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Question Number 133470 by rs4089 last updated on 22/Feb/21

Answered by mr W last updated on 22/Feb/21

$$x=a\cos \theta$$$$y=b\sin \theta$$$$dV=-2\pi (2a-a\cos \theta )2ydx$$$$=4\pi a(2-\cos \theta )b\sin \theta a\sin \theta d\theta$$$$=4\pi a^{2}b(2-\cos \theta ){\sin }^2\theta d\theta$$$$V=4\pi a^{2}b{\int }_0^{\pi }(2-\cos \theta ){\sin }^2\theta d\theta$$$$V=4\pi a^{2}b{\int }_0^{\pi }(1-\cos 2\theta -\cos \theta {\sin }^2\theta )d\theta$$$$V=4\pi a^{2}b{[\theta -\frac{\sin 2\theta }{2}-\frac{{\sin }^3\theta }{3}]}_0^{\pi }$$$$V=4{\pi }^2{a}^{2}b$$$$V=2\pi \left(2a\right)A_{ellipse}$$

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