设区域D={(x,y)|x²+y²≤1,x≥0},计算二重积分I=∫∫(1+xy)/(1+x²+y²)dxdy
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设区域D={(x,y)|x²+y²≤1,x≥0},计算二重积分I=∫∫(1+xy)/(1+x²+y²)dxdy设区域D={(x,y)|x²+y
设区域D={(x,y)|x²+y²≤1,x≥0},计算二重积分I=∫∫(1+xy)/(1+x²+y²)dxdy
设区域D={(x,y)|x²+y²≤1,x≥0},计算二重积分I=∫∫(1+xy)/(1+x²+y²)dxdy
设区域D={(x,y)|x²+y²≤1,x≥0},计算二重积分I=∫∫(1+xy)/(1+x²+y²)dxdy
原式=∫(-π/2,π/2)dθ∫(0,1)[(1+r²sinθcosθ)/(1+r²)]rdr (极坐标变换)
=1/2∫(-π/2,π/2)dθ∫(0,1)[(1+rsinθcosθ)/(1+r)]dr (用r代换r²)
=1/2∫(-π/2,π/2)dθ∫(0,1){1/(1+r)+[1-1/(1+r)]sinθcosθ}dr
=1/2∫(-π/2,π/2){[ln(1+r)+(r-ln(1+r))sinθcosθ]│(0,1)}dθ
=1/2∫(-π/2,π/2)[ln2+(1-ln2)sinθcosθ]dθ
=1/2∫(-π/2,π/2)[ln2+(1-ln2)sin(2θ)/2]dθ
=1/2[θln2-(1-ln2)cos(2θ)/4]│(-π/2,π/2)
=1/2[(π/2)ln2-(1-ln2)cos(π)/4-(-π/2)ln2+(1-ln2)cos(-π)/4]
=(πln2)/2.