I'm working on a question where I have to evaluate, amongst others, the variance of X and Y, where (X,Y) are uniformly distributed in the unit circle centered in (1,2).
Since this is a unit circle, $f_{xy}=\frac{1}{\pi}$.
I started by converting X and Y to polar coordinates: $x=rcos(\theta)+1$. I then calculated $E(X)$ and $E(X^2)$
$$E(X)=\frac{1}{\pi}\int_0^{2\pi}\int_0^{1}rcos(\theta)+1 drd\theta =2$$ $$E(X^2)=\frac{1}{\pi}\int_0^{2\pi}\int_0^{1}(rcos(\theta)+1)^2 drd\theta =\frac{7}{3}$$
I checked my integrals, and they give the correct result.
As you might have noticed, $E(X^2)-E(X)^2=-\frac{5}{3}$. Thus, the variance is negative. Where is my mistake? I have a feeling it might be related to my polar coordinates...
As you are concerned with variance you might as well have the circle centered at the origin. Then, \begin{align} Var(X)=E(X^2) &= {1\over2\pi}\int_{x=-1}^1\int_{y=-\sqrt{1-x^2}}^{\sqrt{1-x^2}}x^2\,dx\,dy\\ &= {1\over2\pi}\int_{x=-1}^1 x^2\,2\sqrt{1-x^2}\,dx\\ &= {1\over8}. \end{align}