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Question Number 104703 by abony1303 last updated on 23/Jul/20

Commented by abony1303 last updated on 23/Jul/20

$pls help too :)$
	Answered by bemath last updated on 23/Jul/20
	$vol =π∫_0 ^a {(sin x)^2 −(((2x)/π))^2 }dx = π ∫_0 ^a {(1/2)−(1/2)cos 2x−((4x^2 )/π^2 )} dx = π { (x/2)−((sin 2x)/4)−((4x^3 )/(3π^2 ))}_0 ^a = π{(a/2)−((sin 2a)/4)−((4a^3 )/(3π^2 ))} where a is solution from ⇒sin x = ((2x)/π)$
	$v o l = π \underset{0}{\int^{a}} {{(\sin x)}^{2} - {(\frac{2 x}{π})}^{2}} d x$ $= π \underset{0}{\int^{a}} {\frac{1}{2} - \frac{1}{2} \cos 2 x - \frac{4 x^{2}}{π^{2}}} d x$ $= π {\frac{x}{2} - \frac{\sin 2 x}{4} - \frac{4 x^{3}}{3 π^{2}}}_{0}^{a}$ $= π {\frac{a}{2} - \frac{\sin 2 a}{4} - \frac{4 a^{3}}{3 π^{2}}}$ $w h e r e a i s s o l u t i o n f r o m$ $\Rightarrow \sin x = \frac{2 x}{π}$
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