换元法x=rcosa x^2+y^2≤1 所以0<=r<=1 0<=a<=2π y=rcosa ∫ ∫D e^(x^2+y^2) dxdy∫[0,2π] ∫[0,1] e^(r^2) rdrda=2π*1/2∫[0,1] e^(r^2) d(r^2)=π*e^(r^2) [0,1]= π(e-1)
