-axis, standard curve |
| Difficulty | Standard +0.3 This is a structured three-part question where part (i) provides the integration result needed for part (ii). The volume of revolution about the y-axis uses the standard formula V = π∫x²dy, requiring rearrangement of y = ½e^(x²) to x² = ln(2y), then direct integration using the given result. Part (iii) is a simple subtraction of volumes. While it involves multiple steps, each component is routine and the question scaffolds the solution clearly, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt use of product rule to produce the form \(\ln 2y + y \times \frac{a}{by}\) | M1 | Note that product rule may be applied to expression in form \(y(\ln 2y - 1)\) |
| Obtain correct \(\ln 2y + y \times \frac{2}{2y}\) | A1 | or equiv |
| Obtain complete \(\ln 2y + 1 - 1\) and confirm | A1 | AG; necessary detail needed |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt to rearrange equation to \(x = \ldots\) or \(x^2 = \ldots\) | M1 | obtaining form \(p\ln qy\) |
| Obtain \(x = \sqrt{\ln 2y}\) or \(x^2 = \ln 2y\) | A1 | |
| State or imply volume is \(\int \pi \ln 2y \, dy\) | A1ft | following their \(x = \ldots\) or \(x^2 = \ldots\); condone absence of \(dy\); condone presence of \(dx\); no need for limits here; \(\pi\) may be implied by its first appearance later in solution |
| Integrate using result of part (i) | M1 | |
| Attempt to use limits \(\frac{1}{2}\) and \(\frac{1}{2}e^4\) correctly with expression involving \(y\) | M1 | |
| Obtain \(\frac{1}{2}\pi(3e^4 + 1)\) | A1 | or equiv involving two terms; dependent on correct work throughout part (ii) |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Subtract answer to part (ii) from \(2\pi e^4 \ldots\) | M1 | \(\ldots\) or its decimal equivalent |
| Obtain \(\frac{1}{2}\pi(e^4 - 1)\) | A1 | or exact equiv involving two terms |
| [2] |
## Question 9:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt use of product rule to produce the form $\ln 2y + y \times \frac{a}{by}$ | M1 | Note that product rule may be applied to expression in form $y(\ln 2y - 1)$ |
| Obtain correct $\ln 2y + y \times \frac{2}{2y}$ | A1 | or equiv |
| Obtain complete $\ln 2y + 1 - 1$ and confirm | A1 | AG; necessary detail needed |
| **[3]** | | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt to rearrange equation to $x = \ldots$ or $x^2 = \ldots$ | M1 | obtaining form $p\ln qy$ |
| Obtain $x = \sqrt{\ln 2y}$ or $x^2 = \ln 2y$ | A1 | |
| State or imply volume is $\int \pi \ln 2y \, dy$ | A1ft | following their $x = \ldots$ or $x^2 = \ldots$; condone absence of $dy$; condone presence of $dx$; no need for limits here; $\pi$ may be implied by its first appearance later in solution |
| Integrate using result of part (i) | M1 | |
| Attempt to use limits $\frac{1}{2}$ and $\frac{1}{2}e^4$ correctly with expression involving $y$ | M1 | |
| Obtain $\frac{1}{2}\pi(3e^4 + 1)$ | A1 | or equiv involving two terms; dependent on correct work throughout part (ii) |
| **[6]** | | |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Subtract answer to part (ii) from $2\pi e^4 \ldots$ | M1 | $\ldots$ or its decimal equivalent |
| Obtain $\frac{1}{2}\pi(e^4 - 1)$ | A1 | or exact equiv involving two terms |
| **[2]** | | |9 (i) Show that the derivative with respect to $y$ of
$$y \ln ( 2 y ) - y$$
is $\ln ( 2 y )$.\\
(ii)\\
\begin{tikzpicture}[>=stealth, thick, scale=1]
\def\xsc{1.2}
\def\ysc{0.12}
% --- Shaded region under curve from x=0 to x=2 ---
\fill[gray!60]
plot[domain=0:1, samples=80, smooth]
({\x}, {1+\x*\x*\x*\x})
-- (0, 2)
-- (0, 1)
-- cycle;
% --- The curve y = 0.5*exp(x^2) ---
\draw[thick]
plot[domain=-1.25:1.25, samples=120, smooth]
({\x}, {1+\x*\x*\x*\x});
% --- Axes ---
\draw[->] (-1.5,0) -- (1.5,0) node[below] {$x$};
\draw[->] (0, -0.3) -- (0, {0.5*exp(4)*\ysc + 0.6}) node[left] {$y$};
% --- Origin label ---
\node[below left] at (0, 0) {$O$};
% --- Vertical line from P down to x-axis (right boundary) ---
\draw (0,2) -- (1,2);
% --- Point P ---
\fill (1, 2) circle (1.5pt);
\node[above left, xshift=2pt] at (1,2) {$P$};
\node[right, yshift=-2pt] at (1,2) {$\left(2,\, \frac{1}{2} e^4\right)$};
\end{tikzpicture}
The diagram shows the curve with equation $y = \frac { 1 } { 2 } \mathrm { e } ^ { x ^ { 2 } }$. The point $P \left( 2 , \frac { 1 } { 2 } \mathrm { e } ^ { 4 } \right)$ lies on the curve. The shaded region is bounded by the curve and the lines $x = 0$ and $y = \frac { 1 } { 2 } e ^ { 4 }$. Find the exact volume of the solid produced when the shaded region is rotated completely about the $y$-axis.\\
(iii) Hence find the volume of the solid produced when the region bounded by the curve and the lines $x = 0$, $x = 2$ and $y = 0$ is rotated completely about the $y$-axis.
\hfill \mbox{\textit{OCR C3 2012 Q9 [11]}}