Question Number 133470 by rs4089 last updated on 22/Feb/21
Answered by mr W last updated on 22/Feb/21
![x=a cos θ y=b sin θ dV=−2π(2a−a cos θ)2ydx =4πa(2−cos θ)bsin θasin θdθ =4πa^2 b(2−cos θ)sin^2 θdθ V=4πa^2 b∫_0 ^π (2−cos θ)sin^2 θdθ V=4πa^2 b∫_0 ^π (1−cos 2θ−cos θsin^2 θ)dθ V=4πa^2 b[θ−((sin 2θ)/2)−((sin^3 θ)/3)]_0 ^π V=4π^2 a^2 b V=2π(2a)A_(ellipse)](https://www.tinkutara.com/question/Q133477.png)
$${x}={a}\:\mathrm{cos}\:\theta \\ $$$${y}={b}\:\mathrm{sin}\:\theta \\ $$$${dV}=−\mathrm{2}\pi\left(\mathrm{2}{a}−{a}\:\mathrm{cos}\:\theta\right)\mathrm{2}{ydx} \\ $$$$=\mathrm{4}\pi{a}\left(\mathrm{2}−\mathrm{cos}\:\theta\right){b}\mathrm{sin}\:\theta{a}\mathrm{sin}\:\theta{d}\theta \\ $$$$=\mathrm{4}\pi{a}^{\mathrm{2}} {b}\left(\mathrm{2}−\mathrm{cos}\:\theta\right)\mathrm{sin}^{\mathrm{2}} \:\theta{d}\theta \\ $$$${V}=\mathrm{4}\pi{a}^{\mathrm{2}} {b}\int_{\mathrm{0}} ^{\pi} \left(\mathrm{2}−\mathrm{cos}\:\theta\right)\mathrm{sin}^{\mathrm{2}} \:\theta{d}\theta \\ $$$${V}=\mathrm{4}\pi{a}^{\mathrm{2}} {b}\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta−\mathrm{cos}\:\theta\mathrm{sin}^{\mathrm{2}} \:\theta\right){d}\theta \\ $$$${V}=\mathrm{4}\pi{a}^{\mathrm{2}} {b}\left[\theta−\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}−\frac{\mathrm{sin}^{\mathrm{3}} \:\theta}{\mathrm{3}}\right]_{\mathrm{0}} ^{\pi} \\ $$$${V}=\mathrm{4}\pi^{\mathrm{2}} {a}^{\mathrm{2}} {b} \\ $$$${V}=\mathrm{2}\pi\left(\mathrm{2}{a}\right){A}_{{ellipse}} \\ $$