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Question Number 183216 by mr W last updated on 23/Dec/22

Commented by mr W last updated on 23/Dec/22

Commented by mr W last updated on 23/Dec/22

$$thegeneralcaseforsphere\left(red\right)cut$$$$byacylinder\left(green\right)$$

Answered by mr W last updated on 23/Dec/22

$$given:R,r,a$$$$findthevolumeoftheportionof$$$$thespherecutbythecylinder$$$$b^2=a^2+r^2-2ar\cos \theta$$$$z=\sqrt{R^2-b^2}=\sqrt{R^2-a^2-r^2+2ar\cos \theta }$$$$a\sin \theta =(a-r\cos \theta )\tan \phi$$$$\Rightarrow \phi ={\tan }^{-1}\left(\frac{a\sin \theta }{a-r\cos \theta }\right)$$$$2bdb=2ar\sin \theta d\theta$$$$dS=2\phi bdb=2ar{\tan }^{-1}\left(\frac{a\sin \theta }{a-r\cos \theta }\right)\sin \theta d\theta$$$$V=2{\int }_0^{\pi }zdS$$$$V=4arR{\int }_0^{\pi }\sqrt{1-{\left(\frac{a}{R}\right)}^2-{\left(\frac{r}{R}\right)}^2+\frac{2ar}{R^2}\cos \theta }{\tan }^{-1}\left(\frac{\sin \theta }{1-\frac{r}{a}\cos \theta }\right)\sin \theta d\theta$$$$let\lambda =\frac{a}{R},\mu =\frac{r}{R}$$$$\Rightarrow V=4\mu \lambda R^3{\int }_0^{\pi }\sqrt{1-{\mu }^2-{\lambda }^2+2\mu \lambda \cos \theta }{\tan }^{-1}\left(\frac{\sin \theta }{1-\frac{\mu }{\lambda }\cos \theta }\right)\sin \theta d\theta$$

Commented by manxsol last updated on 23/Dec/22

$$Thanks,SirW,applause$$$$foryoureffort,veryclear$$$$$$

Commented by mr W last updated on 23/Dec/22

$$thankyoutoo!$$

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