1、y=(2x-x²)^1/2,因此y²=2x-x²,整理后为:(x-1)²+y²=1,这是以(1,0)为圆心,1为半径的圆;不过注意:由于y=(2x-x²)^1/2,说明y为正,因此本题是上半圆.2、极坐标:x²+y²变成r²,x变成rcosθ,y变成rsinθ,dxdy变成rdrdθ原式=∫∫(x²+4y²+9)dxdy=∫∫(x²+4y²)dxdy+9∫∫1dxdy当被积函数为1时,积分结果为区域面积,因此后一个积分结果为:9*π*2²=36π前一个积分用极坐标=∫∫(r²+3r²sin²θ)rdrdθ+9*π*2²=∫∫(r²+3r²sin²θ)rdrdθ+36π=∫ [0--->2π]dθ∫[0--->2] r³(1+3sin²θ)dr+36π=∫ [0--->2π](1+3sin²θ)dθ∫[0--->2] r³dr+36π=(1/4)∫ [0--->2π](1+3sin²θ)r⁴ |[0--->2]dθ+36π=4∫ [0--->2π](1+3sin²θ)dθ+36π=4∫ [0--->2π](1+(3/2)(1-cos2θ))dθ+36π=2∫ [0--->2π](5-3cos2θ))dθ+36π=2(5θ-(3/2)sin2θ) |[0--->2π]+36π=20π+36π=56π