Question Number 73673 by ajfour last updated on 14/Nov/19
Commented by ajfour last updated on 14/Nov/19
$${Q}\:\mathrm{73503}\:\:\:\left({another}\:{way}\:\:{a}\:{try}..\right) \\ $$
Commented by ajfour last updated on 14/Nov/19
Commented by mind is power last updated on 15/Nov/19
$${we}\:{can}\:{use}\:\:\:{parametric}\:{surface} \\ $$$${if}\:\:{f}:{IR}^{\mathrm{2}} \rightarrow{IR}^{\mathrm{3}} \:\:\:{a}\:{surface}\:\:\:{paramtric} \\ $$$${f}\left({u},{v}\right)=\left({a}\left({u},{v}\right),{b}\left({u},{v}\right),{c}\left({u},{v}\right)\right)\:{parametric}\:\:{of}\:{surface} \\ $$$${we}\:{will}\:{ginde}\:{parametrisation}\:{of}\:{ilimunated}\:{light} \\ $$$$\mathcal{A}\left({u}\right)=\int\int_{\mathcal{A}} \:\mid\frac{\partial{f}}{\partial{u}}\left({u},{v}\right)\wedge\frac{\partial{f}}{\partial{v}}\left({u},{v}\right)\mid{dudv} \\ $$$${i}\:{show}\:{steps} \\ $$$$\mathrm{1}\:{put}\:{repert} \\ $$$$\mathrm{2},{equation}\:{of}\:{semis}\:{disc}\:\therefore\:\:{D}\:\:\because{wich}\:{light}\:{pass}\:{throw} \\ $$$$ \\ $$$$\mathrm{3},\forall{a}\in{D}\rightarrow{a}'\in\mathcal{A}\cap{C}\:\:\:{since}\:{light}\:{horizontal} \\ $$$${C}\:{is}\:{Cone} \\ $$$${D}=\left(\mathrm{0},{pcos}\left({t}\right),{psin}\left({t}\right)\right),{t}\in\left[\frac{\pi}{\mathrm{2}},\frac{\mathrm{3}\pi}{\mathrm{2}}\right],{p}\in\left[\mathrm{0},\frac{{a}}{\:\sqrt{\mathrm{3}}}\right] \\ $$$$\Rightarrow{aa}\overset{\rightarrow} {'}={s}.\overset{\rightarrow} {{i}} \\ $$$$\overset{\rightarrow} {{i}}=\left(\mathrm{1},\mathrm{0},\mathrm{0}\right) \\ $$$${use}\:{C}=\left\{\left({x},{y},{z}\right)\in\mathbb{R}^{\mathrm{3}} \right\}\mid{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={z}^{\mathrm{2}} {tg}^{\mathrm{2}} \left(\theta\right) \\ $$$${we}\:{find}\:{s}\:{ther}\:{s}\:\left({t},{p}\right) \\ $$$${a}'\left({t},{p}\right)\:{is}\:{our}\:{paramtrisation}\:{of}\:\mathcal{A}\:{our}\:{f}\left({u},{v}\right)\:{below} \\ $$$${i}\:{will}\:{poste}\:{it}\:\:\:{this}\:{methode}\:{use}\:{somm}\:{topological}\:{object} \\ $$$${surfaces} \\ $$
Answered by ajfour last updated on 15/Nov/19
$${Let}\:{P}\:\:{be}\:{a}\:{point}\:{on}\:{circumference} \\ $$$${of}\:{circle}\:\:{y}^{\mathrm{2}} +\left({z}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${P}\:\left(\mathrm{0},\:{r}\mathrm{sin}\:\theta,\:{r}−{r}\mathrm{cos}\:\theta\right) \\ $$$${Its}\:{shadow}\:{on}\:{slant}\:{surface}\:{of}\:{cone} \\ $$$$\:{S}\left({s}\mathrm{sin}\:\alpha\mathrm{cos}\:\phi,\:{s}\mathrm{sin}\:\alpha\mathrm{sin}\:\phi,\:{a}\sqrt{\mathrm{3}}−{s}\mathrm{cos}\:\alpha\right) \\ $$$$\Rightarrow\:\:{s}\mathrm{sin}\:\alpha\mathrm{sin}\:\phi={r}\mathrm{sin}\:\theta\:\:\:\:\:\:\:\left(\alpha=\mathrm{30}°\right) \\ $$$$\:\:\:\:\:\:\:{a}\sqrt{\mathrm{3}}−{s}\mathrm{cos}\:\alpha={r}−{r}\mathrm{cos}\:\theta \\ $$$$\:\:\:\:{s}=\frac{{a}\sqrt{\mathrm{3}}−{r}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{cos}\:\alpha} \\ $$$$\:\:\:\mathrm{sin}\:\phi=\frac{{r}\mathrm{sin}\:\theta\mathrm{cos}\:\alpha}{\left({a}\sqrt{\mathrm{3}}−{r}+{r}\mathrm{cos}\:\theta\right)\mathrm{sin}\:\alpha} \\ $$$$\:\:\:\mathrm{sin}\:\phi=\frac{\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{3}}−\mathrm{1}+\mathrm{cos}\:\theta} \\ $$$$\:\:{ds}=\frac{−{r}\mathrm{sin}\:\theta{d}\theta}{\mathrm{cos}\:\alpha} \\ $$$$\:\:\frac{{s}}{\mathrm{2}{a}}=\frac{\rho}{{a}}\:\:\Rightarrow\:\:\rho=\frac{{s}}{\mathrm{2}} \\ $$$${A}=\int\rho\left(\mathrm{2}\phi\right)\left(−{ds}\right)\:=\:\int\left({s}\right)\left(−{ds}\right)\phi \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\:\:\pi} \left(\frac{{a}\sqrt{\mathrm{3}}−{r}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{cos}\:\alpha}\right)\left(\frac{{r}\mathrm{sin}\:\theta{d}\theta}{\mathrm{cos}\:\alpha}\right)\left(\mathrm{sin}^{−\mathrm{1}} \left\{\frac{{r}\mathrm{sin}\:\theta\mathrm{cos}\:\alpha}{\left({a}\sqrt{\mathrm{3}}−{r}+{r}\mathrm{cos}\:\theta\right)\mathrm{sin}\:\alpha}\right\}\right) \\ $$$$=\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}}\int_{\mathrm{0}} ^{\:\pi} \left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\right]\left[\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{3}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{cos}\:\theta}{\:\sqrt{\mathrm{3}}}}\right)\right]\mathrm{sin}\:\theta{d}\theta \\ $$$$=\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{9}}\int_{−\mathrm{1}} ^{\:\mathrm{1}} \left(\mathrm{2}+{t}\right)\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{2}+{t}}\right){dt} \\ $$$$\:\:{A}=\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{9}}\left(\frac{\pi}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{3}}{\mathrm{2}}\right)\approx\mathrm{1}.\mathrm{4728}{a}^{\mathrm{2}} \:. \\ $$$$\:\:{A}=\frac{\mathrm{2}{a}^{\mathrm{2}} \left(\mathrm{2}\pi\sqrt{\mathrm{3}}+\mathrm{9}\right)}{\mathrm{27}}\:. \\ $$$$\left({see}\:{Q}.\:\mathrm{73689}\:{on}\:{how}\:{to}\:{solve}\right. \\ $$$${the}\:{integral},{special}\:{thanks}\:{to}\: \\ $$$$\left.\:\:{MjS}\:{Sir}\right) \\ $$
Commented by mr W last updated on 14/Nov/19
$${nice}\:{solution}\:{sir}! \\ $$
Answered by liki last updated on 14/Nov/19
Commented by ajfour last updated on 14/Nov/19
$${wrong}\:{place}\:! \\ $$