Answer
$\dfrac{1688 \pi}{15}$
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, $\rho=\sqrt {x^2+y^2+z^2}$; $\phi =\cos^{-1} [\dfrac{z}{\rho}]; \theta=\cos^{-1}[\dfrac{x}{\rho \sin \phi}]$
Here, we have $(x^2+y^2+z^2)=9 $ and $\rho=3$
Now, consider the integral such as:
$I=\int_0^{\pi} \int_{0}^{2 \pi}\int_{2}^{3} (\rho^2 \sin^2 \phi) \times (\rho^2 \sin \phi) \times d\rho d\theta d\phi=\int_0^{\pi} \int_{0}^{2 \pi} [\dfrac{\rho^5}{5} \times \sin^3 \phi]_2^3 d\theta d\phi$
$=\dfrac{211}{5} \times \int_0^{\pi} (2 \pi \sin^3 \phi ] d\phi$
$=\dfrac{422\pi}{5} \times \int_0^{\pi} \sin \phi(1-\cos^2 \phi) d\phi$
Now, plug $\cos \phi =a $ and $ da=-\sin \phi d\phi$
$I= \dfrac{422\pi}{5} \times \int_1^{-1} 1-a^2 da=[\dfrac{422\pi}{5}][a-\dfrac{a^3}{3}]_1^{-1} =\dfrac{1688 \pi}{15}$
