Answer
$\dfrac{312,500 \pi}{7}$
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, $\rho=\sqrt {x^2+y^2+z^2}$; $\phi =\cos^{-1} [\dfrac{z}{\rho}]; \theta=\cos^{-1}[\dfrac{x}{\rho \sin \phi}]$
Now, $(x^2+y^2+z^2)^2=\rho^4$
Consider $I=\int_0^{\pi} \int_{0}^{2 \pi}\int_{0}^{5} (\rho^4) (\rho^2) \sin \phi d\rho d\theta d\phi=\int_0^{\pi} \int_{0}^{2 \pi}\int_{0}^{5} \rho^6 \times \sin \phi d\rho d\theta d\phi$
$[\int_0^{\pi} \sin \phi d\phi] \times [\int_{0}^{2 \pi}d\theta] \times [\int_{0}^{5} \rho^6 d\rho] =[-\cos \phi]_0^{\pi} \times [\theta]_{0}^{2 \pi} \times [\dfrac{\rho^7}{7}]_0^5$
Hence, we have $I=(2 \pi) (1+1)
[ \dfrac{78125}{7}]=\dfrac{312,500 \pi}{7}$
