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I understand that $1/\pi$ is the pdf for uniform probability on the unit disk (with suitable constraints on $x^2 + y^2 =1$), because the area for radius 1 is just $\pi$.
However, how is the double integral set up to get a value of 1 when integrating?
I have worked with various combinations of 0,1, $\sqrt{1-x^2}$, etc. as the limits of integration and haven't been able to get a combination that works yet so that the calculation of the double integral comes to 1.

cumin
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  • This is a single integral, with respect to the uniform measure on the disk, not a double integral with respect to the Lebesgue measure on $\mathbb R^2$. – Xi'an Jun 08 '21 at 15:33
  • with a single integral over either x or y, of the joint pdf 1/pi, with limits either [-1,1] or [sqrt{1-x^2}], I can't get the result to be 1. – cumin Jun 08 '21 at 15:46
  • The single integral is over the disk, not in $x$ or $y$. – Xi'an Jun 08 '21 at 15:52
  • thanks, but I don't know how to set up that integral then. The pdf is a constant; what is the variable that the single integral is integrating over? – cumin Jun 08 '21 at 15:58
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    Without using measure theory, you can consider the curve $x^2+y^2=1$ (ie, the unit circle) as parameterised by its angle $\vartheta\in(0,2\pi)$ and integrate out in $\vartheta$. – Xi'an Jun 08 '21 at 16:15
  • "without using measure theory" :-) Thanks. I thought there would be a way to integrate the pdf as the other examples show in the class. – cumin Jun 08 '21 at 16:18
  • The technique of reducing a double integral to a sequence of two single integrals derives from Fubini's Theorem. It's a basic element in any second course or textbook in Calculus. You can find many examples here and at [math.se]. – whuber Jun 08 '21 at 21:55

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