QCE Vault / Specialist Maths Applications of integral calculus — Question 209 Original QCE Vault practice · 5 marks
Browse all questions Original practice exam Report an issue Q209 · Practice question Technology-free Complex familiar 5 marks
QUESTION 209 (5 marks) For k>0, the region below y = k sin x y=k\sin x y = k sin x and above the x-axis on 0 ≤ x ≤ π 0\le x\le\pi 0 ≤ x ≤ π is rotated about the x-axis. a) Determine the area and rotated volume in terms of k.
[3 marks] b) If k changes so the area doubles, by what factor does volume change?
[2 marks] WORKED SOLUTION
Practice marking scheme 5 marks ANSWER (a) A = 2 k A=2k A = 2 k , V = k 2 π 2 / 2 V=k^2\pi^2/2 V = k 2 π 2 /2 . (b) 4 4 4 . Worked solution
(a) Area is k ∫ 0 π sin x d x = 2 k k\int_0^\pi\sin x\,dx=2k k ∫ 0 π sin x d x = 2 k . The disc method gives V = π k 2 ∫ 0 π sin 2 x d x V=\pi k^2\int_0^\pi\sin^2x\,dx V = π k 2 ∫ 0 π sin 2 x d x . Using sin 2 x = ( 1 − cos 2 x ) / 2 \sin^2x=(1-\cos2x)/2 sin 2 x = ( 1 − cos 2 x ) /2 , the integral is pi/2, giving the stated volume. (b) Area is proportional to k, so doubling it doubles k. Volume is proportional to k squared, hence becomes four times as large. Horizontal extent is unchanged.
Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: Form the disc integral.
[1 mark] Part a: Use the identity and evaluate volume.
[1 mark] Part b: Identify doubled k.
[1 mark] Part b: Explain squared scaling and factor 4.
[1 mark] Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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Applications of integral calculus