Answer
The volume of the region is $16\pi $.
Work Step by Step
From Figure 23 we see that the region is bounded below by $z = {x^2} + {y^2}$ and bounded above by $z = 8 - {x^2} - {y^2}$. Thus, it can be described by ${\cal W}:{x^2} + {y^2} \le z \le 8 - {x^2} - {y^2}$.
We project ${\cal W}$ onto the $xy$-plane to obtain the domain ${\cal D}$. To find the boundary curve of ${\cal D}$, we find the intersection of the surfaces $z = {x^2} + {y^2}$ and $z = 8 - {x^2} - {y^2}$.
Substituting $z = {x^2} + {y^2}$ in $z = 8 - {x^2} - {y^2}$ gives
$z = 8 - z$
So, $z=4$. Thus, the two surfaces intersect at the plane $z=4$.
Substituting $z=4$ in $z = {x^2} + {y^2}$ we obtain ${x^2} + {y^2} = 4$. Thus, ${\cal D}$ is a disk of radius $2$, namely ${x^2} + {y^2} \le 4$. So, the domain description of ${\cal D}$ in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 2,0 \le \theta \le 2\pi } \right\}$
Since ${x^2} + {y^2} \le z \le 8 - {x^2} - {y^2}$, in cylindrical coordinates, we get ${r^2} \le z \le 8 - {r^2}$.
Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 2,0 \le \theta \le 2\pi ,{r^2} \le z \le 8 - {r^2}} \right\}$
Evaluate the volume of ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 \mathop \smallint \limits_{z = {r^2}}^{8 - {r^2}} r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 r\left( {z|_{{r^2}}^{8 - {r^2}}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 r\left( {8 - 2{r^2}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 \left( {8r - 2{r^3}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {4{r^2} - \frac{1}{2}{r^4}} \right)|_0^2} \right){\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {16 - 8} \right){\rm{d}}\theta $
$ = 8\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = 16\pi $
Thus, the volume of the region is $16\pi $.
