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Question Number 195046 by Mathstar last updated on 22/Jul/23

Commented by Mathstar last updated on 23/Jul/23

$$$$$$\mathrm{Given}\mathrm{curve}\mathrm{ysin}\left(\mathrm{x}\right)\mathrm{and}\mathrm{a}\mathrm{green}$$$$\mathrm{square};\mathrm{solve}\mathrm{for}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{square}.$$

Answered by a.lgnaoui last updated on 23/Jul/23

$$\mathrm{aire}\left(\mathrm{partie}\mathrm{verte}\right)={\int }_{\frac{\mathit{\pi }}{2}-\mathrm{a}}^{\frac{\mathit{\pi }}{2}+\mathrm{a}}\mathrm{ydx}=2\mathbf{a}\sin \mathbf{a}$$$$\mathbf{pour}\mathrm{a}=\frac{\mathit{\pi }}{3}\mathrm{Aire}=\frac{\mathit{\pi }\sqrt{3}}{6}?$$

Commented by Mathstar last updated on 23/Jul/23

Thank you.

Answered by mr W last updated on 23/Jul/23

$$s=sidelengthofsquare$$$$\sin (\frac{\pi }{2}-\frac{s}{2})=s$$$$\cos \frac{s}{2}=s$$$$\Rightarrow s\approx 0.900367$$$$areaofsquare=s^2\approx 0.81$$

Commented by Mathstar last updated on 23/Jul/23

Commented by mr W last updated on 23/Jul/23

Commented by mr W last updated on 23/Jul/23

$$atpointA:$$$$x_A=\frac{\pi }{2}-\frac{s}{2}$$$$y_A=s$$$$pointAisonthecurcey=\sin x,hence$$$$s=\sin (\frac{\pi }{2}-\frac{s}{2})$$

Commented by Mathstar last updated on 23/Jul/23

Thank you so much, Mr W. It is indisputable that you are one of the best Mathematicians on this platform.

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