2015 H2 Mathematics Paper 1 Question 10

Definite Integrals: Areas and Volumes

Answers

(iii)

π3232+π224−2π{\frac{\pi^3\sqrt{2}}{32} + \frac{\pi^2 \sqrt{2}}{4} - \sqrt{2}\pi}

Full solutions

(i)

Using a GC,
A1=∫0π4sin⁡x  dx+∫π4π2cos⁡x  dx=[−cos⁡x]0π4+[sin⁡x]π4π2=−122+1+1−122=2−2\begin{align*} A_1 & = \int_0^{\frac{\pi}{4}} \sin x \; \mathrm{d}x + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \; \mathrm{d}x \\ &= \Big[ -\cos x \Big]_0^{\frac{\pi}{4}} + \Big[ \sin x \Big]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \\ &= -\frac{1}{2} \sqrt{2} + 1 + 1 - \frac{1}{2} \sqrt{2} \\ &= 2-\sqrt{2} \end{align*}
A2=∫0π4cos⁡x  dx−∫0π4sin⁡x  dx=[sin⁡x]0π4−[−cos⁡x]0π4=122−0+122−1=2−1\begin{align*} A_2 & = \int_0^{\frac{\pi}{4}} \cos x \; \mathrm{d}x - \int_0^{\frac{\pi}{4}} \sin x \; \mathrm{d}x \\ &= \Big[ \sin x \Big]_0^{\frac{\pi}{4}} - \Big[ -\cos x \Big]_{0}^{\frac{\pi}{4}} \\ &= \frac{1}{2} \sqrt{2} - 0 + \frac{1}{2} \sqrt{2} - 1 \\ &= \sqrt{2} - 1 \end{align*}
A1A2=2−22−1=2−22−1×2+12+1=22−2+2−22−1=2  ■\begin{align*} \frac{A_1}{A_2} &= \frac{2-\sqrt{2}}{\sqrt{2}-1} \\ &= \frac{2-\sqrt{2}}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} \\ &= \frac{2\sqrt{2} - 2 + 2 - \sqrt{2} }{2-1} \\ &= \sqrt{2} \; \blacksquare \end{align*}

(ii)

Volume of solid formed=π∫0122x2  dy=π∫0122(sin⁡−1y)2  dy  ■\begin{align*} & \textrm{Volume of solid formed} \\ & = \pi \int_0^{\frac{1}{2} \sqrt{2}} x^2 \; \mathrm{d}y \\ & = \pi \int_0^{\frac{1}{2} \sqrt{2}} \left( \sin^{-1}y \right)^2 \; \mathrm{d}y \; \blacksquare \end{align*}

(iii)

dydu=cos⁡uu=sin⁡−1y\begin{gather*} \frac{\mathrm{d}y}{\mathrm{d}u} = \cos u \\ u = \sin^{-1} y \end{gather*}
When y=0,{y = 0, } u=0{u = 0}
When y=122,{y=\frac{1}{2} \sqrt{2}, } u=π4{u = \frac{\pi}{4}}
Exact volume=π∫0122(sin⁡−1y)2  dy=π∫0π4u2cos⁡u  du  ■=π([u2sin⁡u]0π4−∫0π42usin⁡u  du)=π(π21622−0−([−2ucos⁡u]0π4−∫0π4−2cos⁡u  du))=π(π2232+2(π4)22−0−2[sin⁡u]0π4)=π(π2232+π24−2(22)+0)=(π3232+π224−2π) units3  ■\begin{align*} & \textrm{Exact volume} \\ & = \pi \int_0^{\frac{1}{2} \sqrt{2}} \left( \sin^{-1}y \right)^2 \; \mathrm{d}y \\ & = \pi \int_0^{\frac{\pi}{4}} u^2 \cos u \; \mathrm{d}u \; \blacksquare \\ & = \pi \left( \Big[ u^{2} \sin u \Big]_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}} 2u \sin u \; \mathrm{d}u \right) \\ & = \pi \left( \frac{\pi^2}{16} \frac{\sqrt{2}}{2} - 0 - \left( \Big[ -2u \cos u \Big]_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}} -2 \cos u \; \mathrm{d}u \right) \right) \\ & = \pi \left( \frac{\pi^2\sqrt{2}}{32} + 2 \left( \frac{\pi}{4} \right) \frac{\sqrt{2}}{2} - 0 - 2 \Big[ \sin u \Big]_0^{\frac{\pi}{4}} \right) \\ & = \pi \left( \frac{\pi^2\sqrt{2}}{32} + \frac{\pi \sqrt{2}}{4} - 2 \left( \frac{\sqrt{2}}{2} \right) + 0 \right) \\ & = \left( \frac{\pi^3\sqrt{2}}{32} + \frac{\pi^2 \sqrt{2}}{4} - \sqrt{2}\pi \right) \textrm{ units}^3 \; \blacksquare \end{align*}