Question

求由曲面x²=a²-az，x²+y²=（a／2）²，z=0(a＞0)所围立体的体积

一道大一高数题，重积分的

Answer 1

解：∵x²=a²-az，x²+y²=（a／2）²，z=0(a＞0)
∴所围立体是一个下底以a/2为半径的圆，上底以曲面az=y²+3a²/4为顶的圆柱体
故所围立体的体积=4∫(0,a/2)dx∫(0,√(a²/4-x²))[y²/a+3a/4]dy (∫(0,a/2)表示从0到a/2积分，其他类同)
=4∫(0,π/2)dθ∫(0,a/2)(r²sin²θ/a+3a/4)rdr (进行极坐标变换)
=4∫(0,π/2)dθ∫(0,a/2)(r³sin²θ/a+3ar/4)dr
=∫(0,π/2)dθ*(r^4*sin²θ/a+3ar²/2)│(0,a/2)
=(a³/16)∫(0,π/2)(sin²θ+6)dθ
=(a³/32)∫(0,π/2)(13-cos(2θ))dθ
=(a³/32)(13θ-sin(2θ))│(0,π/2)
=(a³/32)(13π/2)
=13πa³/64。