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
2024-10-13 广告