二重积分 ∫∫（(x^2+y^2)arctan(x/y)dxdy,D=﹛0≦√3x≦y,4≤x^2+y^2≤8﹜

Question

如题所述

Answer 1

u=x^2+y^2 v=x/y,于是：0≤v ≤1/√3 ,4≤u≤8u=y^2(1+v^2),x=vy,求出y对u,v的偏导数,x对u,v的偏导数,计算|J|=1/(2(1+v^2))∫∫（(x^2+y^2)arctan(x/y)dxdy=∫∫uarctanv*(1/(2(1+v^2)))dudv=(1/4)(64-16)(1/2)[arct...