The radius X of a circle is uniformly distributed between 1 and 4. Find the probability density function of the area of the circle. Then find the mean and variance of the area of the circle.

\begin{align} The \; p.d.f \; of \; X \; is \; f(x) &= \frac{1}{3} \\ \\ and \; X \; is \; a \; random \; variable, \; so \\ \\ 1 &= \int_{1}^{4} \frac{1}{3}dx \quad \enclose{circle}{\color{black}{1}} \\ \\ Let \; A \; be \; the \; area \; of \; the \; circle. \; Then \\ \\ \pi x^2 &= a \\ \\ x^2 &= \frac{a}{\pi} \\ \\ x &= \left(\frac{a}{\pi} \right)^{\frac{1}{2}} \\ \\ differentiating \; gives \\ dx &= \frac{1}{2\sqrt{\pi}}a^{- \frac{1}{2}} da \quad \enclose{circle}{\color{black}{2}} \\ \\ and \; the \; limits \; [1,4] \; for \; X \\ become \; [\pi,16\pi] \; for \; A. \\ Then \; substituting \; \enclose{circle}{\color{black}{2}} \; into \; \enclose{circle}{\color{black}{1}} \; gives \\ \\ 1 &= \int_{1}^{16} \left(\frac{1}{3} \right) \frac{1}{2\sqrt{\pi}}a^{- \frac{1}{2}} da \\ \\ so \; the \; p.d.f. \; of \; A \; is \\ \\ g(a) &= \frac{1}{6\sqrt\pi}a^{- \frac{1}{2}} \;, \; \pi\leqslant \; a \; \leqslant \; 16\pi \\ \\ \\ To \; find \; the \; mean, \; we \; know \; that \\ \\ E(X) &= \int_{\pi}^{16\pi}a \frac{1}{6\sqrt\pi}a^{- \frac{1}{2}} da \\ \\ &= \frac{1}{6\sqrt\pi} \int_{\pi}^{16\pi}a^{\frac{1}{2}} \\ \\ &= \frac{1}{6\sqrt\pi} \left[ \frac{2}{3}a^{\frac{3}{2}} \right]_{\pi}^{16\pi} \\ \\ &= \frac{2}{18\sqrt\pi} \left[a^{\frac{3}{2}} \right]_{\pi}^{16\pi} \\ \\ &= \frac{2}{18\sqrt\pi} \left[(16\pi)^{\frac{3}{2}}-(\pi)^{\frac{3}{2}} \right] \\ \\ &= 7\pi \quad square \; units \\ \\ To \; find \; the \; variance, \; we \; know \; that \\ \\ Var(X) &= E(X^2)-E^2(X) \\ \\ Using \; the \; p.d.f. \; of \; A \;, \\ \\ E(X^2) &= \int_{\pi}^{16\pi} a^2\frac{1}{6\sqrt\pi}a^{- \frac{1}{2}} \\ \\ &=\frac{1}{6\sqrt\pi} \int_{\pi}^{16\pi}a^{\frac{3}{2}} \\ \\ &=\frac{1}{6\sqrt\pi} \left[\frac{2}{5}a^{\frac{5}{2}} \right]_{\pi}^{16\pi}\\ \\ &=\frac{2}{30\sqrt\pi} \left[(16\pi)^{\frac{5}{2}}-(\pi)^{\frac{5}{2}} \right]_{\pi}^{16\pi}\\ \\ &= 673.11 \quad to \; 2 \; d.p.\\ \\ So \quad Var(X)&= 673.11-(7\pi)^2 \\ \\ &=189.50 \quad to \; 2 \; d.p. \\ \end{align}