Answer
$\dfrac{\pi}{8}$
Work Step by Step
The conversion of rectangular coordinates to spherical coordinates is given as:
$x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$
Here, $\rho=\sqrt {x^2+y^2+z^2}$; $\phi =\cos^{-1} [\dfrac{z}{\rho}]; \theta=\cos^{-1}[\dfrac{x}{\rho \sin \phi}]$
$I=\int_0^{\pi/2} \int_{0}^{\pi/2}\int_{0}^{1} (\rho \sin \phi \cos \theta) e^{\rho^2} \times (\rho^2 \sin \phi) d\rho d\theta d\phi$
$=\int_0^{\pi/2} ( \sin^2 \phi d\phi) \times \int_{0}^{\pi/2} \cos \theta d\theta \times \int_{0}^{1} \rho^3 e^{\rho^2} d\rho $
$=(1-0) \times [\dfrac{\phi}{2}-\dfrac{\sin 2\phi}{4}]_0^{\pi/2} \times [ \dfrac{1}{2} e^{(1)^2} (1-1)- \dfrac{1}{2} e^{(0)^2} (0^2-1)]$
Hence, $I=[\dfrac{\phi}{2}-\dfrac{\sin 2\phi}{4}]_0^{\pi/2} \times [0- \dfrac{1}{2} e^{(0)^2} (0^2-1)]=\dfrac{\pi}{4}[0-\dfrac{1}{2}(-1)]=\dfrac{\pi}{8}$
