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The semicircle [tex] \mbox{f(x) = }\sqrt{a^2-x^2} \mbox{ -a} <=\mbox{ x }<= \mbox{a }[/tex], ( see my last thread) has the parametric equations [tex]x= }a cos\theta\mbox{, y=} a sin\theta, 0 <= \theta <= \pi[/tex], show that the mean value with respect to [tex]\theta[/tex] of the ordinates of the semicircle is [tex] 2a/\pi(.64a)[/tex].

Can someone show how you can get an expression in [tex]\theta[/tex], so I can integrate it, I 'm new to parametric equations. Thanks for the help.

My attempt: [tex] a sin\theta = \sqrt{a^2-a^2cos^2\theta}\mbox{ which gives } a sin\theta = a sin\theta\mbox {which is what you would expect}[/tex]

Can someone show how you can get an expression in [tex]\theta[/tex], so I can integrate it, I 'm new to parametric equations. Thanks for the help.

My attempt: [tex] a sin\theta = \sqrt{a^2-a^2cos^2\theta}\mbox{ which gives } a sin\theta = a sin\theta\mbox {which is what you would expect}[/tex]

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