If a random circle has a radius that is uniformly distributed on the interval $(0,2)$, what are the mean and variance of the area of the circle?
we have $A=\pi r^2$ and $r$ is uniformly distributed on $(0,2)$, thus, $F(A)=1$ on $(0,2)$ with $A$ is defined on $[0,\pi]$. SO,
$$F(A)=Pr(A<a)=Pr(\pi r^2<a)=Pr(-\sqrt{\frac{a}{\pi}}<r<\sqrt{\frac{a}{\pi}})$$ $$F(A)=\int_{0}^{\sqrt{\frac{a}{\pi}}}1\,dx=\sqrt{\frac{a}{\pi}}\qquad\text{then I will have} \;\;\;f(a)=\frac{1}{2a\pi}$$ When I calculate the mean, I didn't get $4\pi/3$. Where I did wrong?
If $r$ is uniformly distributed on $[0,2]$ then the density function for $r$ is $1/2$. Thus the expected value if the area is $$ \int_0^2 \pi r^2\, 1/2\, dr =4\pi/3 $$ and the variance is $$ \int_0^2 (\pi r^2- 4\pi/3)^2\, 1/2\, dr = 64\pi^2/45. $$
Where did you go wrong? Usually the notation $F$ stands for the cumulative distribution function, you probably meant the density function which is $f(r)=1/2$. If you wanted the CDF for the area, then $$ F(a)=P(\pi r^2\leq a)=P(0\leq r\leq \sqrt{a/\pi})=1/2 \sqrt{a/\pi} $$ (for $0\leq r\leq 2$, i.e. $0\leq a \leq \pi\, 2^2=4\pi$). You can then get the mean as $$ \int_0^{4\pi}a \, F'(a)\, da = \int_0^{4\pi}\, \frac{1}{\sqrt{16\, a\, \pi}}\, da=4\pi/3. $$