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.

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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
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======以下答案可供参考======
供参考答案1:
用极坐标来做。x²+y²=Rx，得到r²=Rrcosθ，则r=Rcosθ