Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 218349 by Hanuda354 last updated on 07/Apr/25

Answered by vnm last updated on 07/Apr/25

$$$$$$\phi \left(\theta \right)=\theta -{\sin }^{-1}\frac{\sin \theta }{5}$$$$s\left(\alpha \right)=\frac{5}{2}(5\alpha -\sin \alpha )$$$$S=s\left(\phi \left(\frac{\pi }{3}\right)\right)-s\left(\phi \left(\frac{\pi }{2}\right)\right)+\frac{19\pi }{2}=$$$$$$$$24.174737721...$$

Commented by Hanuda354 last updated on 07/Apr/25

$$\mathrm{Thanks}$$

Answered by mr W last updated on 08/Apr/25

Commented by mr W last updated on 08/Apr/25

$$R^2={\rho }^2+b^2-2b\rho \cos (\frac{\pi }{2}+\theta )$$$${\rho }^2+2b\rho \sin \theta -(R^2-b^2)=0$$$$\rho =-b\sin \theta +\sqrt{R^2-b^2{\cos }^2\theta }$$$$A_1=\frac{1}{2}{\int }_0^{\theta }{\rho }^{2}d\theta$$$$=\frac{1}{2}{\int }_0^{\theta }(R^2-b^2\cos 2\theta -2b\sin \theta \sqrt{R^2-b^2{\cos }^2\theta })d\theta$$$$=\frac{R^2\theta }{2}-\frac{b^2\sin 2\theta }{4}+R^2{\int }_0^{\theta }\sqrt{1-{\left(\frac{b\cos \theta }{R}\right)}^2}d\left(\frac{b\cos \theta }{R}\right)$$$$=\frac{R^2\theta }{2}-\frac{b^2\sin 2\theta }{4}+\frac{R^2}{2}{[{\sin }^{-1}\frac{b\cos \theta }{R}+\frac{b\cos \theta }{R}\sqrt{1-{\left(\frac{b\cos \theta }{R}\right)}^2}]}_0^{\theta }$$$$=\frac{R^2\theta }{2}-\frac{b^2\sin 2\theta }{4}-\frac{R^2}{2}[{\sin }^{-1}\frac{b}{R}-{\sin }^{-1}\frac{b\cos \theta }{R}+\frac{b}{R}\sqrt{1-\frac{b^2}{R^2}}-\frac{b\cos \theta }{R}\sqrt{1-{\left(\frac{b\cos \theta }{R}\right)}^2}]$$$$=\frac{R^2}{2}(\theta -{\sin }^{-1}\frac{b}{R}+{\sin }^{-1}\frac{b\cos \theta }{R})-\frac{b^2\sin 2\theta }{4}-\frac{bR}{2}[\sqrt{1-\frac{b^2}{R^2}}-\cos \theta \sqrt{1-{\left(\frac{b\cos \theta }{R}\right)}^2}]$$$$withb=1,R=5,\theta =\frac{\pi }{6}$$$$A_1=\frac{25}{2}(\frac{\pi }{6}-{\sin }^{-1}\frac{1}{5}+{\sin }^{-1}\frac{\sqrt{3}}{10})-\frac{\sqrt{3}}{8}-\frac{5}{2}(\sqrt{1-\frac{1}{25}}-\frac{\sqrt{3}}{2}\sqrt{1-\frac{3}{100}})$$$$\approx 5.670392$$$$$$$$h^2=a(b+R)=4\times (1+5)=24$$$$\Rightarrow r=\frac{h}{2}=\sqrt{6}$$$$A_{shade}=\frac{\pi R^2}{2}-\frac{\pi r^2}{2}-A_1$$$$=\frac{\pi \times 5^2}{2}-\frac{\pi \times 6}{2}-A_1$$$$\approx 24.174738$$

Commented by Hanuda354 last updated on 08/Apr/25

$$\mathrm{Nice}!!!\mathrm{Thnks}$$

Answered by mr W last updated on 09/Apr/25

Commented by mr W last updated on 09/Apr/25

$$\frac{\sin (\theta -\phi )}{b}=\frac{\sin \theta }{R}$$$$\Rightarrow \phi =\theta -{\sin }^{-1}\frac{b\sin \theta }{R}$$$$A=\frac{R^2\phi }{2}-\frac{bR\sin \phi }{2}=\frac{R^2}{2}(\phi -\frac{b\sin \phi }{R})$$$$A=\frac{R^2}{2}[\theta -{\sin }^{-1}\frac{b\sin \theta }{R}-\frac{b}{R}\sin (\theta -{\sin }^{-1}\frac{b\sin \theta }{R})]$$$$A_1=A{\mid }_{\theta =\frac{\pi }{2}}-A{\mid }_{\theta =\frac{\pi }{3}}$$$$=\frac{5^2}{2}[\frac{\pi }{2}-{\sin }^{-1}\frac{1}{5}-\frac{1}{5}\sin (\frac{\pi }{2}-{\sin }^{-1}\frac{1}{5})]$$$$-\frac{5^2}{2}[\frac{\pi }{3}-{\sin }^{-1}\frac{\sqrt{3}}{10}-\frac{1}{5}\sin (\frac{\pi }{3}-{\sin }^{-1}\frac{\sqrt{3}}{10})]$$$$=\frac{5^2}{2}[\frac{\pi }{6}-{\sin }^{-1}\frac{1}{5}+{\sin }^{-1}\frac{\sqrt{3}}{10}-\frac{1}{5}\cos \left({\sin }^{-1}\frac{1}{5}\right)+\frac{1}{5}\sin (\frac{\pi }{3}-{\sin }^{-1}\frac{\sqrt{3}}{10})]$$$$\approx 5.679392$$$$A_{shade}=\frac{\pi \times 5^2}{2}-\frac{\pi \times {\left(\sqrt{6}\right)}^2}{2}-A_1$$$$\approx 24.174738$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com