Question

∫∫e^(x+y)dxdy，积分区域为x=0,y=0,x+y=1所围成的区域

Answer 1

这题要用到二重积分的换元法……
设x-y=u,x+y=v,得x=(v+u)/2,y=(v-u)/2,则
在此变换下，积分区域边界曲线化为了
v=1,u=2v,u=-v,新的积分区域为
d'={(u,v)|0≤v≤1,-v≤u≤2v}
其雅克比行列式j=
|αx/αu
αx/αv|
|αy/αu
αy/αv|
=
|1/2
1/2|
|-1/2
1/2|
=-1/2
所以
∫∫(d)e^[(x-y)/(x+y)]dxdy=∫∫(d')e^(u/v)*(-1/2)dudv
=(-1/2)∫(0~1)dv∫(-v~2v)e^(u/v)du
=(1/e-e^2)/4

Answer 2

∫∫e^(x+y)dxdy=∫[0,1]
dx
∫[0,1-x]
e^x*e^y
dy=∫[0,1]
e^x
dx
∫[0,1-x]
e^y
dy=∫[0,1]
e^x
dx
(e^y
|[0,1-x])=∫[0,1]
e^x(e^(1-x)-1)
dx
=∫[0,1]
(e-e^x)
dx
=
(ex-e^x)
|[0,1]
=
e