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Question Number 73503 by ajfour last updated on 13/Nov/19

Commented by ajfour last updated on 13/Nov/19

$$Findtheilluminatedcurved$$$$areaofconewithradiusa,and$$$$altitudea\sqrt{3}.Onlytherighthalf$$$$ofconeistherewithlefthalf$$$$coveredbutwithalargestcircular$$$$openinginit.Lightisincident$$$$fromaplanewallsending$$$$horizontalparallelrays.$$

Commented by mr W last updated on 14/Nov/19

$$howistheanswer,iftheopeningis$$$$asquare?$$

Answered by mr W last updated on 13/Nov/19

Commented by mr W last updated on 14/Nov/19

$${OD}^{\prime }=\sqrt{3}a$$$$OC=OD=2a$$$$C^{\prime }{D}^{\prime }=a\sin \theta$$$${OC}^{\prime }=\sqrt{3a^2+a^2{\sin }^2\theta }=a\sqrt{3+{\sin }^2\theta }$$$$\mathrm{\angle }{C}^{\prime }{OD}^{\prime }=\phi$$$$\tan \phi =\frac{\sin \theta }{\sqrt{3}}$$$$\Rightarrow \cos \phi =\frac{\sqrt{3}}{\sqrt{3+{\sin }^2\theta }}$$$$\Rightarrow {4\cos }^2\phi -3=\frac{3{\cos }^2\theta }{3+{\sin }^2\theta }$$$$r=\frac{a}{\sqrt{3}}$$$$OM=\frac{2a}{\sqrt{3}}$$$$r^2={OA}^{\prime 2}+{OM}^2-2\times {OA}^{\prime }\times OM\cos \phi$$$$\frac{a^2}{3}={OA}^{\prime 2}+\frac{4a^2}{3}-2\times {OA}^{\prime }\times \frac{2a}{\sqrt{3}}\cos \phi$$$${OA}^{\prime 2}-{OA}^{\prime }\times \frac{4a}{\sqrt{3}}\cos \phi +a^2=0$$$${OA}^{\prime }=\frac{2a\cos \phi }{\sqrt{3}}-\sqrt{\frac{4a^2{\cos }^2\phi }{3}-a^2}$$$${OA}^{\prime }=\frac{(2\cos \phi -\sqrt{4{\cos }^2\phi -3})a}{\sqrt{3}}$$$${OB}^{\prime }=\frac{(2\cos \phi +\sqrt{4{\cos }^2\phi -3})a}{\sqrt{3}}$$$$\frac{OA}{{OA}^{\prime }}=\frac{OC}{{OC}^{\prime }}=\frac{2}{\sqrt{3+{\sin }^2\theta }}$$$$\Rightarrow OA=\frac{2(2\cos \phi -\sqrt{4{\cos }^2\phi -3})a}{\sqrt{3(3+{\sin }^2\theta )}}$$$$similarly$$$$\Rightarrow OB=\frac{2(2\cos \phi +\sqrt{4{\cos }^2\phi -3})a}{\sqrt{3(3+{\sin }^2\theta )}}$$$$\stackrel{⌢}{CD}=a\theta =2a\varphi$$$$\Rightarrow \varphi =\frac{\theta }{2}$$$$dA=\frac{1}{2}({OB}^2-{OA}^2)d\varphi$$$$dA=2a^2\left[\frac{{(2\cos \phi +\sqrt{4{\cos }^2\phi -3})}^2-{(2\cos \phi -\sqrt{4{\cos }^2\phi -3})}^2}{3(3+{\sin }^2\theta )}\right]d\varphi$$$$dA=\frac{8a^2}{3}\left[\frac{\cos \phi \sqrt{4{\cos }^2\phi -3}}{3+{\sin }^2\theta }\right]d\theta$$$$dA=\frac{8a^2\cos \theta d\theta }{{(3+{\sin }^2\theta )}^2}$$$$A=2\times 8a^2{\int }_0^{\frac{\pi }{2}}\frac{\cos \theta d\theta }{{(3+{\sin }^2\theta )}^2}$$$$A=16a^2{\int }_0^{\frac{\pi }{2}}\frac{d\left(\sin \theta \right)}{{(3+{\sin }^2\theta )}^2}$$$$A=16a^2{\int }_0^1\frac{dt}{{(3+t^2)}^2}$$$$A=\frac{8a^2}{3}{[\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{t}{\sqrt{3}}+\frac{t}{t^2+3}]}_0^1$$$$A=\frac{8a^2}{3}(\frac{1}{\sqrt{3}}\times \frac{\pi }{6}+\frac{1}{4})$$$$\Rightarrow A=\frac{2(2\sqrt{3}\pi +9)a^2}{27}\approx 1.4728a^2$$$$$$$$Areaofsurfaceofsemi-cone:$$$$\frac{1}{2}\times \frac{\pi }{2}\times {\left(2a\right)}^2=\pi a^2$$$$\frac{1.4728a^2}{\pi a^2}=0.47=47\mathrm{\%}$$$$i.e.47\mathrm{\%}ofthesemiconesurfaceis$$$$lighted.$$

Commented by mr W last updated on 13/Nov/19

Commented by mr W last updated on 13/Nov/19

Commented by mr W last updated on 13/Nov/19

$$thisishowthelightedareareally$$$$lookslike:$$

Commented by mr W last updated on 14/Nov/19

Commented by ajfour last updated on 14/Nov/19

$$ThanksSir,\mathcal{SPLENDID}solution!$$$$(nicereverseswing!).$$

Commented by mr W last updated on 14/Nov/19

$$thanksforviewingsir!{it}^{\prime }savery$$$$nicequestion!$$

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