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
$$\mathrm{halfcircle}:\:{y}=\mathrm{4}−\sqrt{\mathrm{8}{x}−{x}^{\mathrm{2}} } \\ $$$$\mathrm{line}:\:{y}=\frac{{x}}{\mathrm{2}} \\ $$$$\mathrm{intersection}\:=\:\begin{pmatrix}{\frac{\mathrm{8}}{\mathrm{5}}}\\{\frac{\mathrm{4}}{\mathrm{5}}}\end{pmatrix} \\ $$$$\mathrm{red}\:\mathrm{area} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{8}}{\mathrm{5}}} {\int}}\frac{{x}}{\mathrm{2}}{dx}+\underset{\frac{\mathrm{8}\:\:}{\mathrm{5}}} {\overset{\mathrm{4}} {\int}}\left(\mathrm{4}−\sqrt{\mathrm{8}{x}−{x}^{\mathrm{2}} }\right){dx}=\frac{\mathrm{32}}{\mathrm{5}}−\mathrm{4}\pi+\mathrm{16arctan}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\approx\mathrm{1}.\mathrm{25199} \\ $$
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
$$\mathrm{Sir},\:\mathrm{how}\:\mathrm{is}\:\mathrm{half}\:\mathrm{circle}\:\:\:\mathrm{y}\:=\:\mathrm{4}\:−\:\sqrt{\mathrm{8x}\:−\:\mathrm{x}^{\mathrm{2}} } \\ $$
Commented by MJS last updated on 23/Apr/19
$$\mathrm{I}\:\mathrm{put}\:\mathrm{the}\:\mathrm{left}\:\mathrm{handed}\:\mathrm{lower}\:\mathrm{bottom}\:\mathrm{verticle} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle}\:{A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\Rightarrow\:\mathrm{center}\:\mathrm{of}\:\mathrm{circle}\:= \\ $$$$=\begin{pmatrix}{\mathrm{4}}\\{\mathrm{4}}\end{pmatrix}\:\Rightarrow\:\mathrm{circle}:\:\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \:\Rightarrow \\ $$$$\Rightarrow\:{y}=\mathrm{4}\pm\sqrt{\mathrm{8}{x}−{x}^{\mathrm{2}} } \\ $$
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
$$\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{8}}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\theta=\frac{\pi−\mathrm{2}\alpha}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}−\alpha \\ $$$${A}_{{red}\:{shade}} =\mathrm{4}×\mathrm{4}−\frac{\pi\mathrm{4}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{16}−\mathrm{4}\pi \\ $$$${A}_{{green}\:{shade}} =\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{2}\theta−\mathrm{sin}\:\mathrm{2}\theta\right)=\mathrm{16}\left(\theta−\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\right) \\ $$$$=\mathrm{16}\left(\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}×\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\right)=\mathrm{16}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{5}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$ \\ $$$${A}_{{red}} =\frac{\mathrm{4}×\mathrm{8}}{\mathrm{2}}−\mathrm{16}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{5}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right)−\left(\mathrm{16}−\mathrm{4}\pi\right) \\ $$$$=\frac{\mathrm{32}}{\mathrm{5}}+\mathrm{16}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\mathrm{4}\pi \\ $$$$=\mathrm{1}.\mathrm{25199} \\ $$
Commented by Tawa1 last updated on 23/Apr/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$