发布网友 发布时间:2022-05-13 15:18
共1个回答
热心网友 时间:2023-10-14 02:10
解:积分区域如图所示;
(1)由图可见
D={(x, y)|0≤x≤1, 0≤y≤x²}
={(x, y)|0≤y≤1, √y≤x≤1},
所以
∫∫D f(x,y)dσ
=∫[0,1]dx∫[0,x²]f(x,y)dy
=∫[0,1]dy∫[√y,1]f(x,y)dx;
(2)由图可见
D={(x, y)|0≤x≤4, x≤y≤2√x}
={(x, y)|0≤y≤4, y²/4≤x≤y},
所以
∫∫D f(x,y)dσ
=∫[0,4]dx∫[x, 2√x]f(x,y)dy
=∫[0,4]dy∫[y²/4, y]f(x,y)dx.