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Question Number 214020 by ajfour last updated on 24/Nov/24

Commented by ajfour last updated on 24/Nov/24

$$Redarea=Bluearea.Findrin$$$$termsofk,\varphi .$$

Answered by dionigi last updated on 24/Nov/24

$$$$$${kp}^2=(1+\cos \phi )R$$$$p^2=\frac{(1+\cos \phi )R}{k}$$$$A_1=2(k{p}^3-\frac{k{p}^3}{3})=\frac{4k{p}^3}{3}$$$$A_2=\pi R^2-\phi R^2+\sin \phi \cos \phi R^2$$$$A_2=(\pi -\phi +\sin \phi \cos \phi )R^2$$$$A_1=A_2$$$$A_1^2=A_2^2$$$$\frac{16k^2{p}^6}{9}={(\pi -\phi +\sin \phi \cos \phi )}^2{R}^4$$$$\frac{16{(1+\cos \phi )}^3{R}^3}{9k}={(\pi -\phi +\sin \phi \cos \phi )}^2{R}^4$$$$R=\frac{16{(1+\cos \phi )}^3}{9k{(\pi -\phi +\sin \phi \cos \phi )}^2}$$

Commented by ajfour last updated on 24/Nov/24

$$Thankyou.$$

Answered by mr W last updated on 24/Nov/24

$$r(1+\cos \phi )={kp}^2$$$$\Rightarrow p=\sqrt{\frac{r(1+\cos \phi )}{k}}$$$$\frac{r^2(\pi -\phi )}{2}+\frac{r^2\sin \phi \cos \phi }{2}=\frac{2pr(1+\cos \phi )}{3}$$$$\frac{r^2(\pi -\phi )}{2}+\frac{r^2\sin \phi \cos \phi }{2}=\frac{2r(1+\cos \phi )\sqrt{r(1+\cos \phi )}}{3\sqrt{k}}$$$$\Rightarrow r=\frac{16{(1+\cos \phi )}^3}{9k{(\pi -\phi +\sin \phi \cos \phi )}^2}$$

Commented by mr W last updated on 24/Nov/24

Commented by mr W last updated on 24/Nov/24

Commented by ajfour last updated on 24/Nov/24

$$ThanksSir$$$$\Rightarrow for\phi =\frac{\pi }{2},k=1$$$$r=\frac{16}{9{\pi }^2}={\left(\frac{4}{3\pi }\right)}^2(Iamremindedof$$$$c.m.heightofsemi-disc)$$$$$$

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