若点M(x,y)是圆x^2+y^2=2y上的动点,求2x-y的取值范围

用极坐标,x²+y²=2y的极坐标方程为：r=2sinθ
∫∫ xy dxdy
=∫∫ r³cosθsinθ drdθ
=∫[π/4→π/2] cosθsinθ dθ∫[0→2sinθ] r³ dr
=(1/4)∫[π/4→π/2] r^4cosθsinθ |[0→2sinθ] dθ
=4∫[π/4→π/2] cosθ(sinθ)^5 dθ
=4∫[π/4→π/2] (sinθ)^5 d(sinθ)
=(2/3)(sinθ)^6 |[π/4→π/2]
=(2/3)(1-1/8)
=7/12
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圆方程x²+(y-1)²=1
设x=cosθ y=1+sinθ
因 x+y+a=cosθ+1+sinθ+a≥0恒成立
∴ -a≤（cosθ+sinθ+1)的最小值
sinθ+cosθ+1=√2 sin(θ+π/4)+1≥-√2+1
∴-a≤-√2+1
a≥√2-1