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Question Number 98290 by Rio Michael last updated on 12/Jun/20

Commented by Rio Michael last updated on 13/Jun/20

$$\mathrm{The}\mathrm{above}\mathrm{figure}\mathrm{shows}\mathrm{a}\mathrm{square}\mathrm{lamina}\mathrm{of}\mathrm{side}a\mathrm{with}\mathrm{vertex}$$$$\mathrm{inclined}\mathrm{at}\mathrm{an}\mathrm{angle}x,0⩽x⩽\frac{\pi }{2},\mathrm{placed}\mathrm{on}\mathrm{a}\mathrm{line}AB.$$$$\mathrm{show}\mathrm{that}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{lamina}\mathrm{is}\mathrm{at}\mathrm{a}\mathrm{distance}\frac{a\sqrt{2}}{2}\sin (x+\frac{\pi }{4})\mathrm{from}$$$$AB.\mathrm{The}\mathrm{lamina}\mathrm{is}\mathrm{rotated}\mathrm{completely}\mathrm{about}\mathrm{the}\mathrm{line}AB.$$$$\mathrm{use}\mathrm{the}\mathrm{second}\mathrm{law}\mathrm{of}\mathrm{Pappus}\mathrm{or}\mathrm{otherwise}\mathrm{show}\mathrm{that}$$$$\mathrm{the}\mathrm{volume}\mathrm{generated}\mathrm{is}\sqrt{2}\pi a^3\sin (x+\frac{\pi }{4}).$$

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