Question

计算二重积分∫∫1/(x^2+y^2+R^2)dxdy,其中D为x^2+y^2

Answer 1

转化到极坐标系,则

x²+y²=r²,x=rcosθ,y=rsinθ

积分域D={(x,y)|x²+y²≤R²}={(r,θ)|0≤r≤R,0≤θ≤2π}

∫∫1/(x²+y²+R²)dxdy

=∫∫ r/(r²+R²)drdθ

=∫(0,2π)dθ ∫(0,R)r/(r²+R²)dr

=2π*[ln(r²+R²)/2]|(0,R)

=π[ln(2R²)-ln(R²)]

=πln2