Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2} + {z^2}{\rm{d}}V = \frac{{124}}{5}\pi $
Work Step by Step
We have $f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}$ and the region $1 \le {x^2} + {y^2} + {z^2} \le 4$.
Using ${\rho ^2} = {x^2} + {y^2} + {z^2}$, the inequality in spherical coordinates is $1 \le \rho \le 2$.
Thus, the region description of ${\cal W}$ in spherical coordinates is given by
${\cal W} = \left\{ {\left( {\rho ,\phi ,\theta } \right)|1 \le \rho \le 2,0 \le \phi \le \pi ,0 \le \theta \le 2\pi } \right\}$
Evaluate the triple integral of $f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}$ over ${\cal W}$ in spherical coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2} + {z^2}{\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi \mathop \smallint \limits_{\rho = 1}^2 {\rho ^4}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
$ = \left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{\phi = 0}^\pi \sin \phi {\rm{d}}\phi } \right)\left( {\mathop \smallint \limits_{\rho = 1}^2 {\rho ^4}{\rm{d}}\rho } \right)$
$ = - \frac{2}{5}\pi \left( {\cos \phi |_0^\pi } \right)\left( {{\rho ^5}|_1^2} \right) = \frac{4}{5}\pi \cdot31 = \frac{{124}}{5}\pi $
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {x^2} + {y^2} + {z^2}{\rm{d}}V = \frac{{124}}{5}\pi $.
