Solve Parametric Eq: Semicircle Mean Value w/ Respect to Theta

[John O' Meara]

The semicircle $$\mbox{f(x) = }\sqrt{a^2-x^2} \mbox{ -a} <=\mbox{ x }<= \mbox{a }$$, ( see my last thread) has the parametric equations $$x= }a cos\theta\mbox{, y=} a sin\theta, 0 <= \theta <= \pi$$, show that the mean value with respect to $$\theta$$ of the ordinates of the semicircle is $$2a/\pi(.64a)$$.
Can someone show how you can get an expression in $$\theta$$, so I can integrate it, I 'm new to parametric equations. Thanks for the help.
My attempt: $$a sin\theta = \sqrt{a^2-a^2cos^2\theta}\mbox{ which gives } a sin\theta = a sin\theta\mbox {which is what you would expect}$$

 

[HallsofIvy]

Well, you can't get what you wrote- it isn't true! For one thing, when $\theta= 0$, a sin(0)= 0 but $\sqrt{a^2- a^2 sin^2(0)}= a$. Did you mean $a cos\theta = \sqrt{a^2- a^2sin^2\theta}$? That true because $\sqrt{a^2- a^2sin^2\theta}= \sqrt{a^2(1- sin^2\theta )}= \sqrt{a^2 cos^2\theta}$.

 

[John O' Meara]

Sorry for the mistake I meant $$a sin\theta = \sqrt{a^2-a^2cos^2\theta}\mbox{ which gives } a sin\theta = a sin\theta$$, which is of no help.

 

[Dick]

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?

 

[John O' Meara]

$$\int_{\theta=0}^{\theta=\pi} \sqrt{a^2-a^2 cos^2\theta}dx$$, I don't know what dx is in terms of $$\theta$$. Can someone show me please? Thanks for the help.

 

[Dick]

Write dx=d(a*cos(theta)). It's just like finding du in a change of variable. Just differentiate!