62问答网 > 求函数z=(R^2-x^2-y^2)^0.5在圆域x^2+y^2<=r^2上的平均值

求函数z=(R^2-x^2-y^2)^0.5在圆域x^2+y^2<=r^2上的平均值

2025-06-24 15:02:02

推荐回答（2个）

回答1：

z在区域D上的平均值即为z在区域D的二重积分值除以区域的面积

回答2：

化成极坐标形式的积分
x＾2+y＾2=Rx的极坐标方程为r=Rcost (t∈[-π/2,π/2])
又根据对称性有：
原积分=2∫[0->π/2]∫[0->Rcost] (R^2-r^2)^(1/2)rdrdt
=2∫[0->π/2] -(2/3)(R^2-r^2)^(3/2) | [0->Rcost] dt
=2∫[0->π/2] -(2/3)[(Rsint)^3-R^3] dt
= (4/3)∫[0->π/2] R^3-(Rsint)^3 dt
= (4/3)[R^3(π/2-0) - (R^3)∫[0->π/2] (sint)^3dt]
= (2/3)πR^3-(4/3)(1!!/3!!)R^3
= (2/3)πR^3-(4/9)R^3
= (2R^3)/3}(π-4/3)