Answer
The coordinates of the centroid: $\left( {\bar x,\bar y,\bar z} \right) = \left( {0,0,\frac{3}{8}H} \right)$. Thus, it depends on the height $H$ but not on the radius $R$.
Work Step by Step
We have the region ${\cal W}$ defined by the upper half of the ellipsoid ${x^2} + {y^2} + {\left( {\frac{R}{H}z} \right)^2} = {R^2}$, where $z \ge 0$ as is shown in Figure 16.
Write
${x^2} + {y^2} + {\left( {\frac{R}{H}z} \right)^2} = {R^2}$
${\left( {\frac{R}{H}z} \right)^2} = {R^2} - {x^2} - {y^2}$
$z = \frac{H}{R}\sqrt {{R^2} - {x^2} - {y^2}} $
We can consider ${\cal W}$ as a $z$-simple region bounded below by $z=0$ and bounded above by $z = \frac{H}{R}\sqrt {{R^2} - {x^2} - {y^2}} $. The projection of ${\cal W}$ onto the $xy$-plane is ${\cal D}$, a disk of radius $R$, ${x^2} + {y^2} \le R$. Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le R,0 \le \theta \le 2\pi ,0 \le z \le \frac{H}{R}\sqrt {{R^2} - {r^2}} } \right\}$
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}^R \mathop \smallint \limits_{z = 0}^{\frac{H}{R}\sqrt {{R^2} - {r^2}} } r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^{\frac{H}{R}\sqrt {{R^2} - {r^2}} } r{\rm{d}}z{\rm{d}}r} \right)$
$ = 2\pi \mathop \smallint \limits_{r = 0}^R r\left( {z|_0^{\frac{H}{R}\sqrt {{R^2} - {r^2}} }} \right){\rm{d}}r$
$ = 2\pi \mathop \smallint \limits_{r = 0}^R r\left( {\frac{H}{R}\sqrt {{R^2} - {r^2}} } \right){\rm{d}}r$
$ = \frac{{2\pi H}}{R}\mathop \smallint \limits_{r = 0}^R r\sqrt {{R^2} - {r^2}} {\rm{d}}r$
$ = - \frac{{2\pi H}}{{3R}}\left( {{{\left( {{R^2} - {r^2}} \right)}^{3/2}}|_0^R} \right) = \frac{2}{3}\pi H{R^2}$
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{3}{{2\pi H{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^{\frac{H}{R}\sqrt {{R^2} - {r^2}} } {r^2}\cos \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{3}{{2\pi H{R^2}}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^{\frac{H}{R}\sqrt {{R^2} - {r^2}} } {r^2}{\rm{d}}z{\rm{d}}r} \right)$
Since $\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta = \sin \theta |_0^{2\pi } = 0$, 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{3}{{2\pi H{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^{\frac{H}{R}\sqrt {{R^2} - {r^2}} } {r^2}\sin \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{3}{{2\pi H{R^2}}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^{\frac{H}{R}\sqrt {{R^2} - {r^2}} } {r^2}{\rm{d}}z{\rm{d}}r} \right)$
Since $\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta = - \left( {\cos \theta |_0^{2\pi }} \right) = 0$, 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{3}{{2\pi H{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^{\frac{H}{R}\sqrt {{R^2} - {r^2}} } zr{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{3}{{2\pi H{R^2}}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^{\frac{H}{R}\sqrt {{R^2} - {r^2}} } zr{\rm{d}}z{\rm{d}}r} \right)$
$ = \frac{3}{{2\pi H{R^2}}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta } \right)\left( {\frac{1}{2}\mathop \smallint \limits_{r = 0}^R \left( {{z^2}|_0^{\frac{H}{R}\sqrt {{R^2} - {r^2}} }} \right)r{\rm{d}}r} \right)$
$ = \frac{3}{{2H{R^2}}}\mathop \smallint \limits_{r = 0}^R \frac{{{H^2}}}{{{R^2}}}\left( {{R^2} - {r^2}} \right)r{\rm{d}}r$
$ = \frac{{3H}}{{2{R^4}}}\mathop \smallint \limits_{r = 0}^R \left( {r{R^2} - {r^3}} \right){\rm{d}}r$
$ = \frac{{3H}}{{2{R^4}}}\left( {\frac{1}{2}\left( {{r^2}|_0^R} \right){R^2} - \frac{1}{4}\left( {{r^4}|_0^R} \right)} \right)$
$ = \frac{{3H}}{{2{R^4}}}\left( {\frac{1}{2}{R^4} - \frac{1}{4}{R^4}} \right) = \frac{3}{8}H$
So, the coordinates of the centroid: $\left( {\bar x,\bar y,\bar z} \right) = \left( {0,0,\frac{3}{8}H} \right)$. Thus, it depends on the height $H$ but not on the radius $R$.
