| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| 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 Part (a) is a standard bookwork derivation of the volume of revolution formula that appears in most textbooks. Part (b) requires algebraic manipulation to simplify the rational function before integration, then integration using substitution or partial fractions, followed by logarithm manipulation. While it involves several steps, these are all standard A-level techniques with no novel insight required. The question is slightly easier than average due to part (a) being pure recall and part (b) following a predictable pattern for this topic. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Draws or describes thin strip(s) under graph | B1 | May fit curve exactly or may be rectangular |
| Volume of thin disc \(\approx \pi y^2\,\delta x\) | M1 | Obtains expression for approximate volume of a thin disc. Condone expression for volume of a cylinder of radius \(y\) or \(f(x)\) and of any height |
| Total volume of discs \(= \sum_{x=a}^{b} \pi y^2\,\delta x\) | A1 | Obtains expression for approximate total volume of discs |
| Volume of solid \(= \lim_{\delta x \to 0}\left(\sum_{x=a}^{b}\pi y^2\,\delta x\right) = \pi\int_a^b y^2\,dx = \pi\int_a^b(f(x))^2\,dx\) | R1 | Completes correct argument including taking the limit as \(\delta x \to 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(f(x)\right)^2 = \frac{(x+3)^2}{x(x+1)^2}\) | B1 | Obtains correct expression |
| \(\frac{(x+3)^2}{x(x+1)^2} \equiv \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2}\), so \(x^2+6x+9 = A(x+1)^2 + Bx(x+1) + Cx\) | M1 | Expresses \((f(x))^2\) as partial fractions in correct format |
| \(x=0 \Rightarrow A=9\); \(x=-1 \Rightarrow C=-4\); Compare \(x^2\)-coeff: \(B=-8\) | A1 | Obtains correct expression in partial fractions |
| \(\frac{(x+3)^2}{x(x+1)^2} \equiv \frac{9}{x} - \frac{8}{x+1} - \frac{4}{(x+1)^2}\); \(V = \pi\int_1^2 \frac{(x+3)^2}{x(x+1)^2}\,dx\) | M1 | Integrates to obtain two logarithmic terms and one algebraic fraction; condone missing \(\pi\) |
| \(V = \pi\int_1^2\left(\frac{9}{x} - \frac{8}{x+1} - \frac{4}{(x+1)^2}\right)dx = \pi\left[9\ln x - 8\ln(x+1) + \frac{4}{x+1}\right]_1^2\) | A1F | Correct result of integration; follow through their numerators; condone missing \(\pi\) |
| \(= \pi\left\{\left(9\ln 2 - 8\ln 3 + \frac{4}{3}\right) - \left(9\ln 1 - 8\ln 2 + \frac{4}{2}\right)\right\}\) | M1 | Substitutes limits into three-term integrated expression; must include \(\pi\) |
| \(= \pi\left(17\ln 2 - 8\ln 3 - \frac{2}{3}\right) = \pi\left(\ln\left(\frac{2^{17}}{3^8}\right) - \frac{2}{3}\right)\) | R1 | Completes reasoned argument to show required result, including correct re-arrangement of log terms |
## Question 12(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Draws or describes thin strip(s) under graph | B1 | May fit curve exactly or may be rectangular |
| Volume of thin disc $\approx \pi y^2\,\delta x$ | M1 | Obtains expression for approximate volume of a thin disc. Condone expression for volume of a cylinder of radius $y$ or $f(x)$ and of any height |
| Total volume of discs $= \sum_{x=a}^{b} \pi y^2\,\delta x$ | A1 | Obtains expression for approximate total volume of discs |
| Volume of solid $= \lim_{\delta x \to 0}\left(\sum_{x=a}^{b}\pi y^2\,\delta x\right) = \pi\int_a^b y^2\,dx = \pi\int_a^b(f(x))^2\,dx$ | R1 | Completes correct argument including taking the limit as $\delta x \to 0$ |
## Question 12(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(f(x)\right)^2 = \frac{(x+3)^2}{x(x+1)^2}$ | B1 | Obtains correct expression |
| $\frac{(x+3)^2}{x(x+1)^2} \equiv \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2}$, so $x^2+6x+9 = A(x+1)^2 + Bx(x+1) + Cx$ | M1 | Expresses $(f(x))^2$ as partial fractions in correct format |
| $x=0 \Rightarrow A=9$; $x=-1 \Rightarrow C=-4$; Compare $x^2$-coeff: $B=-8$ | A1 | Obtains correct expression in partial fractions |
| $\frac{(x+3)^2}{x(x+1)^2} \equiv \frac{9}{x} - \frac{8}{x+1} - \frac{4}{(x+1)^2}$; $V = \pi\int_1^2 \frac{(x+3)^2}{x(x+1)^2}\,dx$ | M1 | Integrates to obtain two logarithmic terms and one algebraic fraction; condone missing $\pi$ |
| $V = \pi\int_1^2\left(\frac{9}{x} - \frac{8}{x+1} - \frac{4}{(x+1)^2}\right)dx = \pi\left[9\ln x - 8\ln(x+1) + \frac{4}{x+1}\right]_1^2$ | A1F | Correct result of integration; follow through their numerators; condone missing $\pi$ |
| $= \pi\left\{\left(9\ln 2 - 8\ln 3 + \frac{4}{3}\right) - \left(9\ln 1 - 8\ln 2 + \frac{4}{2}\right)\right\}$ | M1 | Substitutes limits into three-term integrated expression; must include $\pi$ |
| $= \pi\left(17\ln 2 - 8\ln 3 - \frac{2}{3}\right) = \pi\left(\ln\left(\frac{2^{17}}{3^8}\right) - \frac{2}{3}\right)$ | R1 | Completes reasoned argument to show required result, including correct re-arrangement of log terms |
---12 The shaded region shown in the diagram below is bounded by the $x$-axis, the curve $y = \mathrm { f } ( x )$, and the lines $x = a$ and $x = b$\\
\includegraphics[max width=\textwidth, alt={}, center]{74b8239a-1f46-45e7-ad20-2dce7bf4baf6-16_661_721_406_662}
The shaded region is rotated through $2 \pi$ radians about the $x$-axis to form a solid.\\
12
\begin{enumerate}[label=(\alph*)]
\item Show that the volume of this solid is
$$\pi \int _ { a } ^ { b } ( \mathrm { f } ( x ) ) ^ { 2 } \mathrm {~d} x$$
12
\item In the case where $a = 1 , b = 2$ and
$$f ( x ) = \frac { x + 3 } { ( x + 1 ) \sqrt { x } }$$
show that the volume of the solid is
$$\pi \left( \ln \left( \frac { 2 ^ { m } } { 3 ^ { n } } \right) - \frac { 2 } { 3 } \right)$$
where $m$ and $n$ are integers.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2022 Q12 [11]}}