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Parametric equation

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]
 
Last edited:

Answers and Replies

HallsofIvy
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Well, you can't get what you wrote- it isn't true! For one thing, when [itex]\theta= 0[/itex], a sin(0)= 0 but [itex]\sqrt{a^2- a^2 sin^2(0)}= a[/itex]. Did you mean [itex]a cos\theta = \sqrt{a^2- a^2sin^2\theta}[/itex]? That true because [itex]\sqrt{a^2- a^2sin^2\theta}= \sqrt{a^2(1- sin^2\theta )}= \sqrt{a^2 cos^2\theta}[/itex].
 
Sorry for the mistake I meant [tex] a sin\theta = \sqrt{a^2-a^2cos^2\theta}\mbox{ which gives } a sin\theta = a sin\theta[/tex], which is of no help.
 
Dick
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It also shows y=sqrt(a^2-x^2), which shows your parametric representation is correct. Now just write your integral of y dx in terms of theta. What's dx?
 
[tex]\int_{\theta=0}^{\theta=\pi} \sqrt{a^2-a^2 cos^2\theta}dx[/tex], I don't know what dx is in terms of [tex]\theta[/tex]. Can someone show me please? Thanks for the help.
 
Dick
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Write dx=d(a*cos(theta)). It's just like finding du in a change of variable. Just differentiate!
 

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