Answer
$e-\dfrac{5}{2} $
Work Step by Step
Given: $ f(x, y, z)=e^z$
The iterated triple integral can be calculated as:
$\iiint_{\mathcal{w}} e^z \ d V =\int_0^1 \int_0^{1-y} \int_0^{1-x-y} e^z \ dz dx dy \\=\int_0^1 (\int_0^{1-y} [e^{1-x-y}-1] \ dx) \ dy\\=\int_0^1 [e^{-1-x-y}]_0^{1-y} \ dy\\=\int_0^1 e^{1-y}+y-2) \ dy \\=[-e^{1-y}+\dfrac{y^2}{2}-2y]_0^1\\=e-\dfrac{5}{2} $
