(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

2. Relevant equations

I have that [tex]p(\theta)d\theta = \frac{1}{\pi}d\theta[/tex], this is definitely correct.

Also y=r*sin(theta) so dy=r*cos(theta)*d(theta).

Substituting d(theta) in above and simplifying, I have:

[tex] p(y)dy = \frac{1}{pi} \frac{dy}{\sqrt{r^2-y^2}} [/tex]

The problem is, integrating this from y=0 to y=r, I obtain 1/2, and not 1, as expected. I don't understand why this is.... anyone?

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# Homework Help: Probability and expectation values

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