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Question Number 132564 by abony1303 last updated on 15/Feb/21

Answered by abony1303 last updated on 15/Feb/21

$$\mathrm{pls}\mathrm{help}.\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{shaded}\mathrm{region}$$

Answered by MJS_new last updated on 15/Feb/21

$$\mathrm{rotate}\mathrm{the}\mathrm{picture}\mathrm{by}-45°$$$$\Rightarrow$$$$\mathrm{big}\mathrm{upper}\mathrm{half}\mathrm{circle}$$$$y=\sqrt{100-x^2}$$$$\mathrm{small}\mathrm{upper}\mathrm{half}\mathrm{circle}$$$$y=5\sqrt{2}+\sqrt{25-x^2}$$$$\mathrm{they}\mathrm{intersect}\mathrm{at}x=\pm \frac{5\sqrt{14}}{4}$$$$\frac{1}{4}\mathrm{of}\mathrm{blue}\mathrm{region}$$$$\underset{0}{\stackrel{\frac{5\sqrt{14}}{4}}{\int }}(5\sqrt{2}+\sqrt{25-x^2}-\sqrt{100-x^2})dx=$$$$={[5\sqrt{2}x+\frac{x\sqrt{25-x^2}}{2}+\frac{25\arcsin \frac{x}{5}}{2}-\frac{x\sqrt{100-x^2}}{2}-50\arcsin \frac{x}{10}]}_0^{\frac{5\sqrt{14}}{4}}=$$$$=\frac{25}{4}(\sqrt{7}+2\arcsin \frac{\sqrt{14}}{4}-8\arcsin \frac{\sqrt{14}}{8})=$$$$=\frac{25}{4}(\sqrt{7}-\arctan \frac{1541\sqrt{7}}{393})\approx 7.31906$$$$\Rightarrow \mathrm{blue}\mathrm{area}=25(\sqrt{7}-\arctan \frac{1541\sqrt{7}}{393})\approx 29.2763$$

Commented by mr W last updated on 15/Feb/21

$$ithinkthisisthebestpath.$$

Answered by mr W last updated on 15/Feb/21

Commented by mr W last updated on 15/Feb/21

$$\cos \alpha =\frac{{\left(2r\right)}^2+{\left(\sqrt{2}r\right)}^2-r^2}{2\times 2r\times \sqrt{2}r}=\frac{5\sqrt{2}}{8}$$$$\cos \gamma =\frac{{\left(2r\right)}^2+r^2-{\left(\sqrt{2}r\right)}^2}{2\times 2r\times r}=\frac{3}{4}$$$$\beta =\alpha +\gamma$$$$orangeshadedsegment:$$$$A_{Or}=\frac{r^2}{2}(2\beta -\sin 2\beta )$$$$greenshadedsegment:$$$$A_{Gr}=\frac{{\left(2r\right)}^2}{2}(2\alpha -\sin 2\alpha )$$$$blueshadedarea:$$$$A_{blue}=2(A_{Or}-A_{Gr})$$$$=[2\beta -\sin 2\beta -8\alpha +4\sin 2\alpha ]r^2$$$$=[10\alpha +2\gamma -\sin (2\alpha +2\gamma )+4\sin 2\alpha ]r^2$$$$=[10\alpha +2\gamma +\sin 2\alpha (4-\cos 2\gamma )-\cos 2\alpha \sin 2\gamma ]r^2$$$$=2[\gamma -3\alpha +\sin \alpha \cos \alpha (5-2{\cos }^2\gamma )-(2{\cos }^2\alpha -1)\sin \gamma \cos \gamma ]r^2$$$$=2[\gamma -3\alpha +\frac{\sqrt{14}}{8}\times \frac{5\sqrt{2}}{8}(5-2\times \frac{3^2}{4^2})-(2\times \frac{25\times 2}{64}-1)\times \frac{\sqrt{7}}{4}\times \frac{3}{4}]r^2$$$$=2(\gamma -3\alpha +\frac{\sqrt{7}}{2})r^2$$$$=2({\cos }^{-1}\frac{3}{4}-3{\cos }^{-1}\frac{5\sqrt{2}}{8}+\frac{\sqrt{7}}{2})r^2$$$$\approx 1.171050076r^2\approx 29.276252$$

Commented by Tawa11 last updated on 06/Nov/21

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

Answered by mr W last updated on 15/Feb/21

Commented by mr W last updated on 15/Feb/21

$$r_1=a=2r=10$$$$r^2=r_2^2+{\left(\sqrt{2}r\right)}^2-2r_2\sqrt{2}r\cos \theta$$$$r_2^2-2\sqrt{2}\cos \theta {rr}_2+r^2=0$$$$r_2=r(\sqrt{2}\cos \theta +\sqrt{\cos 2\theta })$$$$r(\sqrt{2}\cos \alpha +\sqrt{\cos 2\alpha })=2r$$$$\sqrt{\cos 2\alpha }=2-\sqrt{2}\cos \alpha$$$$\cos 2\alpha =4-4\sqrt{2}\cos \alpha +2{\cos }^2\theta$$$$\cos \alpha =\frac{5\sqrt{2}}{8}$$$$\frac{1}{4}{A}_{blue}={\int }_0^{\alpha }\frac{(r_2^2-r_1^2)d\theta }{2}$$$$A_{blue}=2{\int }_0^{\alpha }(r_2^2-r_1^2)d\theta$$$$=2r^2{\int }_0^{\alpha }[{(\sqrt{2}\cos \theta +\sqrt{\cos 2\theta })}^2-4]d\theta$$$$=2r^2{\int }_0^{\alpha }({2\cos }^2\theta +\cos 2\theta +2\cos \theta \sqrt{2\cos 2\theta }-4)d\theta$$$$=2r^2{\int }_0^{\alpha }(2\cos 2\theta +2\cos \theta \sqrt{2(1-{2\sin }^2\theta )}-3)d\theta$$$$=2r^2{[\sin 2\theta -3\theta +{\sin }^{-1}\left(\sqrt{2}\sin \theta \right)+\sqrt{2}\sin \theta \sqrt{2{\cos }^2\theta -1}]}_0^{\alpha }$$$$=2r^2[\sin 2\alpha -3\alpha +{\sin }^{-1}\left(\sqrt{2}\sin \alpha \right)+\sqrt{2}\sin \alpha \sqrt{2{\cos }^2\alpha -1}]$$$$=2r^2[\frac{5\sqrt{7}}{16}-{3\cos }^{-1}\frac{5\sqrt{2}}{8}+{\sin }^{-1}\frac{\sqrt{7}}{4}+\frac{3\sqrt{7}}{16}]$$$$=[\sqrt{7}-{6\cos }^{-1}\frac{5\sqrt{2}}{8}+2{\sin }^{-1}\frac{\sqrt{7}}{4}]r^2$$$$=1.171050r^2=29.276252$$

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