Question

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,求解答过程。。。。。。

答案 {( R^3)/3} (π-4/3)

Answer 1

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