91Ó°ÊÓ

X is the radius of a circle which is a random variable having pdf

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{18\left( {3x + 1} \right)\,\,\,\,for\,0 < x < 2}\\{0\,\,\,otherwise}\end{array}} \right.\)

Let Y be the area of the circle then\(Y = \pi {X^2}\)

In real value,

\(y = \pi {x^2}\)

Differentiating both sides w.r.t x

\(\frac{{dy}}{{dx}} = 2\pi x\)which is\( > 0\)for\(x > 0\)

Then the pdf of Y is given by the following transformation formula

\({f_Y}\left( y \right) = {f_X}\left( x \right)\left| {\frac{{dx}}{{dy}}} \right|\)

So,

\(\begin{aligned}{}{f_Y}\left( y \right) &= {f_X}\left( x \right)\left| {\frac{{dx}}{{dy}}} \right|\\ &= \frac{1}{8}\left( {3x + 1} \right)\frac{1}{{2\pi x}}\\ &= \frac{{\left( {3x + 1} \right)}}{{16\pi x}}\\ &= \frac{{3\sqrt {\pi y} + 1}}{{16\pi \sqrt {\pi y} }}\end{aligned}\)

Where \(4\pi < y < 0\)

\(\therefore {f_Y}\left( y \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{{3\sqrt {\pi y} + 1}}{{16\pi \sqrt {\pi y} }}\,\,\,\,,\,\,0 < y < 4\pi \,\,\,\,}\\{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,otherwise}\end{array}} \right.\,\,\,\,\,\,\,\,\)

is the required pdf of the area of the circle.