Question Number 208896 by efronzo1 last updated on 26/Jun/24
Answered by MM42 last updated on 27/Jun/24
$${s}_{\mathrm{1}} =\mathrm{32}−\int_{\mathrm{0}} ^{\mathrm{4}} \sqrt{\mathrm{64}−{x}^{\mathrm{2}} }{dx}\:\:\:\:\:;\:\:{x}=\mathrm{8}{sin}\theta \\ $$$$\Rightarrow=\mathrm{32}−\mathrm{64}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{cos}^{\mathrm{2}} \theta{d}\theta \\ $$$$\Rightarrow{s}_{\mathrm{1}} =\mathrm{32}−\mathrm{8}\sqrt{\mathrm{3}}−\frac{\mathrm{16}\pi}{\mathrm{3}}=\:{y} \\ $$$$\int_{\mathrm{4}} ^{\mathrm{8}} \sqrt{\mathrm{64}−{x}^{\mathrm{2}} }\:{dx}=\frac{\mathrm{32}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}} \\ $$$$\int_{\mathrm{4}} ^{\mathrm{8}} \sqrt{\mathrm{16}−\left({x}−\mathrm{4}\right)^{\mathrm{2}} }{dx}\:\:\:\:;\:\:{x}−\mathrm{4}={u} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{4}} \sqrt{\mathrm{16}−{u}^{\mathrm{2}} }\:{du}=\mathrm{4}\pi \\ $$$$\Rightarrow{s}_{\mathrm{2}} =\frac{\mathrm{20}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}}\:={x} \\ $$$$\Rightarrow{Ans}=\frac{\mathrm{32}−\mathrm{8}\sqrt{\mathrm{3}}−\frac{\mathrm{16}\pi}{\mathrm{3}}}{\frac{\mathrm{20}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}}}\:\:\:\checkmark \\ $$$$ \\ $$
Commented by A5T last updated on 26/Jun/24
$$\mathrm{4}=\mathrm{8}{sin}\theta\Rightarrow{sin}\theta=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\theta=\frac{\pi}{\mathrm{6}} \\ $$$${Shouldn}'{t}\:{the}\:{second}\:{line}\:{be}:\:\mathrm{32}−\mathrm{64}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} {cos}^{\mathrm{2}} \theta{d}\theta? \\ $$
Commented by A5T last updated on 26/Jun/24
$${Also},\:{shouldn}'{t}\:{s}_{\mathrm{2}} ={x}=\int_{\mathrm{4}} ^{\mathrm{8}} \sqrt{\mathrm{64}−{x}^{\mathrm{2}} }{dx}−\frac{\mathrm{16}\pi}{\mathrm{4}} \\ $$$$=\frac{\mathrm{20}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}}\:? \\ $$
Answered by mr W last updated on 26/Jun/24
$${big}\:{circle}: \\ $$$${y}=\sqrt{\mathrm{8}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$${small}\:{circle}: \\ $$$${y}=\sqrt{\mathrm{4}^{\mathrm{2}} −\left({x}−\mathrm{4}\right)^{\mathrm{2}} }=\sqrt{\mathrm{8}{x}−{x}^{\mathrm{2}} } \\ $$$${Y}=\int_{\mathrm{0}} ^{\mathrm{4}} \left(\mathrm{8}−\sqrt{\mathrm{8}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right){dx}=\mathrm{32}−\mathrm{8}\sqrt{\mathrm{3}}−\frac{\mathrm{16}\pi}{\mathrm{3}} \\ $$$${X}=\int_{\mathrm{4}} ^{\mathrm{8}} \left(\sqrt{\mathrm{8}^{\mathrm{2}} −{x}^{\mathrm{2}} }−\sqrt{\mathrm{8}{x}−{x}^{\mathrm{2}} }\right){dx}=\frac{\mathrm{20}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}} \\ $$$$\frac{{Y}}{{X}}=\frac{\mathrm{32}−\mathrm{8}\sqrt{\mathrm{3}}−\frac{\mathrm{16}\pi}{\mathrm{3}}}{\frac{\mathrm{20}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}}}\approx\mathrm{0}.\mathrm{196} \\ $$$${or} \\ $$$${X}=\mathrm{4}×\mathrm{8}−\left(\mathrm{8}×\mathrm{8}−\frac{\mathrm{8}^{\mathrm{2}} \pi}{\mathrm{4}}−{Y}\right)−\frac{\mathrm{4}^{\mathrm{2}} \pi}{\mathrm{4}} \\ $$$$\:\:\:\:=−\mathrm{32}+\mathrm{12}\pi+\mathrm{32}−\mathrm{8}\sqrt{\mathrm{3}}−\frac{\mathrm{16}\pi}{\mathrm{3}} \\ $$$$\:\:\:\:=\frac{\mathrm{20}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}} \\ $$