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 μ=(b/a) x=a cos θ y=b sin θ dx=−a sin θ dθ dy=b cos θ dθ ds=(√((dx)^2 +(dy)^2 ))=(√(a^2 sin^2 θ+b^2 cos^2 θ))dθ S=2×2π∫_0 ^(π/2) b sin θ (√(a^2 sin^2 θ+b^2 cos^2 θ))dθ =4πab∫_0 ^(π/2) sin θ (√(sin^2 θ+μ^2 cos^2 θ))dθ =4πab∫_(π/2) ^0 (√(1−(1−μ^2 )cos^2 θ))d(cos θ) =4πab∫_0 ^1 (√(1−kt^2 ))dt with k=1−μ^2 , t=cos θ =... =4πab[((sin^(−1) (√k)t)/(2(√k)))+((t(√(1−kt^2 )))/2)]_0 ^1 =4πab[((sin^(−1) (√k))/(2(√k)))+((√(1−k))/2)] =2πab[((sin^(−1) (√(1−μ^2 )))/( (√(1−μ^2 ))))+μ] with μ=(b/a) with a=b ⇒sphere a=b=R,μ=1 S=2πR^2 (1+1)=4πR^2](https://www.tinkutara.com/question/Q87355.png)
$${let}\:\mu=\frac{{b}}{{a}} \\ $$$${x}={a}\:\mathrm{cos}\:\theta \\ $$$${y}={b}\:\mathrm{sin}\:\theta \\ $$$${dx}=−{a}\:\mathrm{sin}\:\theta\:{d}\theta \\ $$$${dy}={b}\:\mathrm{cos}\:\theta\:{d}\theta \\ $$$${ds}=\sqrt{\left({dx}\right)^{\mathrm{2}} +\left({dy}\right)^{\mathrm{2}} }=\sqrt{{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}{d}\theta \\ $$$${S}=\mathrm{2}×\mathrm{2}\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {b}\:\mathrm{sin}\:\theta\:\sqrt{{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}{d}\theta \\ $$$$=\mathrm{4}\pi{ab}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\:\theta\:\sqrt{\mathrm{sin}^{\mathrm{2}} \:\theta+\mu^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}{d}\theta \\ $$$$=\mathrm{4}\pi{ab}\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \sqrt{\mathrm{1}−\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \:\theta}{d}\left(\mathrm{cos}\:\theta\right) \\ $$$$=\mathrm{4}\pi{ab}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{kt}^{\mathrm{2}} }{dt}\:\:{with}\:{k}=\mathrm{1}−\mu^{\mathrm{2}} ,\:{t}=\mathrm{cos}\:\theta \\ $$$$=… \\ $$$$=\mathrm{4}\pi{ab}\left[\frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{{k}}{t}}{\mathrm{2}\sqrt{{k}}}+\frac{{t}\sqrt{\mathrm{1}−{kt}^{\mathrm{2}} }}{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{4}\pi{ab}\left[\frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{{k}}}{\mathrm{2}\sqrt{{k}}}+\frac{\sqrt{\mathrm{1}−{k}}}{\mathrm{2}}\right] \\ $$$$=\mathrm{2}\pi{ab}\left[\frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{\mathrm{1}−\mu^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−\mu^{\mathrm{2}} }}+\mu\right]\:{with}\:\mu=\frac{{b}}{{a}} \\ $$$$ \\ $$$${with}\:{a}={b}\:\Rightarrow{sphere}\:{a}={b}={R},\mu=\mathrm{1} \\ $$$${S}=\mathrm{2}\pi{R}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{1}\right)=\mathrm{4}\pi{R}^{\mathrm{2}} \\ $$
Commented by ajfour last updated on 04/Apr/20
$${Exactly},\:{thanks}\:{Sir}. \\ $$$${S}=\mathrm{2}\pi{b}^{\mathrm{2}} +\mathrm{2}\pi{ab}\left(\frac{\mathrm{sin}^{−\mathrm{1}} \lambda}{\lambda}\right) \\ $$$$\:\:\:\lambda=\frac{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{a}}\:. \\ $$