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Question Number 181903 by Shrinava last updated on 01/Dec/22

Answered by mr W last updated on 02/Dec/22

$$atz:$$$$thecross-sectionis$$$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1+\frac{z^2}{c^2}$$$$whichisanellipsewithsemi-axes:$$$$A=\frac{a}{\sqrt{1+\frac{z^2}{c^2}}},B=\frac{b}{\sqrt{1+\frac{z^2}{c^2}}}.$$$$theareaofthecross-sectionisthen$$$$S\left(z\right)=\pi AB=\frac{\pi ab}{1+\frac{z^2}{c^2}}$$$$thevolumeforz\in [-h,h]is$$$$V=2{\int }_0^{h}S\left(z\right)dz$$$$=2\pi ab{\int }_0^h\frac{dz}{1+\frac{z^2}{c^2}}$$$$=2\pi abc{\int }_0^h\frac{d\left(\frac{z}{c}\right)}{1+{\left(\frac{z}{x}\right)}^2}$$$$=2\pi abc{\left[{\tan }^{-1}\frac{z}{c}\right]}_0^h$$$$=2\pi abc{\tan }^{-1}\frac{h}{c}✓$$$$withh=2inthequestion.$$

Commented by Shrinava last updated on 05/Dec/22

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}\mathrm{cool}$$

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