| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2020 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area under curve with fractional/negative powers or roots |
| Difficulty | Standard +0.3 This is a standard volumes of revolution question requiring students to set up and evaluate ∫π y² dx for a semicircle. While it involves the circle equation x² + y² = r², substitution, and careful algebraic manipulation, it's a well-known textbook result with clear setup and straightforward integration steps, making it slightly easier than average. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2+y^2=r^2\) | B1 | May be implied by correct integral |
| \(\{V\}=\pi\displaystyle\int_{-r}^{r} r^2-x^2\,\mathrm{d}x\) or \(\{V\}=2\pi\displaystyle\int_{0}^{r} r^2-x^2\,\mathrm{d}x\) | B1 | Limits and d\(x\) may be implied; if using limits \(r\) and \(0\), the \(2\) could appear later with reasoning |
| Integrates to form \(\alpha x \pm \beta x^3\); correct integration gives \(r^2x - \dfrac{1}{3}x^3\) | M1 | Do not award if \(r^2 \to \lambda r^3\) |
| Substitutes limits \(-r\) and \(r\): \(\left(r^2(r)-\dfrac{1}{3}(r)^3\right)-\left(r^2(-r)-\dfrac{1}{3}(-r)^3\right)\) or limits \(0\) and \(r\) with twice the volume: \(\left(r^2(r)-\dfrac{1}{3}(r)^3\right)-(0)\) | dM1 | Dependent on previous M; correct use of limits; limit of 0 can be implied |
| \(V=\dfrac{4}{3}\pi r^3\) * cso | A1* | Note: rotation about \(y\)-axis all marks available, but final accuracy mark must refer to symmetry |
# Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2+y^2=r^2$ | B1 | May be implied by correct integral |
| $\{V\}=\pi\displaystyle\int_{-r}^{r} r^2-x^2\,\mathrm{d}x$ or $\{V\}=2\pi\displaystyle\int_{0}^{r} r^2-x^2\,\mathrm{d}x$ | B1 | Limits and d$x$ may be implied; if using limits $r$ and $0$, the $2$ could appear later with reasoning |
| Integrates to form $\alpha x \pm \beta x^3$; correct integration gives $r^2x - \dfrac{1}{3}x^3$ | M1 | Do not award if $r^2 \to \lambda r^3$ |
| Substitutes limits $-r$ and $r$: $\left(r^2(r)-\dfrac{1}{3}(r)^3\right)-\left(r^2(-r)-\dfrac{1}{3}(-r)^3\right)$ **or** limits $0$ and $r$ with twice the volume: $\left(r^2(r)-\dfrac{1}{3}(r)^3\right)-(0)$ | dM1 | Dependent on previous M; correct use of limits; limit of 0 can be implied |
| $V=\dfrac{4}{3}\pi r^3$ * cso | A1* | Note: rotation about $y$-axis all marks available, but final accuracy mark must refer to symmetry |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{09bd14c0-c368-4ae1-bee0-cc8bf82abecc-06_582_588_255_758}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a circle with radius $r$ and centre at the origin.\\
The region $R$, shown shaded in Figure 1, is bounded by the $x$-axis and the part of the circle for which $y > 0$\\
The region $R$ is rotated through $360 ^ { \circ }$ about the $x$-axis to create a sphere with volume $V$\\
Use integration to show that $V = \frac { 4 } { 3 } \pi r ^ { 3 }$
\hfill \mbox{\textit{Edexcel CP AS 2020 Q3 [5]}}