Given a circle of radius 1 with uniform distribution, let X and Y be the random variables that associate a point $(x,y)$ to $x$ and $y$ respectively. Find the expected value of X, $E(X)$, its variance $Var(X)$ and the covariance $cov(X,Y)$.
I have tried solving this problem, but I don't know if I'm doing it right. My attempt is as follows:
Knowing that the area of the circle is $\pi$, and supposing my circle is centered in the origin I could define: $$ p_{X,Y}(x,y)= \begin{cases} 1/\pi & \text{if } x^2+y^2\leq 1\\ 0 & \text{if } x^2 + y^2>1 \end{cases} $$ Using that calculate $p_X(x)$ $$ p_{X}(x) = \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\frac{1}{\pi}dy = \frac{y}{\pi}\Big|_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} = \frac{\sqrt{1-x^2}}{\pi} + \frac{\sqrt{1-x^2}}{\pi} = \frac{2}{\pi}\sqrt{1-x^2} $$
For the expected value I get: $$ E[X] = \int_{-\infty}^{\infty}xp_X(x)dx = \int_{-1}^{1}\frac{2}{\pi}x\sqrt{1-x^2}dx = \frac{2}{\pi}\int_{-1}^{1}x\sqrt{1-x^2}dx = 0 $$ For the variance I get: $$ Var(X) = E[X^2] - E[X]^2 = E[X^2] = \int_{-1}^{1}x^2p_X(x)dx = \int_{-1}^{1}x^2\frac{2}{\pi}\sqrt{1-x^2}dx=\frac{1}{4} $$ And finally, for the covariance: $$ cov(X,Y) = E[XY] - E[X]E[Y] = E[XY] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}xy\cdot p_{X,Y}dxdy = \int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}xy\cdot p_{X,Y}dxdy = \int_{-1}^{1}y\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} x\frac{1}{\pi}dxdy = \int_{-1}^{1} y\left( \frac{x^2}{2\pi}\Big|_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\right) dy = \int_{-1}^{1}y\left(0\right) = 0 $$
I found it a little weird getting these result such as the ones in the variance and covariance. I got $cov(X,Y)=0$ even though $X$ and $Y$ are definitely not independent (I know that if $cov(X,Y)=0$ it doesn't necessarily imply that $X$ and $Y$ are independent, but I still can't get why this happens or if it's correct at all). As for the remaining results, I'm just starting with this topic, so I still don't know if I'm doing these exercises correctly. Any help is appreciated, thanks.