Answer
$\mathop \smallint \limits_{ - 2}^2 \mathop \smallint \limits_{ - \sqrt {4 - {x^2}} }^{\sqrt {4 - {x^2}} } \mathop \smallint \limits_0^{\sqrt {4 - {x^2} - {y^2}} } {{\rm{e}}^{ - {{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}}{\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /2} \mathop \smallint \limits_{\rho = 0}^2 {{\rm{e}}^{ - {\rho ^3}}}{\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta = 2.1$
Work Step by Step
From the triple integral $\mathop \smallint \limits_{ - 2}^2 \mathop \smallint \limits_{ - \sqrt {4 - {x^2}} }^{\sqrt {4 - {x^2}} } \mathop \smallint \limits_0^{\sqrt {4 - {x^2} - {y^2}} } {{\rm{e}}^{ - {{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}}{\rm{d}}z{\rm{d}}y{\rm{d}}x$, we obtain the domain of integration:
$ - 2 \le x \le 2$, ${\ \ \ }$ $ - \sqrt {4 - {x^2}} \le y \le \sqrt {4 - {x^2}} $, ${\ \ \ }$ $0 \le z \le \sqrt {4 - {x^2} - {y^2}} $
Let ${\cal W}$ denote the region. We see that ${\cal W}$ is the upper hemisphere of radius $2$. Its projection onto the $xy$-plane is the disk ${x^2} + {y^2} \le 4$.
Using $x = \rho \sin \phi \cos \theta $, $y = \rho \sin \phi \sin \theta $, in spherical coordinates, the description of ${\cal W}$:
${\cal W} = \left\{ {\left( {\rho ,\phi ,\theta } \right)|0 \le \rho \le 2,0 \le \phi \le \pi /2,0 \le \theta \le 2\pi } \right\}$
Evaluate $\mathop \smallint \limits_{ - 2}^2 \mathop \smallint \limits_{ - \sqrt {4 - {x^2}} }^{\sqrt {4 - {x^2}} } \mathop \smallint \limits_0^{\sqrt {4 - {x^2} - {y^2}} } {{\rm{e}}^{ - {{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}}{\rm{d}}z{\rm{d}}y{\rm{d}}x$ in spherical coordinates:
$\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /2} \mathop \smallint \limits_{\rho = 0}^2 {{\rm{e}}^{ - {\rho ^3}}}{\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = - \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /2} \left( {{{\rm{e}}^{ - {\rho ^3}}}|_0^2} \right)\sin \phi {\rm{d}}\phi {\rm{d}}\theta $
$ = - \frac{{{{\rm{e}}^{ - 8}} - 1}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /2} \sin \phi {\rm{d}}\phi {\rm{d}}\theta $
$ = \frac{{{{\rm{e}}^{ - 8}} - 1}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\cos \phi |_0^{\pi /2}} \right){\rm{d}}\theta $
$ = \frac{{1 - {{\rm{e}}^{ - 8}}}}{3}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{2}{3}\pi \left( {1 - {{\rm{e}}^{ - 8}}} \right) \simeq 2.1$
So,
$\mathop \smallint \limits_{ - 2}^2 \mathop \smallint \limits_{ - \sqrt {4 - {x^2}} }^{\sqrt {4 - {x^2}} } \mathop \smallint \limits_0^{\sqrt {4 - {x^2} - {y^2}} } {{\rm{e}}^{ - {{\left( {{x^2} + {y^2} + {z^2}} \right)}^{3/2}}}}{\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /2} \mathop \smallint \limits_{\rho = 0}^2 {{\rm{e}}^{ - {\rho ^3}}}{\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta = 2.1$
