Question 4 - A-Level Maths

CAIE FP1 2010 June — Question 4 7 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2010
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Parametric integration
Type	Parametric surface area of revolution
Difficulty	Challenging +1.8 This is a Further Maths parametric surface area of revolution question requiring multiple calculus techniques: finding dy/dx or the arc length element from parametric equations, setting up the surface area integral (2πy√(1+(dy/dx)²) or 2πy ds), and evaluating a non-trivial integral. The parametric functions involve products of trigonometric and linear terms, making differentiation and simplification substantial. While the technique is standard for FP1, the algebraic complexity and multi-step nature elevate it above average difficulty.
Spec	4.08d Volumes of revolution: about x and y axes

4 The parametric equations of a curve are $$x = \cos t + t \sin t , \quad y = \sin t - t \cos t$$ The arc of the curve joining the point where $t = 0$ to the point where $t = \frac { 1 } { 2 } \pi$ is rotated about the $x$-axis through one complete revolution. Find the area of the surface generated, leaving your result in terms of $\pi$.

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Answer	Marks	Guidance
$\frac{dx}{dt} = t\cos t$, $\frac{dy}{dt} = t\sin t$ (both)	B1
$\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = t$	B1
$S = 2\pi \int_0^{\pi/2} (t\sin t - t^2 \cos t) dt$ (AEF)	M1A1
$\int t \sin t \, dt = -t\cos t + \sin t$	B1
$\int t^2 \cos t \, dt = t^2 \sin t + 2t \cos t - 2 \sin t$	B1
$S = (\pi/2)[2 - \pi^2]$	A1	[7]

Accept forms such as $6\pi - \pi^3/2$, etc.

OR for lines 4 and 5:

Answer	Marks
$2\pi \left[(-t\cos t) + \int \cos t \, dt - \left[(t^2 \sin t) - \int 2t \sin t \, dt\right]\right]_0^{\pi/2}$ (LNR)	M1
$= 2\pi \left[-t\cos t + \sin t - t^2 \sin t - 2t\cos t + 2\sin t\right]_0^{\pi/2}$ (LNR)	A1

$\frac{dx}{dt} = t\cos t$, $\frac{dy}{dt} = t\sin t$ (both) | B1 |

$\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = t$ | B1 |

$S = 2\pi \int_0^{\pi/2} (t\sin t - t^2 \cos t) dt$ (AEF) | M1A1 |

$\int t \sin t \, dt = -t\cos t + \sin t$ | B1 |

$\int t^2 \cos t \, dt = t^2 \sin t + 2t \cos t - 2 \sin t$ | B1 |

$S = (\pi/2)[2 - \pi^2]$ | A1 | [7]

Accept forms such as $6\pi - \pi^3/2$, etc.

**OR for lines 4 and 5:**

$2\pi \left[(-t\cos t) + \int \cos t \, dt - \left[(t^2 \sin t) - \int 2t \sin t \, dt\right]\right]_0^{\pi/2}$ (LNR) | M1 |

$= 2\pi \left[-t\cos t + \sin t - t^2 \sin t - 2t\cos t + 2\sin t\right]_0^{\pi/2}$ (LNR) | A1 |

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4 The parametric equations of a curve are

$$x = \cos t + t \sin t , \quad y = \sin t - t \cos t$$

The arc of the curve joining the point where $t = 0$ to the point where $t = \frac { 1 } { 2 } \pi$ is rotated about the $x$-axis through one complete revolution. Find the area of the surface generated, leaving your result in terms of $\pi$.

\hfill \mbox{\textit{CAIE FP1 2010 Q4 [7]}}

This paper (13 questions)

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Q1 4 Q2 5 Q3 6 Q4 7 Q5 8 Q6 8 Q7 8 Q8 9 Q9 10 Q10 10 Q11 12 Q12 EITHER Q12 OR

Answer	Marks	Guidance
\(\frac{dx}{dt} = t\cos t\), \(\frac{dy}{dt} = t\sin t\) (both)	B1
\(\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = t\)	B1
\(S = 2\pi \int_0^{\pi/2} (t\sin t - t^2 \cos t) dt\) (AEF)	M1A1
\(\int t \sin t \, dt = -t\cos t + \sin t\)	B1
\(\int t^2 \cos t \, dt = t^2 \sin t + 2t \cos t - 2 \sin t\)	B1
\(S = (\pi/2)[2 - \pi^2]\)	A1	[7]

Answer	Marks
\(2\pi \left[(-t\cos t) + \int \cos t \, dt - \left[(t^2 \sin t) - \int 2t \sin t \, dt\right]\right]_0^{\pi/2}\) (LNR)	M1
\(= 2\pi \left[-t\cos t + \sin t - t^2 \sin t - 2t\cos t + 2\sin t\right]_0^{\pi/2}\) (LNR)	A1

CAIE FP1 2010 June — Question 4 7 marks

Accept forms such as \(6\pi - \pi^3/2\), etc.

OR for lines 4 and 5:

This paper (13 questions)