Answer
$\dfrac{1688 \pi}{15}$
Work Step by Step
Set up the integral.
$I=\int_0^{\pi} \int_{0}^{2 \pi}\int_{2}^{3} (\rho^2 \sin^2 \phi) \times (\rho^2 \sin \phi) d\rho d\theta d\phi$
$=\int_0^{\pi} \int_{0}^{2 \pi} [\dfrac{\rho^5}{5}\sin^3 \phi]_2^3 d\theta d\phi$
$=( \dfrac{211}{5}) \int_0^{\pi} (2 \pi \sin^3 \phi ] d\phi$
$=\dfrac{422\pi}{5} \int_0^{\pi} \sin \phi \times (1-\cos^2 \phi) d\phi$
Substitute $\cos \phi=a; da=-\sin \phi d\phi$
$I= \dfrac{422\pi}{5} \int_1^{-1} (1-a^2)da=[\dfrac{422\pi}{5}][a-\dfrac{a^3}{3}]_1^{-1}$
$ =\dfrac{1688 \pi}{15}$
