Question

求二重积分∫∫根号下（R^2 -X^2-Y^2）dxdy,其中积分区域D为圆周X^2+Y^2=RX. 简单 谢谢！！！！

Answer 1

极坐标标
∫∫ √(R²-x²-y²) dxdy
=∫∫ r√(R²-r²) drdθ
=∫[-π/2→π/2] dθ∫[0→Rcosθ] r√(R²-r²) dr
=(1/2)∫[-π/2→π/2] dθ∫[0→Rcosθ] √(R²-r²) d(r²)
=-(1/2)(2/3)∫[-π/2→π/2] (R²-r²)^(3/2) |[0→Rcosθ] dθ
=(1/3)∫[-π/2→π/2] (R³-R³|sinθ|³) dθ
=(2R³/3)∫[0→π/2] (1-sin³θ) dθ
=(2R³/3)[∫[0→π/2] 1 dθ - ∫[0→π/2] sin³θ dθ]
=(2R³/3)[π/2 + ∫[0→π/2] sin²θ d(cosθ)]
=(2R³/3)[π/2 + ∫[0→π/2] (1-cos²θ) d(cosθ)]
=(2R³/3)[π/2 + θ - (1/3)cos³θ] |[0→π/2]
=(2R³/3)[π/2 + π/2 - 0]
=2πR³/3

【数学之美】团队为您解答，若有不懂请追问，如果解决问题请点下面的“选为满意答案”。

追问

你错了，答案是1/3R^3(π-4/3)

追答

倒数第三步时出错了，从倒数第四步开始
=(2R³/3)[π/2 + ∫[0→π/2] (1-cos²θ) d(cosθ)]
=(2R³/3)[π/2 + cosθ - (1/3)cos³θ] |[0→π/2]
=(2R³/3)[π/2 -1 + 1/3]
=(2R³/3)[π/2 - 2/3]
=(R³/3)[π - 4/3]

Answer 2

用极坐标来做。x²+y²=Rx，得到r²=Rrcosθ，则r=Rcosθ