2023 H2 Mathematics Paper 1 Question 6 Definite Integrals: Areas and Volumes
Answers
81 π 8 ( π − 9 8 3 ) . {{\frac{81 \pi}{8} \left( \pi - \frac{9}{8} \sqrt{3} \right).}} 8 81 π ( π − 8 9 3 ) . Full solutions
(a)
cos 4 θ = ( cos 2 θ ) 2 = ( cos 2 θ + 1 2 ) 2 = cos 2 2 θ + 2 cos 2 θ + 1 4 = cos 4 θ + 1 2 + 2 cos 2 θ + 1 4 = 1 8 ( cos 4 θ + 1 + 4 cos 2 θ + 2 ) = 1 8 ( cos 4 θ + 4 cos 2 θ + 3 ) ■ \begin{align*}
\cos^4 \theta &= \left( \cos^2 \theta \right)^2 \\
&= \left( \frac{\cos 2 \theta + 1}{2} \right)^2 \\
&= \frac{\cos^2 2\theta + 2 \cos 2 \theta + 1}{4} \\
& = \frac{\frac{\cos 4\theta +1}{2} + 2 \cos 2 \theta + 1}{4} \\
& = \frac{1}{8} \left( \cos 4 \theta + 1 + 4 \cos 2 \theta + 2 \right) \\
&= \frac{1}{8} \left( \cos 4\theta + 4 \cos 2 \theta + 3 \right) \; \blacksquare
\end{align*} cos 4 θ = ( cos 2 θ ) 2 = ( 2 cos 2 θ + 1 ) 2 = 4 cos 2 2 θ + 2 cos 2 θ + 1 = 4 2 c o s 4 θ + 1 + 2 cos 2 θ + 1 = 8 1 ( cos 4 θ + 1 + 4 cos 2 θ + 2 ) = 8 1 ( cos 4 θ + 4 cos 2 θ + 3 ) ■
(b)
x = 3 sin θ d x d θ = 3 cos θ \begin{align*}
x &= 3 \sin \theta \\
\frac{\operatorname{d}\!x}{\operatorname{d}\!\theta} &= 3 \cos \theta
\end{align*} x d θ d x = 3 sin θ = 3 cos θ When x = 1.5 , {{x=1.5,}} x = 1.5 , θ = 1 6 π {{\theta = \frac{1}{6} \pi}} θ = 6 1 π x = 3 , {{x=3,}} x = 3 , θ = 1 2 π {{\theta = \frac{1}{2} \pi}} θ = 2 1 π
Volume = π ∫ 1.5 3 y 2 d x = π ∫ 1.5 3 ( 9 − x 2 ) 3 2 d x = π ∫ 1 6 π 1 2 π ( 9 − 9 sin 2 θ ) 3 2 3 cos θ d θ = 9 3 2 ( 3 π ) ∫ 1 6 π 1 2 π ( 1 − sin 2 θ ) 3 2 cos θ d θ = 81 π ∫ 1 6 π 1 2 π ( cos 2 θ ) 3 2 cos θ d θ = 81 π ∫ 1 6 π 1 2 π cos 4 θ d θ = 81 π 8 ∫ 1 6 π 1 2 π ( cos 4 θ + 4 cos 2 θ + 3 ) d θ = 81 π 8 [ sin 4 θ 4 + 2 sin 2 θ + 3 θ ] 1 6 π 1 2 π = 81 π 8 ( 0 + 0 + 3 π 2 − ( sin 2 π 3 4 + 2 sin π 3 + π 2 ) ) = 81 π 8 ( π − 3 8 − 3 ) = 81 π 8 ( π − 9 8 3 ) units 2 ■ \begin{align*}
&\text{Volume} \\
&= \pi \int_{1.5}^{3} y^2 \, \mathrm{d}x \\
&= \pi \int_{1.5}^{3} ( 9 - x^2 )^{\frac{3}{2}} \, \mathrm{d}x \\
&= \pi \int_{\frac{1}{6} \pi}^{\frac{1}{2} \pi} \left( 9 - 9 \sin^2 \theta \right)^{\frac{3}{2}} 3 \cos \theta \, \mathrm{d}\theta \\
&= 9^{\frac{3}{2}} (3\pi) \int_{\frac{1}{6} \pi}^{\frac{1}{2} \pi} \left( 1 - \sin^2 \theta \right)^{\frac{3}{2}} \cos \theta \, \mathrm{d}\theta \\
&= 81 \pi \int_{\frac{1}{6} \pi}^{\frac{1}{2} \pi} \left( \cos^2 \theta \right)^{\frac{3}{2}} \cos \theta \, \mathrm{d}\theta \\
&= 81 \pi \int_{\frac{1}{6} \pi}^{\frac{1}{2} \pi} \cos^4 \theta \, \mathrm{d}\theta \\
&= \frac{81 \pi}{8} \int_{\frac{1}{6} \pi}^{\frac{1}{2} \pi} \left( \cos 4\theta + 4 \cos 2 \theta + 3 \right) \, \mathrm{d}\theta \\
&= \frac{81 \pi}{8} \left[ \frac{\sin 4 \theta}{4} + 2 \sin 2 \theta + 3\theta \right]_{\frac{1}{6} \pi}^{\frac{1}{2} \pi} \\
&= \frac{81 \pi}{8} \left( 0 + 0 + \frac{3\pi}{2} - \left( \frac{\sin \frac{2\pi}{3}}{4} + 2 \sin \frac{\pi}{3} + \frac{\pi}{2} \right) \right) \\
&= \frac{81 \pi}{8} \left( \pi - \frac{\sqrt{3}}{8} - \sqrt{3} \right) \\
&= \frac{81 \pi}{8} \left( \pi - \frac{9}{8} \sqrt{3} \right) \text{ units}^2 \; \blacksquare
\end{align*} Volume = π ∫ 1.5 3 y 2 d x = π ∫ 1.5 3 ( 9 − x 2 ) 2 3 d x = π ∫ 6 1 π 2 1 π ( 9 − 9 sin 2 θ ) 2 3 3 cos θ d θ = 9 2 3 ( 3 π ) ∫ 6 1 π 2 1 π ( 1 − sin 2 θ ) 2 3 cos θ d θ = 81 π ∫ 6 1 π 2 1 π ( cos 2 θ ) 2 3 cos θ d θ = 81 π ∫ 6 1 π 2 1 π cos 4 θ d θ = 8 81 π ∫ 6 1 π 2 1 π ( cos 4 θ + 4 cos 2 θ + 3 ) d θ = 8 81 π [ 4 sin 4 θ + 2 sin 2 θ + 3 θ ] 6 1 π 2 1 π = 8 81 π ( 0 + 0 + 2 3 π − ( 4 sin 3 2 π + 2 sin 3 π + 2 π ) ) = 8 81 π ( π − 8 3 − 3 ) = 8 81 π ( π − 8 9 3 ) units 2 ■
Question Commentary
The first part is rather tedious and requires the use of the double angle formula
twice. Subsequently, the second part is more straightforward once we apply our
volume of solid of revolution formula and using the result from the first part.