Answer
The volume of the region is $6\pi $.
Work Step by Step
Let ${\cal W}$ denote the region $4 - {x^2} - {y^2} \le z \le 10 - 4{x^2} - 4{y^2}$.
We find the boundary curve between the two surfaces $z = 4 - {x^2} - {y^2}$ and $z = 10 - 4{x^2} - 4{y^2}$:
$z = 4 - {x^2} - {y^2} = 10 - 4{x^2} - 4{y^2}$
$3{x^2} + 3{y^2} = 6$
${x^2} + {y^2} = 2$
So, the boundary curve is the circle of radius $\sqrt 2 $ defined by ${x^2} + {y^2} = 2$.
Thus, the projection of ${\cal W}$ onto the $xy$-plane is the disk ${x^2} + {y^2} \le 2$. Let ${\cal D}$ denote the disk.
The description of ${\cal D}$ in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le \sqrt 2 ,0 \le \theta \le 2\pi } \right\}$
In cylindrical coordinates ${\cal W}$ is bounded below by $z = 4 - {r^2}$ and bounded above by $z = 10 - 4{r^2}$. So, the description:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le \sqrt 2 ,0 \le \theta \le 2\pi ,4 - {r^2} \le z \le 10 - 4{r^2}} \right\}$
Using Eq. (5) of Theorem 2 in Section 16.4, 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 = 4 - {r^2}}^{10 - 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|_{4 - {r^2}}^{10 - 4{r^2}}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^{\sqrt 2 } r\left( {10 - 4{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( {6r - 3{r^3}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {3{r^2} - \frac{3}{4}{r^4}} \right)|_0^{\sqrt 2 }} \right){\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {6 - 3} \right){\rm{d}}\theta = 3\cdot2\pi = 6\pi $
So, the volume of the region is $6\pi $.
