Question

cos(x+y)的绝对值且区域由y=x和y=0.x=兀/2构成的二重积分

Answer 1

直线x+y=π将积分区间分成2部分,左下部分记为D1,右上部分记为D2
则先积D1里的积分,在D1内,由于x+yπ] dx∫[0--->π-x] sin(x+y) dy
=-∫[0--->π] cos(x+y) |[0--->π-x] dx
=∫[0--->π] (cosx-cosπ) dx
=∫[0--->π] (cosx+1) dx
=sinx+x |[0--->π]
=π
对于D2,ππ] sin(x+y) dy
=∫[0--->π] cos(x+y) |[π-x--->π] dx
=∫[0--->π] (cos(π+x)-cosπ) dx
=∫[0--->π] (-cosx+1) dx
=-sinx+x |[0--->π]

Answer 2

解 记D＝{(x，y)：0≤y≤x，0≤x≤π2}，
D1＝{(x，y)∈D：x+y≤π2}，D2＝{(x，y)∈D：x+y＞π2}，
则?Dcos（x+y）dxdy=?D1[-cos（x+y）
]dxdy+?D2|cos（x+y）|dxdy=∫π40dy∫π2?yycos(x+y)dx+∫π2π4dx∫xπ2?x[?cos(x+y)]dy=∫π40(1?sin2y)dy+∫π2π4(1?sin2x)dx=∫

Answer 3

参看
http://zhidao.baidu.com/question/1755618597484023708

Answer 4

解 记D＝{(x，y)：0≤y≤x，0≤x≤π2}，D1＝{(x，y)∈D：x+y≤π2}，D2＝{(x，y)∈D：x+y＞π2}，则?Dcos（x+y）dxdy=?D1[-cos（x+y）]dxdy+?D2|cos（x+y）|dxdy=∫π40dy∫π2?yycos(x+y)dx+∫π2π4dx∫xπ2?x[?cos(x+y)]dy=∫π40(1?sin2y)dy+∫π2π4(1?sin2x)dx=∫