## Calculus 8th Edition

$\dfrac{162\pi}{5}$
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, $i=\int_0^{\pi} \int_{0}^{\pi}\int_{0}^{3} (\rho^2 \sin^2 \phi \sin^2 \theta) \times (\rho^2 \sin \phi) \times d\rho d\theta d\phi=\int_0^{\pi} \int_{0}^{\pi} [\dfrac{\rho^5}{5} \times \sin^3 \phi \sin^2 \theta]_0^3 d\theta d\phi$ $\int_0^{\pi} \int_0^{\pi} \dfrac{243}{5} \times ( \sin^3 \phi \sin^2 \theta) d\theta d\phi$ $=\int_0^{\pi} \int_0^{\pi} \sin^3 \phi d \phi \times (\dfrac{1}{2}-\dfrac{1}{2} \cos 2 \theta) d\theta \times \dfrac{243}{5}$ $=\int_0^{\pi} \pi \sin^3 \phi d\phi \times \dfrac{243}{10}$ $=\int_0^{\pi} \sin \phi -\sin \phi\cos^2 \phi \times \dfrac{243 \pi}{10}$ Hence, $I=\dfrac{162\pi}{5}$