Integration and applications of integration — Question 17

QCAA 2022, Paper 1 · 5 marks

Q17 · 2022 · Technology-freeComplex familiar5 marks

QUESTION 17 (5 marks)

The region between the

x

-axis and the curve of the function

y=1+\sin(2x)

for

0\le x\le\frac{\pi}{2}

is rotated about the

x

-axis to form a solid of revolution. Determine the volume of this solid. Express your answer in simplest form.

WORKED SOLUTION

5 marks

ANSWER

\pi\left(\frac{3\pi}{4}+2\right)

units

^3

Worked solution

Using the volume of a solid of revolution rule,

V=\pi\int_0^{\pi/2}(1+\sin(2x))^2\,dx.

Expand the integrand and use

\sin^2(2x)=\frac12(1-\cos(4x))

V=\pi\int_0^{\pi/2}\left(\frac32+2\sin(2x)-\frac12\cos(4x)\right)dx.

Hence

V=\pi\left[\frac32x-\cos(2x)-\frac18\sin(4x)\right]_0^{\pi/2} =\pi\left(\frac{3\pi}{4}+2\right)\text{ units}^3.

correctly substitutes into the appropriate volume of a solid of revolution rule

[1 mark]

expands an integrand

[1 mark]

uses a suitable double-angle identity to enable an integration process to be completed

[1 mark]

integrates an expression

[1 mark]

determines volume in simplest form

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

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