Question

这道二重积分怎么算 丿丿（x+y）^2*(x-y)e^(x-y)dxdy 0

Answer 1

设x+y=u
x-y=v
x=(u+v)/2
y=(u-v)/2
求矩阵的det
dx/du dx/dv
dy/du dy/dv
=-(1/2)^2-(1/2)^2=-1/2
用代换法 dxdy=|det|dudv 注意这里是用det的绝对值
∫(0~1)∫(0~1) u^2v(e^v)|det| dudv
=(1/2) ∫(0~1)∫(0~1) u^2v(e^v) dudv
=(1/2) ∫(0~1)u^2du ∫(0~1)v(e^v)dv
∫ve^vdv=∫ve^v+e^v-e^v dv=ve^v-e^v+C=(v-1)e^v+C
带回原式
=(1/2) {u^3/3 (0~1)} {(v-1)(e^v) (0~1)}
=(1/2)(1/3)(0-(-1))
=1/6