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Question-104703




Question Number 104703 by abony1303 last updated on 23/Jul/20
Commented by abony1303 last updated on 23/Jul/20
pls help too :)
$$\left.\mathrm{pls}\:\mathrm{help}\:\mathrm{too}\::\right) \\ $$
Answered by bemath last updated on 23/Jul/20
vol =π∫_0 ^a  {(sin x)^2 −(((2x)/π))^2 }dx  = π ∫_0 ^a  {(1/2)−(1/2)cos 2x−((4x^2 )/π^2 )} dx  = π { (x/2)−((sin 2x)/4)−((4x^3 )/(3π^2 ))}_0 ^a   = π{(a/2)−((sin 2a)/4)−((4a^3 )/(3π^2 ))}  where a is solution from  ⇒sin x = ((2x)/π)
$${vol}\:=\pi\underset{\mathrm{0}} {\overset{{a}} {\int}}\:\left\{\left(\mathrm{sin}\:{x}\right)^{\mathrm{2}} −\left(\frac{\mathrm{2}{x}}{\pi}\right)^{\mathrm{2}} \right\}{dx} \\ $$$$=\:\pi\:\underset{\mathrm{0}} {\overset{{a}} {\int}}\:\left\{\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\pi^{\mathrm{2}} }\right\}\:{dx} \\ $$$$=\:\pi\:\left\{\:\frac{{x}}{\mathrm{2}}−\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{4}}−\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}\pi^{\mathrm{2}} }\right\}_{\mathrm{0}} ^{{a}} \\ $$$$=\:\pi\left\{\frac{{a}}{\mathrm{2}}−\frac{\mathrm{sin}\:\mathrm{2}{a}}{\mathrm{4}}−\frac{\mathrm{4}{a}^{\mathrm{3}} }{\mathrm{3}\pi^{\mathrm{2}} }\right\} \\ $$$${where}\:{a}\:{is}\:{solution}\:{from} \\ $$$$\Rightarrow\mathrm{sin}\:{x}\:=\:\frac{\mathrm{2}{x}}{\pi} \\ $$

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