Radius of a circle is approximately measured. Considering that its size is uniformly distributed on an interval $[a,b],(a,b>0)$, find the distribution of probabilities of the circle area, its expectation and variance.
My attempt:
$1)$ $F_A(A)=?$ $$p_R(R)=\frac{1}{b-a}$$ $$A=\pi\frac{R^2}{4}$$ $$R=2\sqrt{\frac{A}{\pi}}$$ $$\frac{dR}{dA}=\frac{1}{\sqrt{\pi A}}$$ $$F_A(A)=p_R(R)\frac{dR}{dA}$$ $$F_A(A)=\frac{1}{\sqrt{\pi A}(b-a)},A\in [\pi\frac{a^2}{4},\pi\frac{b^2}{4}]$$
I am not quite sure if this value for $F_A(A)$ should actually be the density function $f_A(A)$, or is it the distribution function (in that case, this evaluation is correct). Could someone tell?
From here, it is easy to evaluate expectation and variance by using the density function $f_A(A)$.
The more straightforward way to do it is to go from the cdf of radius to the cdf of area, which involves no derivatives at all: \begin{align} \Pr[R < x] &= \frac{x-a}{b-a} & x \in [a,b]\\ \Pr[\pi R^2 < \pi x^2] &= \frac{x-a}{b-a} \\ \Pr[\pi R^2 < y] &= \frac{\sqrt{y/\pi}-a}{b-a} & (\pi x^2 = y \iff x = \sqrt{y/\pi}) \\ F_A(y) &= \frac{\sqrt{y/\pi}-a}{b-a} & y \in [\pi a^2, \pi b^2] \end{align} From here, if you want, you can take the derivative and get $f_A(y) = \frac{1}{2\sqrt{\pi y} (b-a)}$.
This density function matches the function you got, except that you seem to be using the formula for the area of a circle in terms of the diameter, rather than the radius.