Answer
$\dfrac{e^2-2e}{4}$
Work Step by Step
$ \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$
This implies that
$\int_{0}^1 \int_{0}^{1} [xye^{2-x^2-y^2}-xye^0] dy dx=(e^2) \int_{0}^{1} ye^{-y^2} dy \int_0^1 xe^{-x^2} dx-(\dfrac{x^2}{2})_0^1[\dfrac{y^2}{2}]_0^1 $
Now consider $p=-x^2; dp=(2x) dx$
$ (e^2) [ \int_0^1 xe^{-x^2} dx-(\dfrac{x^2}{2})]-(\dfrac{1}{2})(\dfrac{1}{2}) = (e^2) [(-\dfrac{1}{2}) \int_0^{-1}e^p dp]^2-\dfrac{1}{4} = (e^2) [\dfrac{1-(1/e)}{2}]^2-(1/4)$
Hence, $\int_{0}^1 \int_{0}^{1} \int_{0}^{2-x^2-y^2} xye^z dz dy dx =\dfrac{(e-1)^2-1}{4}=\dfrac{e^2-2e}{4}$
