Answer
$\dfrac{1688 \pi}{15}$
Work Step by Step
Conversion of rectangular to spherical coordinates is as follows:
$x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$
and
$\rho=\sqrt {x^2+y^2+z^2}$;
$\cos \phi =\dfrac{z}{\rho}$; $\cos \theta=\dfrac{x}{\rho \sin \phi}$\
Here, we have:
$(x^2+y^2+z^2)=9 \implies \rho=3$
We need to set up the integral.
$I=\int_0^{\pi} \int_{0}^{2 \pi}\int_{2}^{3} (\rho^2 \sin^2 \phi) (\rho^2 \sin \phi) d\rho d\theta d\phi$
or, $I=\int_0^{\pi} \int_{0}^{2 \pi} [\dfrac{\rho^5}{5}\sin^3 \phi]_2^3 d\theta d\phi$
or, $I=\int_0^{\pi} \dfrac{211}{5}(2 \pi \sin^3 \phi ] d\phi$
or, $I=[\dfrac{422\pi}{5} \int_0^{\pi} \sin \phi(1-\cos^2 \phi) d\phi$
Plug in:
$p=\cos \phi \implies dp=-\sin \phi d\phi$
Thus, we have $I= [\dfrac{422\pi}{5} \int_1^{-1} 1-p^2 dp$
or, $[\dfrac{422\pi}{5}](p-\dfrac{p^3}{3}]_1^{-1} =\dfrac{1688 \pi}{15}$
