Answer
(a) ${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 2,0 \le \theta \le 2\pi ,2 \le z \le 6 - {r^2}} \right\}$
(b) the volume of ${\cal W}$ is $8\pi $.
Work Step by Step
We have the region ${\cal W}$ above the plane $z=2$ and below the paraboloid $z = 6 - \left( {{x^2} + {y^2}} \right)$. So, $2 \le z \le 6 - \left( {{x^2} + {y^2}} \right)$. The region is illustrated in the figure attached.
(a) To find the projection of ${\cal W}$ onto the $xy$-plane we solve for the intersection of the plane $z=2$ and the paraboloid $z = 6 - \left( {{x^2} + {y^2}} \right)$.
Substituting $z=2$ in the equation of the paraboloid gives
$2 = 6 - \left( {{x^2} + {y^2}} \right)$
${x^2} + {y^2} = 4$
Notice that this a circle of radius $2$.
Thus, the projection of ${\cal W}$ onto the $xy$-plane is ${\cal D}$, a disk ${x^2} + {y^2} \le 4$.
So, the polar description of ${\cal D}$:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 2,0 \le \theta \le 2\pi } \right\}$
In cylindrical coordinates, the paraboloid becomes $z = 6 - {r^2}$.
Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 2,0 \le \theta \le 2\pi ,2 \le z \le 6 - {r^2}} \right\}$
(b) Using Eq. (5) of Theorem 2, we evaluate the volume of ${\cal W}$:
$V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 \mathop \smallint \limits_{z = 2}^{6 - {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|_2^{6 - {r^2}}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 \left( {4r - {r^3}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {2{r^2} - \frac{1}{4}{r^4}} \right)|_0^2} \right){\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {8 - 4} \right){\rm{d}}\theta $
$ = 4\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = 8\pi $
Thus, the volume of ${\cal W}$ is $8\pi $.
