Answer
(a) we show that the projection of ${\cal W}$ on the $xy$-plane is the disk ${x^2} + {y^2} \le 2$.
(b) the volume of ${\cal W}$ is $4.824$.
Work Step by Step
We have a region ${\cal W}$ above the sphere ${x^2} + {y^2} + {z^2} = 6$ and below the paraboloid $z = 4 - {x^2} - {y^2}$ as is shown in Figure 19 and the figure attached.
(a) Expressing in cylindrical coordinates, the sphere and the paraboloid are given by ${r^2} + {z^2} = 6$ and $z = 4 - {r^2}$, respectively. To find the projection of ${\cal W}$ onto the $xy$-plane we solve for the intersection of the two surfaces.
Write the paraboloid: ${r^2} = 4 - z$. Substituting it in the equation of the sphere gives
$4 - z + {z^2} = 6$
${z^2} - z - 2 = 0$
$\left( {z + 1} \right)\left( {z - 2} \right) = 0$
So, the two surfaces intersect at the planes $z=-1$ and $z=2$. Since the region is above the $xy$-plane, we choose $z=2$.
Substituting $z=2$ in $z = 4 - {x^2} - {y^2}$ we obtain the boundary of the region:
$2 = 4 - {x^2} - {y^2}$
${x^2} + {y^2} = 2$
Notice that this a circle of radius $\sqrt 2 $.
Thus, the projection of ${\cal W}$ on the $xy$-plane is the disk ${x^2} + {y^2} \le 2$.
(b) In polar coordinates, the disk ${\cal D}:{x^2} + {y^2} \le 2$ is given by
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le \sqrt 2 ,0 \le \theta \le 2\pi } \right\}$
Expressing in cylindrical coordinates, the region ${\cal W}$ is bounded below by the sphere ${r^2} + {z^2} = 6$ and bounded above by the paraboloid $z = 4 - {r^2}$. Therefore, $z$ ranges from $z = \sqrt {6 - {r^2}} $ to $z = 4 - {r^2}$. Thus, the region description of ${\cal W}$:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le \sqrt 2 ,0 \le \theta \le 2\pi ,\sqrt {6 - {r^2}} \le z \le 4 - {r^2}} \right\}$
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}^{\sqrt 2 } \mathop \smallint \limits_{z = \sqrt {6 - {r^2}} }^{4 - {r^2}} r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt 2 } r\left( {z|_{\sqrt {6 - {r^2}} }^{4 - {r^2}}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt 2 } \left( {4r - {r^3} - r\sqrt {6 - {r^2}} } \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {2{r^2} - \frac{1}{4}{r^4} + \frac{1}{3}{{\left( {6 - {r^2}} \right)}^{3/2}}} \right)|_0^{\sqrt 2 }} \right){\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {4 - 1 + \frac{8}{3} - 2\sqrt 6 } \right){\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{{17}}{3} - 2\sqrt 6 } \right){\rm{d}}\theta $
$ = \left( {\frac{{34}}{3} - 4\sqrt 6 } \right)\pi \simeq 4.824$
Thus, the volume of ${\cal W}$ is $4.824$.
