A2 June 2025 Paper 1 Q3

OCR ACurrent spec6 marksIntegration

3 The region \(R_1\) is bounded by the curve \(y = \dfrac{80}{\sqrt{25 - x^2}}\), the line \(x = k\) and the coordinate axes, as shown in Fig. 1.

Fig. 1: U-shaped curve y = 80 over root(25 - x squared) above the x-axis, with the vertical line x = k to the right of the y-axis; the region R1 between the curve, the axes and x = k is shaded — Fig. 1

(a) In this question you must show detailed reasoning.
Given that the area of \(R_1\) is \(\dfrac{40}{3}\pi\), determine the value of \(k\). [3]

The region \(R_2\) is bounded by the curve \(y = \dfrac{80}{\sqrt{25 - x^2}}\), the line \(y = 20\) and the \(y\)-axis as shown in Fig. 2.

Fig. 2: the same curve with the horizontal line y = 20 crossing it on both sides; the region R2 between the y-axis, the curve and y = 20, to the right of the y-axis, is shaded — Fig. 2

(b) Find, in an exact form, the volume of the solid formed when \(R_2\) is rotated by \(2\pi\) radians about the y-axis. [3]

Mark scheme (a)

Scheme	Marks	AO
DR \(\displaystyle\int \frac{80}{\sqrt{25 - x^2}}\,\mathrm{d}x = 80\sin^{-1}\left(\frac{1}{5}x\right)\)	B1	1.1
\(\left(\displaystyle\int_0^k \frac{80}{\sqrt{25 - x^2}}\,\mathrm{d}x = \right) \quad 80\sin^{-1}\left(\dfrac{1}{5}k\right) = \dfrac{40}{3}\pi\)	M1	2.1
\(\left(\sin^{-1}\left(\dfrac{1}{5}k\right) = \dfrac{1}{6}\pi \Rightarrow\right) \quad k = \dfrac{5}{2}\)	A1	1.1
	[3]

Notes

B1: Correct integration – this mark can be implied by seeing \(80\sin^{-1}\left(\frac{1}{5}k\right)\) but www (so if \(x\) missing from the integrated expression before the limits were applied then B0) – this mark can also be awarded for \(\sin^{-1}\left(\frac{1}{5}x\right)\) from \(\displaystyle\int \frac{1}{\sqrt{25 - x^2}}\,\mathrm{d}x\)

M1: For setting up an equation of the form \(a\sin^{-1}(bk) - a\sin^{-1}(0) = \frac{40}{3}\pi\) or \(a\sin^{-1}(bk) = \frac{40}{3}\pi\) where \(a \neq 0, b \neq 0\) or 1 oe (e.g. may have divided both sides by 80). Must have come from their \(a\sin^{-1}(bx)\)

A1: cao – oe e.g. 2.5 - do not need to explicitly see \(\sin^{-1}(0) = 0\) for full marks

Mark scheme (b)

Scheme	Marks	AO
\(x^2 = -\dfrac{80^2}{y^2} + 25\)	M1	1.1
\(\pi\displaystyle\int_{16}^{20} \left(-\frac{6400}{y^2} + 25\right)\mathrm{d}y\)	M1	3.1a
\(= 20\pi\)	A1	1.1
	[3]

Notes

M1: Rearranging to make \(x^2\) the subject – allow sign errors only when re-arranging

M1: Integral of the form \(\pi\displaystyle\int_{16}^{20} \left(\frac{k_1}{y^2} + k_2\right)\mathrm{d}y\) for any non-zero \(k_1, k_2\) with correct limits, M0 if \(\pi\) missing but condone missing \(\mathrm{d}y\)

A1: BC – must be exact (ISW if correct exact answer seen and then replaced with non-exact). A correct answer with no working (or no incorrect working) can score full marks (as this part is not DR)