Question

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

Answer 1

u=x^2+y^2 v=x/y,于是：0≤v ≤1/√3 ,4≤u≤8

u=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)[arctanv]^2 | (0,1/√3)

=6[arctan(1/√3)]^2

=(π^2)/6