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Question Number 58447 by Tawa1 last updated on 23/Apr/19

Commented by Tawa1 last updated on 23/Apr/19

$$\mathrm{Find}\mathrm{the}\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{Red}\mathrm{portion}$$

Answered by MJS last updated on 23/Apr/19

$$\mathrm{halfcircle}:y=4-\sqrt{8x-x^2}$$$$\mathrm{line}:y=\frac{x}{2}$$$$\mathrm{intersection}=\left(\begin{array}{c}\frac{8}{5}\\ \frac{4}{5}\end{array}\right)$$$$\mathrm{red}\mathrm{area}$$$$\underset{0}{\stackrel{\frac{8}{5}}{\int }}\frac{x}{2}dx+\underset{\frac{8}{5}}{\stackrel{4}{\int }}(4-\sqrt{8x-x^2})dx=\frac{32}{5}-4\pi +16\arctan \frac{1}{2}$$$$\approx 1.25199$$

Commented by Tawa1 last updated on 23/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by Tawa1 last updated on 23/Apr/19

$$\mathrm{Sir},\mathrm{how}\mathrm{is}\mathrm{half}\mathrm{circle}\mathrm{y}=4-\sqrt{8\mathrm{x}-{\mathrm{x}}^2}$$

Commented by MJS last updated on 23/Apr/19

$$\mathrm{I}\mathrm{put}\mathrm{the}\mathrm{left}\mathrm{handed}\mathrm{lower}\mathrm{bottom}\mathrm{verticle}$$$$\mathrm{of}\mathrm{the}\mathrm{rectangle}A=\left(\begin{array}{c}0\\ 0\end{array}\right)\Rightarrow \mathrm{center}\mathrm{of}\mathrm{circle}=$$$$=\left(\begin{array}{c}4\\ 4\end{array}\right)\Rightarrow \mathrm{circle}:{(x-4)}^2+{(y-4)}^2=4^2\Rightarrow$$$$\Rightarrow y=4\pm \sqrt{8x-x^2}$$

Commented by Tawa1 last updated on 23/Apr/19

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 23/Apr/19

Commented by mr W last updated on 23/Apr/19

$$\alpha ={\tan }^{-1}\frac{4}{8}={\tan }^{-1}\frac{1}{2}$$$$\theta =\frac{\pi -2\alpha }{2}=\frac{\pi }{2}-\alpha$$$$A_{redshade}=4\times 4-\frac{\pi 4^2}{4}=16-4\pi$$$$A_{greenshade}=\frac{4^2}{2}(2\theta -\sin 2\theta )=16(\theta -\sin \theta \cos \theta )$$$$=16(\frac{\pi }{2}-{\tan }^{-1}\frac{1}{2}-\frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}})=16(\frac{\pi }{2}-\frac{2}{5}-{\tan }^{-1}\frac{1}{2})$$$$$$$$A_{red}=\frac{4\times 8}{2}-16(\frac{\pi }{2}-\frac{2}{5}-{\tan }^{-1}\frac{1}{2})-(16-4\pi )$$$$=\frac{32}{5}+16{\tan }^{-1}\frac{1}{2}-4\pi$$$$=1.25199$$

Commented by Tawa1 last updated on 23/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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