用極坐標(biāo),積分區(qū)域被y = x分開(kāi)為兩部分
D₁是個(gè)等腰三角形：y = 0、x = 1、y = x
D₂是個(gè)弓形：y = x,y = √(2x - x²)
化為極坐標(biāo),
D₁：θ：0→π/4,x = 1 ==> rcosθ = 1 ==> r = secθ
D₂：θ：π/4→π/2,y = √(2x - x²) ==> r² = 2rcosθ ==> r = 2cosθ
所以∫∫D f(x,y) dxdy
= ∫∫D₁ f(x,y) dxdy + ∫∫D₂ f(x,y) dxdy
= ∫(0→π/4) ∫(0→secθ) f(rcosθ,rsinθ) rdrdθ + ∫(π/4→π/2) ∫(0→2cosθ) f(rcosθ,rsinθ) rdrdθ
