$$ { 解：根据三次积分可知：\left\{ \left( x,y,z \right) |0≥ x≥ 1,0≥ z≥ 1-x,0≥ y≥ 1-x-z \right\} ；} \\ { 通过作图知，这个区域为\Omega ：\left\{ \left( x,y,z \right) |x+y+z≥ 1,x≤ 0,y≤ 0,z≤ 0 \right\} ；} \\ { 因此：I=\int_0^1{\mathrm{d}x}\int_0^{1-x}{\mathrm{d}z}\int_0^{1-x-z}{\left( 1-y \right)}\mathrm{e}^{-\left( 1-y-z \right) ^2}\mathrm{d}y=\iiint_{\Omega}{\left( 1-y \right)}\mathrm{e}^{-\left( 1-y-z \right) ^2}\mathrm{d}V} \\ { 因为被积函数\left( 1-y \right) \mathrm{e}^{-\left( 1-y-z \right) ^2}中有1-y-z，并且被积函数中没有变量x，} \\ { 恰好x上限表示为x=1-y-z；} \\ { 因此考虑首先对x进行积分；} \\ { 又因为被积函数中有1-y，恰好z的上限可表示为：z=1-y；} \\ { 因此其次考虑对z进行积分；} \\ { 综上所述，三重积分的积分次序见下：} \\ { \iiint_{\Omega}{\left( 1-y \right)}\mathrm{e}^{-\left( 1-y-z \right) ^2}\mathrm{d}V\begin{array}{l} =\int_0^1{\mathrm{d}y}\int_0^{1-y}{\mathrm{d}z}\int_0^{1-y-z}{\left( 1-y \right) \mathrm{e}^{-\left( 1-y-z \right) ^2}}\mathrm{d}x\\ =\int_0^1{\mathrm{d}y}\int_0^{1-y}{\mathrm{d}z}\left( 1-y \right) \mathrm{e}^{-\left( 1-y-z \right) ^2}\cdot x\mid_{0}^{1-y-z}\\ =\int_0^1{\mathrm{d}y}\int_0^{1-y}{\left( 1-y \right) \left( 1-y-z \right) \mathrm{e}^{-\left( 1-y-z \right) ^2}\mathrm{d}z}\\ =\int_0^1{\left( 1-y \right) \mathrm{d}y}\int_0^{1-y}{-\frac{1}{2}\mathrm{e}^{-\left( 1-y-z \right) ^2}\mathrm{d}\left( 1-y-z \right) ^2}\\ =\int_0^1{\left( 1-y \right) \mathrm{d}y}\cdot \frac{1}{2}\int_0^{1-y}{-\mathrm{e}^{-\left( 1-y-z \right) ^2}\mathrm{d}\left( 1-y-z \right) ^2}\\ =\int_0^1{\left( 1-y \right) \mathrm{d}y}\cdot \frac{1}{2}\int_0^{1-y}{\mathrm{e}^{-\left( 1-y-z \right) ^2}\mathrm{d}\left[ -\left( 1-y-z \right) ^2 \right]}\\ =\int_0^1{\left( 1-y \right) \mathrm{d}y}\cdot \frac{1}{2}\mathrm{e}^{-\left( 1-y-z \right) ^2}\mid_{0}^{1-y}\\ =\frac{1}{2}\int_0^1{\left( 1-y \right)}\mathrm{d}y\cdot \left[ \mathrm{e}^0-\mathrm{e}^{-\left( 1-y \right) ^2} \right]\\ =\frac{1}{2}\int_0^1{\left( 1-y \right) \left[ 1-\mathrm{e}^{-\left( 1-y \right) ^2} \right]}\mathrm{d}y\\ =\frac{1}{2}\int_0^1{\left( 1-y \right) \mathrm{d}y}-\frac{1}{2}\int_0^1{\left( 1-y \right) \mathrm{e}^{-\left( 1-y \right) ^2}}\mathrm{d}y\\ =\frac{1}{2}\left( y-\frac{1}{2}y^2 \right) \mid_{0}^{1}-\frac{1}{2}\cdot \frac{1}{2}\int_0^1{-\mathrm{e}^{-\left( 1-y \right) ^2}}\mathrm{d}\left( 1-y \right) ^2\\ =\frac{1}{2}\left( 1-\frac{1}{2}\cdot 1^2 \right) -0-\frac{1}{4}\int_0^1{\mathrm{e}^{-\left( 1-y \right) ^2}}\mathrm{d}\left[ -\left( 1-y \right) ^2 \right]\\ =\frac{1}{4}-\frac{1}{4}\mathrm{e}^{-\left( 1-y \right) ^2}\mid_{0}^{1}=\frac{1}{4}-\frac{1}{4}\left( \mathrm{e}^0-\mathrm{e}^{-1} \right)\\ =\frac{1}{4}-\frac{1}{4}+\frac{1}{4\mathrm{e}}=\frac{1}{4\mathrm{e}}.\\ \end{array}} $$