Answer
$\dfrac{e^2-2e}{4}$
Work Step by Step
We need to integrate the given integral first with respect to $z$, then $y$, and then $x$.
$ \int_{0}^1 \int_{0}^{1} \int_{0}^{2-x^2-y^2} xye^z dz dy dx= \int_{0}^1 \int_{0}^{1} [xye^z]_{0}^{2-x^2-y^2} dy dx$
or, $=\int_{0}^1 \int_{0}^{1} [xye^{2-x^2-y^2}-xye^0] dy dx$
or, $=(e^2) \int_{0}^{1} ye^{-y^2} dy \int_0^1 xe^{-x^2} dx-(x^2/2)_0^1(y^2/2)_0^1 $
or, $= (e^2) [ \int_0^1 xe^{-x^2} dx-(x^2/2)]-(1/2)(1/2) $
Plug $a=-x^2 \implies da=2x dx$
or, $ (e^2) [(-1/2) \int_0^{-1}e^a da]^2-\dfrac{1}{4} = (e^2) [\dfrac{1-(1/e)}{2}]^2-\dfrac{1}{4}$
or, $\dfrac{(e-1)^2-1}{4}=\dfrac{e^2-2e}{4}$
