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Question Number 16613 by ajfour last updated on 24/Jun/17

Commented by ajfour last updated on 24/Jun/17

$$\mathrm{From}\mathrm{the}\mathrm{endpoint}\mathrm{of}\mathrm{a}\mathrm{diameter}$$$$\mathrm{of}\mathrm{a}\mathrm{sphere}\mathrm{draw}\mathrm{a}\mathrm{chord}\mathrm{so}\mathrm{that}$$$$\mathrm{the}\mathrm{surface}\mathrm{generated}\mathrm{by}\mathrm{a}\mathrm{rotation}$$$$\mathrm{about}\mathrm{this}\mathrm{diameter}\mathrm{divides}\mathrm{the}$$$$\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{sphere}\mathrm{into}\mathrm{two}$$$$\mathrm{equal}\mathrm{parts}.\mathrm{Determine}\mathrm{the}\mathrm{angle}$$$$\mathrm{between}\mathrm{the}\mathrm{chord}\mathrm{and}\mathrm{the}\mathrm{diameter}.$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 24/Jun/17

$$mrAjfour!wayyoursphereissounhappy?$$

Answered by mrW1 last updated on 26/Jun/17

$$\mathrm{with}\mathrm{h}=\mathrm{height}\mathrm{of}\mathrm{cap}$$$$$$$$\pi {\mathrm{h}}^2(\mathrm{R}-\frac{\mathrm{h}}{3})+\frac{1}{3}\pi \mathrm{h}{(2\mathrm{R}-\mathrm{h})}^2=\frac{1}{2}\times \frac{4}{3}\pi {\mathrm{R}}^3$$$${\mathrm{h}}^2(3\mathrm{R}-\mathrm{h})+\mathrm{h}{(2\mathrm{R}-\mathrm{h})}^2={2\mathrm{R}}^3$$$${3\mathrm{Rh}}^2-{\mathrm{h}}^3+{4\mathrm{R}}^2\mathrm{h}-{4\mathrm{Rh}}^2+{\mathrm{h}}^3-{2\mathrm{R}}^3=0$$$${\mathrm{h}}^2-4\mathrm{Rh}+{2\mathrm{R}}^2=0$$$$\mathrm{h}=\frac{4\mathrm{R}-\mathrm{R}\sqrt{16-4\times 2}}{2}=(2-\sqrt{2})\mathrm{R}$$$$\cos 2\theta =\frac{\mathrm{R}-\mathrm{h}}{\mathrm{R}}=\sqrt{2}-1$$$$\Rightarrow \theta =\frac{1}{2}{\cos }^{-1}(\sqrt{2}-1)=32.7°$$

Commented by ajfour last updated on 26/Jun/17

$$\mathrm{Sir},\mathrm{answer}\mathrm{given}\mathrm{is}\theta ={\cos }^{-1}\left(\frac{1}{\sqrt[4]{2^{}}}\right).$$

Commented by mrW1 last updated on 26/Jun/17

$${\mathrm{That}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{same}:$$$$\theta =\frac{1}{2}{\cos }^{-1}(\sqrt{2}-1)={\cos }^{-1}\frac{1}{{}^4\sqrt{2}}$$$$$$$$\cos 2\theta =\sqrt{2}-1$$$${2\cos }^2\theta -1=\sqrt{2}-1$$$${2\cos }^2\theta =\sqrt{2}$$$${\cos }^2\theta =\frac{1}{\sqrt{2}}$$$$\cos \theta =\frac{1}{{}^4\sqrt{2}}$$$$\theta ={\cos }^{-1}\left(\frac{1}{{}^4\sqrt{2}}\right)$$

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