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Question Number 87296 by ajfour last updated on 03/Apr/20

$$Ifellipse\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$$ $$isrotatedaboutx-axis,findthe$$ $$surfaceofthesolidofrevolution.$$

Answered by mr W last updated on 04/Apr/20

$$let\mu =\frac{b}{a}$$ $$x=a\cos \theta$$ $$y=b\sin \theta$$ $$dx=-a\sin \theta d\theta$$ $$dy=b\cos \theta d\theta$$ $$ds=\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}=\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }d\theta$$ $$S=2\times 2\pi {\int }_0^{\frac{\pi }{2}}b\sin \theta \sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }d\theta$$ $$=4\pi ab{\int }_0^{\frac{\pi }{2}}\sin \theta \sqrt{{\sin }^2\theta +{\mu }^2{\cos }^2\theta }d\theta$$ $$=4\pi ab{\int }_{\frac{\pi }{2}}^0\sqrt{1-(1-{\mu }^2){\cos }^2\theta }d\left(\cos \theta \right)$$ $$=4\pi ab{\int }_0^1\sqrt{1-{kt}^2}dtwithk=1-{\mu }^2,t=\cos \theta$$ $$=...$$ $$=4\pi ab{[\frac{{\sin }^{-1}\sqrt{k}t}{2\sqrt{k}}+\frac{t\sqrt{1-{kt}^2}}{2}]}_0^1$$ $$=4\pi ab[\frac{{\sin }^{-1}\sqrt{k}}{2\sqrt{k}}+\frac{\sqrt{1-k}}{2}]$$ $$=2\pi ab[\frac{{\sin }^{-1}\sqrt{1-{\mu }^2}}{\sqrt{1-{\mu }^2}}+\mu ]with\mu =\frac{b}{a}$$ $$$$ $$witha=b\Rightarrow spherea=b=R,\mu =1$$ $$S=2\pi R^2(1+1)=4\pi R^2$$

Commented byajfour last updated on 04/Apr/20

$$Exactly,thanksSir.$$ $$S=2\pi b^2+2\pi ab\left(\frac{{\sin }^{-1}\lambda }{\lambda }\right)$$ $$\lambda =\frac{\sqrt{a^2-b^2}}{a}.$$

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