题目

计算二重积分.∫∫根下（R^2-x^2-y^2）dσ,D是由圆周x＾2+y＾2=Rx所围成的区域,
{( R^3)/3} (π-4/3)

提问时间：2020-06-13

答案

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)
其中用到了∫[0->π/2] (sint)^ndt=(n-1)!/n!当n为奇数时
(π/2)*(n-1)!/n!当n为偶数时
我算出的结果和你给的结果有点出入,也许是我算错了吧,不过方法就是这样的