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Suppose a point has a random location in the circle of radius 1 around the origin. The coordinates $(X,Y)$ of the point have a joint density

$f_{X,Y}(x,y) = \begin{cases}c(x^2+y^2)&\mathrm{\ if \ } x^2+y^2\le1\\ 0&\mathrm{\ otherwise\ }\end{cases}$

How do I find $c$? Currently, I have the following.

$\int_{-1}^1\int_{-1}^{\sqrt{1-y^2}} c(x^2+y^2)dxdy = 1$

Is this correct?

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    @Andrew I think you meant over $[-\sqrt{1-y^2},\sqrt{1-y^2}]$, maybe?2012-09-28

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An efficient way to do it is to switch to polar coordinates. We have $x^2+y^2=r^2$, so our integral is equal to $\int_{\theta=0}^{2\pi}\left(\int_{r=0}^1 (cr^2)r\,dr\right)d\theta.$. The integration is exceptionally easy. Then, as you did, you will be setting the result equal to $1$ to evaluate $c$.

Your procedure will also succeed, with somewhat more work. A small change needs to be made, for correctness.

If one does it more or less your way, the inner integral needs to be from $x=-\sqrt{1-y^2}$ to $x=\sqrt{1-y^2}$. Out of habit, other things being equal, I slightly prefer integrating first with respect to $y$, from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$.

I would suggest exploiting symmetry . Integrate over the first quadrant, and multiply the result by $4$.

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    If you do the integration the long way, then after a fair bit of effort you will find that the constant is $\frac{2}{\pi}$.2012-09-28