What is the pdf of $\pi R^2$ given the pdf of $R$?

Question

Let $f_{R}(r) = 6r(1-r), 0 < r < 1$ be the pdf of a circle's measured radius. What is the pdf of $Y = \pi R^{2}$?

Textbook says the answer is $\frac{3(\pi^{\frac{1}{2}} - y^{\frac{1}{2}})}{\pi^{\frac{3}{2}}}$, but I cannot get this answer using either the CDF method or transformation of one-to-one.

Answer 1

$$ R=\sqrt{\frac Y\pi}\;, $$

so

$$ \mathrm dr=\frac1{2\sqrt{\pi y}}\mathrm dy\;. $$

Thus

$$ f_R(r)\mathrm dr=f_Y(y)\mathrm dy $$

yields

$$ f_Y(y)=f_R(r)\frac{\mathrm dr}{\mathrm dy}=\frac{6r(1-r)}{2\sqrt{\pi Y}}=\frac3\pi\left(1-\sqrt\frac y\pi\right)\;, $$

which coincides with the answer you were expecting.