Answer
The coordinates of the centroid: $\left( {\bar x,\bar y,\bar z} \right) = \left( {0,0,2.82} \right)$.
Work Step by Step
We have a region ${\cal W}$ lying 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 15 and the figure attached.
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 is a circle of radius $\sqrt 2 $.
Thus, the projection of ${\cal W}$ onto the $xy$-plane is the disk ${x^2} + {y^2} \le 2$.
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 ${\cal W}$:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le \sqrt 2 ,0 \le \theta \le 2\pi ,\sqrt {6 - {r^2}} \le z \le 4 - {r^2}} \right\}$
Evaluate the volume of ${\cal W}$:
$V = \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}^{\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 $4.824$.
Evaluate the average of $x$-coordinate:
$\bar x = \frac{1}{V}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} x{\rm{d}}V$
$ = \frac{1}{{\left( {\frac{{34}}{3} - 4\sqrt 6 } \right)\pi }}\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^2}\cos \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\left( {\frac{{34}}{3} - 4\sqrt 6 } \right)\pi }}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^{\sqrt 2 } \mathop \smallint \limits_{z = \sqrt {6 - {r^2}} }^{4 - {r^2}} {r^2}{\rm{d}}z{\rm{d}}r} \right)$
Since $\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta = \left( {\sin \theta } \right)|_0^{2\pi } = 0$, so $\bar x = 0$.
Evaluate the average of $y$-coordinate:
$\bar y = \frac{1}{V}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y{\rm{d}}V$
$ = \frac{1}{{\left( {\frac{{34}}{3} - 4\sqrt 6 } \right)\pi }}\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^2}\sin \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\left( {\frac{{34}}{3} - 4\sqrt 6 } \right)\pi }}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^{\sqrt 2 } \mathop \smallint \limits_{z = \sqrt {6 - {r^2}} }^{4 - {r^2}} {r^2}{\rm{d}}z{\rm{d}}r} \right)$
Since $\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta = - \left( {\cos \theta } \right)|_0^{2\pi } = 0$, so $\bar y = 0$.
Evaluate the average of $z$-coordinate:
$\bar z = \frac{1}{V}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V$
$ = \frac{1}{{\left( {\frac{{34}}{3} - 4\sqrt 6 } \right)\pi }}\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}} zr{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\left( {\frac{{34}}{3} - 4\sqrt 6 } \right)\pi }}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^{\sqrt 2 } \mathop \smallint \limits_{z = \sqrt {6 - {r^2}} }^{4 - {r^2}} zr{\rm{d}}z{\rm{d}}r} \right)$
$ = \frac{1}{{\left( {\frac{{34}}{3} - 4\sqrt 6 } \right)}}\mathop \smallint \limits_{r = 0}^{\sqrt 2 } r\left( {{z^2}|_{\sqrt {6 - {r^2}} }^{4 - {r^2}}} \right){\rm{d}}r$
$ = \frac{1}{{\left( {\frac{{34}}{3} - 4\sqrt 6 } \right)}}\mathop \smallint \limits_{r = 0}^{\sqrt 2 } r\left( {16 - 8{r^2} + {r^4} - 6 + {r^2}} \right){\rm{d}}r$
$ = \frac{1}{{\left( {\frac{{34}}{3} - 4\sqrt 6 } \right)}}\mathop \smallint \limits_{r = 0}^{\sqrt 2 } \left( {10r - 7{r^3} + {r^5}} \right){\rm{d}}r$
$ = \frac{1}{{\left( {\frac{{34}}{3} - 4\sqrt 6 } \right)}}\left( {5{r^2} - \frac{7}{4}{r^4} + \frac{1}{6}{r^6}} \right)|_0^{\sqrt 2 }$
$ = \frac{1}{{\left( {\frac{{34}}{3} - 4\sqrt 6 } \right)}}\left( {10 - 7 + \frac{4}{3}} \right) = \frac{{13}}{{3\left( {\frac{{34}}{3} - 4\sqrt 6 } \right)}} \simeq 2.82$
Thus, the coordinates of the centroid: $\left( {\bar x,\bar y,\bar z} \right) = \left( {0,0,2.82} \right)$.
