五、设圆的直径X在[1,3]上服从均匀分布，求圆面积的数学期望和方差.

设圆的直径 $X$ 在 $[1, 3]$ 上服从均匀分布，圆面积 $Y = \frac{\pi X^2}{4}$。 **数学期望：** \[ E(Y) = E\left(\frac{\pi X^2}{4}\right) = \frac{\pi}{4} E(X^2) \] 其中， \[ E(X^2) = \frac{b^3 - a^3}{3(b - a)} = \frac{27 - 1}{6} = \frac{13}{3} \] 故 \[ E(Y) = \frac{\pi}{4} \times \frac{13}{3} = \frac{13\pi}{12} \] **方差：** \[ D(Y) = E(Y^2) - [E(Y)]^2 \] \[ E(Y^2) = E\left(\frac{\pi^2 X^4}{16}\right) = \frac{\pi^2}{16} E(X^4) \] 其中， \[ E(X^4) = \frac{b^5 - a^5}{5(b - a)} = \frac{243 - 1}{10} = \frac{121}{5} \] 故 \[ E(Y^2) = \frac{\pi^2}{16} \times \frac{121}{5} = \frac{121\pi^2}{80} \] \[ [E(Y)]^2 = \left(\frac{13\pi}{12}\right)^2 = \frac{169\pi^2}{144} \] \[ D(Y) = \frac{121\pi^2}{80} - \frac{169\pi^2}{144} = \frac{61\pi^2}{180} \] **答案：** \[ \boxed{\frac{13\pi}{12}, \frac{61\pi^2}{180}} \]