急求: 二重积分∫∫(3x^2+y^2)^2dxdy D的区域x

令x=rcosθ,y=rsinθ,积分区域变为0≤r≤1,π≤θ≤3π/2
∴原积分=∫∫(3r²cos²θ+r²sin²θ)²rdrdθ
=∫∫r^5(1+2cos²θ)²drdθ
=∫r^5dr·∫(1+2cos²θ)²dθ
=(r^6)/6|[0->1] ·∫(2+cos2θ)²dθ
=(1/6)∫(4+4cos2θ+cos²2θ)dθ
=(1/6)(2π+∫(1+cos4θ)/2dθ)
=π/3+(1/6)(π/4)
=9π/24