Question

计算二重积分∫∫|y-x^2|dxdy,其中区域D={(x,y)|0<=x<=1,0<=y<=1}

Answer 1

区域D={(x,y)|0<=x<=1,0<=y<=1}分为2部分：
D1={(x,y)|0<=x<=1,0<=y<=x²}
D2={(x,y)|0<=x<=1,x²<=y<=1}
∫∫[D] |y-x²| dxdy
=∫∫【0<=x<=1,0<=y<=x²】（-y+x²）dxdy+∫∫【0<=x<=1,x²<=y<=1】（y-x²）dxdy
=∫【0,1】dx∫【0,x²】（-y+x²）dy+∫【0,1】dx ∫【x²,1】（y-x²）dy
=1/2

追问

D1={(x,y)|0<=x<=1,0<=y<=x²}
D2={(x,y)|0<=x<=1,x²<=y<=1}我就是y 的范围我弄不出来 ，不知道怎么出来的 ？？？？？

追答

被积函数|y-x²|，要去绝对值。

Answer 2

简单分析一下，答案如图所示