mathS~~~~~~`

a) V = π∫(y = 0 → h) x2 dy

= π∫(y = 0 → h) y dy

= π[y2/2](y = 0 → h)

= πh2/2

bi) 代入 a 部的結果, 其中 h = 4/5:

水體積 = π(4/5)2/2

= 8π/25 立方單位

ii) 以 dV/dt = -π(t - 15)/100 計:

dV/dt = π(15 - t)/100

V = (π/100)(15t - t2/2) + C 其中 C 為常數.

t = 0 時, V = 8π/25, 即:

C = 8π/25

所以 V = (π/100)(15t - t2/2) + 8π/25

= (π/100)(32 + 15t - t2/2)

iii) (1) V = 0 時:

32 + 15t - t2/2 = 0

64 + 30t - t2 = 0

(32 - t)(2 + t)

t = 32 s 或 -2 s (捨棄)

(2) 代 h = 1, 整個下半部體積 = π/2 立方單位

所以空間體積 = π/2 - 8π/25 = 9π/50 立方單位

即空間的高度為:

πh2/2 = 9π/50

h = 3/5

所以水深 = 2/5 單位