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Question Number 183165 by universe last updated on 21/Dec/22

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

Commented by mr W last updated on 21/Dec/22

$$r=a\cos \theta$$$$dr=-a\sin \theta d\theta$$$$dA=-2\theta rdr=2a^2\theta \cos \theta \sin \theta d\theta =a^2\theta \sin 2\theta d\theta$$$$z=\sqrt{a^2-r^2}=\sqrt{a^2(1-{\cos }^2\theta )}=a\sin \theta$$$$dV=2zdA=2a^3\theta \sin 2\theta \sin \theta d\theta$$$$V=2a^3{\int }_0^{\frac{\pi }{2}}\theta \sin 2\theta \sin \theta d\theta$$$$V=2a^3\times \frac{3\pi -4}{9}$$$$\Rightarrow V=\frac{2(3\pi -4)a^3}{9}\approx 1.2055a^3$$

Commented by universe last updated on 22/Dec/22

$$thankssir$$

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

$$iftheradiusofthesphereisR>a,$$$$andlet\frac{a}{R}=\lambda <1,$$$$thenwehave$$$$z=\sqrt{R^2-r^2}=R\sqrt{1-{\lambda }^2{\cos }^2\theta }$$$$dA={\lambda }^2{R}^2\theta \sin 2\theta d\theta$$$$dV=2zdA=2{\lambda }^2{R}^3\theta \sin 2\theta \sqrt{1-{\lambda }^2{\cos }^2\theta }d\theta$$$$V=2{\lambda }^2{R}^3{\int }_0^{\frac{\pi }{2}}\theta \sin 2\theta \sqrt{1-{\lambda }^2{\cos }^2\theta }d\theta$$$$thisintegralcannotbeexactlysolved.$$

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