计算二重积分I=∫∫xdxdy其中区域D是x^2+y^2<=x+y？
方法一:
x²＋y²≤x＋y
令x=rcosθ，y=rsinθ
则r²≤r(sinθ＋cosθ)
r≤sinθ＋cosθ
I=∫∫ xdxdy
=∫(－π/4，3π/4) dθ ∫(0，sinθ＋cosθ) r²cosθdr
=1/3 ∫(－π/4，3π/4) (sinθ＋cosθ)³cosθdθ
=1/3 ∫(－π/4，3π/4) (sinθcosθ＋2sin²θcos²θ＋cos²θ＋2sinθcos³θ)dθ
=1/6 ∫(－π/4，3π/4) sin2θdθ＋1/6 ∫(－π/4，3π/4) sin²2θdθ＋1/3 ∫(－π/4，3π/4) cos²θdθ＋2/3 ∫(－π/4，3π/4) sinθcos³θdθ
=1/12 ∫(－π/4，3π/4) sin2θd(2θ) ＋1/12 ∫(－π/4，3π/4) (1－cos4θ)dθ ＋1/6 ∫(－π/4，3π/4) (1＋cos2θ)dθ －2/3 ∫(－π/4，3π/4) cos³θdcosθ
=[－1/12 cos2θ ＋θ/12 －1/48 sin4θ＋θ/6＋＋1/12 sin2θ －1/6 (cosθ)^4] |(－π/4，3π/4)
=π/16＋π/8－1/12 －1/24＋π/48＋π/24＋1/12＋1/24
=(3π＋6π＋π＋2π)/48
=12π/48
=π/4
方法二:
x²＋y²≤x＋y
(x－1/2)²＋(y－1/2)²≤1/2
令x=1/2＋rcosθ，y=1/2＋rsinθ
则|=∫∫ xdxdy
=∫(0，2π) dθ∫(0，∨2/2) (1/2 ＋rcosθ)rdr
=∫(0，2π) (1/8 ＋∨2/12 cosθ)dθ
=(1/8 θ＋∨2/12 sinθ) |(0，2π)
=1/8 ×2π
=π/4
