數學求積法(圓錐+半球體)一問

a) 設半球半徑為 r ,
則圓錐高 = 14 - r圓錐體績 = (1/3) πr² (14 - r)
半球體績 = (1/2) (4/3) πr³ ∴ (1/3) πr² (14 - r) : (1/2) (4/3) πr³ = 2 : 3
(14 - r) : 2r = 2 : 3
4r = 3(14 - r)
r = 6該立體體績
= 圓錐體績 + 半球體績
= (1/3) π(6)² (14 - 6) + (1/2) (4/3) π(6)³
= 96π + 144π
= 240π cm³

b) 該立體表面面績
= 圓錐曲面面績 + 半球曲面面績
= π(6)L + (1/2) 4π(6)² ........... L 是圓錐斜邊
= π(6) √( (14-6)² + 6² ) + (1/2) 4π(6)²
= 60π + 72π
= 132π cm²

2012-11-18 15:04:22 補充：
"績"應是"積"之誤。