急求: 二重積分∫∫(3x^2 y^2)^2dxdy D的區域x網友廻答：
- 令x=rcosθ,y=rsinθ,積分區域變爲0≤r≤1,π≤θ≤3π/2
∴原積分=∫∫(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
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