Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \frac{{61\pi }}{6}$
Work Step by Step
We have $f\left( {x,y,z} \right) = z$.
Let ${\cal W}$ denote the region above the disk ${x^2} + {y^2} \le 1$ in the $xy$-plane and below the surface $z = 4 + {x^2} + {y^2}$ as is illustrated in the figure attached.
Let ${\cal D}$ be the disk ${x^2} + {y^2} \le 1$ as the projection of ${\cal W}$ onto the $xy$-plane. In polar coordinates ${\cal D}$ is described by
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 1,0 \le \theta \le 2\pi } \right\}$
Since ${\cal W}$ is above the $xy$-plane and below the surface $z = 4 + {x^2} + {y^2}$, we have in cylindrical coordinates the range of $z$ given by $0 \le z \le 4 + {r^2}$. Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 1,0 \le \theta \le 2\pi ,0 \le z \le 4 + {r^2}} \right\}$
Using Eq. (5) of Theorem 2, we evaluate the integral of the function $f\left( {x,y,z} \right) = z$ over ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^{4 + {r^2}} zr{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \left( {{z^2}|_0^{4 + {r^2}}} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \left( {16 + 8{r^2} + {r^4}} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \left( {16r + 8{r^3} + {r^5}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {8{r^2} + 2{r^4} + \frac{1}{6}{r^6}} \right)|_0^1} \right){\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {8 + 2 + \frac{1}{6}} \right){\rm{d}}\theta $
$ = \frac{{61}}{{12}}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{{61}}{6}\pi $
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \frac{{61\pi }}{6}$.
