Question Number 87296 by ajfour last updated on 03/Apr/20 | ||
$${If}\:\:{ellipse}\:\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:\left({a}>{b}\right) \\ $$ $${is}\:{rotated}\:{about}\:{x}-{axis},\:{find}\:{the} \\ $$ $${surface}\:{of}\:{the}\:{solid}\:{of}\:{revolution}. \\ $$ | ||
Answered by mr W last updated on 04/Apr/20 | ||
$${let}\:\mu=\frac{{b}}{{a}} \\ $$ $${x}={a}\:\mathrm{cos}\:\theta \\ $$ $${y}={b}\:\mathrm{sin}\:\theta \\ $$ $${dx}=−{a}\:\mathrm{sin}\:\theta\:{d}\theta \\ $$ $${dy}={b}\:\mathrm{cos}\:\theta\:{d}\theta \\ $$ $${ds}=\sqrt{\left({dx}\right)^{\mathrm{2}} +\left({dy}\right)^{\mathrm{2}} }=\sqrt{{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}{d}\theta \\ $$ $${S}=\mathrm{2}×\mathrm{2}\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {b}\:\mathrm{sin}\:\theta\:\sqrt{{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}{d}\theta \\ $$ $$=\mathrm{4}\pi{ab}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\:\theta\:\sqrt{\mathrm{sin}^{\mathrm{2}} \:\theta+\mu^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}{d}\theta \\ $$ $$=\mathrm{4}\pi{ab}\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \sqrt{\mathrm{1}−\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta}{d}\left(\mathrm{cos}\:\theta\right) \\ $$ $$=\mathrm{4}\pi{ab}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{kt}^{\mathrm{2}} }{dt}\:\:{with}\:{k}=\mathrm{1}−\mu^{\mathrm{2}} ,\:{t}=\mathrm{cos}\:\theta \\ $$ $$=... \\ $$ $$=\mathrm{4}\pi{ab}\left[\frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{{k}}{t}}{\mathrm{2}\sqrt{{k}}}+\frac{{t}\sqrt{\mathrm{1}−{kt}^{\mathrm{2}} }}{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$ $$=\mathrm{4}\pi{ab}\left[\frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{{k}}}{\mathrm{2}\sqrt{{k}}}+\frac{\sqrt{\mathrm{1}−{k}}}{\mathrm{2}}\right] \\ $$ $$=\mathrm{2}\pi{ab}\left[\frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{\mathrm{1}−\mu^{\mathrm{2}} }}{\sqrt{\mathrm{1}−\mu^{\mathrm{2}} }}+\mu\right]\:{with}\:\mu=\frac{{b}}{{a}} \\ $$ $$ \\ $$ $${with}\:{a}={b}\:\Rightarrow{sphere}\:{a}={b}={R},\mu=\mathrm{1} \\ $$ $${S}=\mathrm{2}\pi{R}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{1}\right)=\mathrm{4}\pi{R}^{\mathrm{2}} \\ $$ | ||
Commented byajfour last updated on 04/Apr/20 | ||
$${Exactly},\:{thanks}\:{Sir}. \\ $$ $${S}=\mathrm{2}\pi{b}^{\mathrm{2}} +\mathrm{2}\pi{ab}\left(\frac{\mathrm{sin}^{−\mathrm{1}} \lambda}{\lambda}\right) \\ $$ $$\:\:\:\lambda=\frac{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{a}}\:. \\ $$ | ||