AQA Further AS Paper 1 2020 June — Question 15 4 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2020
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeVolume using cone or cylinder formula
DifficultyStandard +0.8 This is a Further Maths question requiring volume of revolution calculus with careful setup. Part (a) is straightforward geometry (k=r/h), but part (b) requires correctly setting up and evaluating the integral βˆ«Ο€yΒ² dx from 0 to h, substituting y=kx, and algebraically simplifying to the given form. While the integration itself is routine, the geometric interpretation and algebraic manipulation across multiple steps makes this moderately challenging, though still a standard Further Maths exercise rather than requiring novel insight.
Spec1.08a Fundamental theorem of calculus: integration as reverse of differentiation4.08d Volumes of revolution: about x and y axes
A segment of the line \(y = kx\) is rotated about the \(x\)-axis to generate a cone with vertex \(O\). The distance of \(O\) from the centre of the base of the cone is \(h\). The radius of the base of the cone is \(r\). \includegraphics{figure_15}
  1. Find \(k\) in terms of \(r\) and \(h\). [1 mark]
  2. Use calculus to prove that the volume of the cone is $$\frac{1}{3}\pi r^2 h$$ [3 marks]
Question 15:
AnswerMarks
15(a)Obtains the correct expression .
π‘Ÿπ‘Ÿ
AnswerMarks Guidance
π‘˜π‘˜ = β„Ž1.1b B1
AnswerMarks
15(b)Uses the formula for volume of revolution .
2
Condone missing and missing or inco𝑉𝑉rr=ecπœ‹πœ‹t l∫imπ‘šπ‘šitπ‘₯π‘₯s. 𝑑𝑑π‘₯π‘₯
AnswerMarks Guidance
πœ‹πœ‹,𝑑𝑑π‘₯π‘₯1.1a M1
β„Ž
2
π‘Ÿπ‘Ÿπ‘₯π‘₯
Volume= πœ‹πœ‹οΏ½οΏ½ οΏ½ 𝑑𝑑π‘₯π‘₯
β„Ž
0
β„Ž
2 2
π‘Ÿπ‘Ÿ π‘₯π‘₯
= πœ‹πœ‹οΏ½ 2 𝑑𝑑π‘₯π‘₯
β„Ž
0
2 3 β„Ž
π‘Ÿπ‘Ÿ π‘₯π‘₯
= πœ‹πœ‹οΏ½ 2οΏ½
3β„Ž 0
2 3
π‘Ÿπ‘Ÿ β„Ž 1 2
=πœ‹πœ‹οΏ½ 2 βˆ’ 0οΏ½ = πœ‹πœ‹π‘Ÿπ‘Ÿ β„Ž
Correctly integrates their , with an expression for in terms of
and . 2
(π‘˜π‘˜π‘₯π‘₯) π‘˜π‘˜ π‘Ÿπ‘Ÿ
AnswerMarks Guidance
β„Ž1.1a M1
Completes a rigorous proof to show that .
1 2
AnswerMarks Guidance
𝑉𝑉 = 3πœ‹πœ‹π‘Ÿπ‘Ÿ β„Ž2.1 R1
Total4 3β„Ž 3
QMarking instructions AO 
Question 15:
--- 15(a) ---
15(a) | Obtains the correct expression .
π‘Ÿπ‘Ÿ
π‘˜π‘˜ = β„Ž | 1.1b | B1 | π‘Ÿπ‘Ÿ
--- 15(b) ---
15(b) | Uses the formula for volume of revolution .
2
Condone missing and missing or inco𝑉𝑉rr=ecπœ‹πœ‹t l∫imπ‘šπ‘šitπ‘₯π‘₯s. 𝑑𝑑π‘₯π‘₯
πœ‹πœ‹,𝑑𝑑π‘₯π‘₯ | 1.1a | M1 | β„Ž
β„Ž
2
π‘Ÿπ‘Ÿπ‘₯π‘₯
Volume= πœ‹πœ‹οΏ½οΏ½ οΏ½ 𝑑𝑑π‘₯π‘₯
β„Ž
0
β„Ž
2 2
π‘Ÿπ‘Ÿ π‘₯π‘₯
= πœ‹πœ‹οΏ½ 2 𝑑𝑑π‘₯π‘₯
β„Ž
0
2 3 β„Ž
π‘Ÿπ‘Ÿ π‘₯π‘₯
= πœ‹πœ‹οΏ½ 2οΏ½
3β„Ž 0
2 3
π‘Ÿπ‘Ÿ β„Ž 1 2
=πœ‹πœ‹οΏ½ 2 βˆ’ 0οΏ½ = πœ‹πœ‹π‘Ÿπ‘Ÿ β„Ž
Correctly integrates their , with an expression for in terms of
and . 2
(π‘˜π‘˜π‘₯π‘₯) π‘˜π‘˜ π‘Ÿπ‘Ÿ
β„Ž | 1.1a | M1
Completes a rigorous proof to show that .
1 2
𝑉𝑉 = 3πœ‹πœ‹π‘Ÿπ‘Ÿ β„Ž | 2.1 | R1
Total | 4 | 3β„Ž 3
Q | Marking instructions | AO | Marks | Typical solution
A segment of the line $y = kx$ is rotated about the $x$-axis to generate a cone with vertex $O$.

The distance of $O$ from the centre of the base of the cone is $h$.

The radius of the base of the cone is $r$.

\includegraphics{figure_15}

\begin{enumerate}[label=(\alph*)]
\item Find $k$ in terms of $r$ and $h$.
[1 mark]

\item Use calculus to prove that the volume of the cone is
$$\frac{1}{3}\pi r^2 h$$
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2020 Q15 [4]}}
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