Answer
The coordinates of the centroid is $\left( { - \frac{{2R}}{{11\pi }},\frac{{2\left( {\sqrt 3 - 2} \right)R}}{{11\pi }},\frac{H}{2}} \right)$.
Work Step by Step
Let ${\cal W}$ denote the solid (A) in Figure 4. The description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le R,\frac{\pi }{6} \le \theta \le 2\pi ,0 \le z \le H} \right\}$
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 = \pi /6}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^H r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \left( {\mathop \smallint \limits_{\theta = \pi /6}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R r{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = 0}^H {\rm{d}}z} \right)$
$ = \frac{{11}}{6}\pi \left( {\frac{1}{2}{r^2}|_0^R} \right)\left( {z|_0^H} \right)$
$ = \frac{{11}}{{12}}\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{{12}}{{11\pi H{R^2}}}\mathop \smallint \limits_{\theta = \pi /6}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^H {r^2}\cos \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{{12}}{{11\pi H{R^2}}}\left( {\mathop \smallint \limits_{\theta = \pi /6}^{2\pi } \cos \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^2}{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = 0}^H {\rm{d}}z} \right)$
$ = \frac{{12}}{{11\pi H{R^2}}}\left( {\sin \theta |_{\pi /6}^{2\pi }} \right)\left( {\frac{1}{3}{r^3}|_0^R} \right)H$
$ = \frac{{12}}{{11\pi H{R^2}}}\left( { - \frac{1}{2}} \right)\left( {\frac{1}{3}{R^3}} \right)H = - \frac{{2R}}{{11\pi }}$
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{{12}}{{11\pi H{R^2}}}\mathop \smallint \limits_{\theta = \pi /6}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^H {r^2}\sin \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{{12}}{{11\pi H{R^2}}}\left( {\mathop \smallint \limits_{\theta = \pi /6}^{2\pi } \sin \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^2}{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = 0}^H {\rm{d}}z} \right)$
$ = \frac{{12}}{{11\pi H{R^2}}}\left( { - \cos \theta |_{\pi /6}^{2\pi }} \right)\left( {\frac{1}{3}{r^3}|_0^R} \right)H$
$ = \frac{{12}}{{11\pi H{R^2}}}\left( { - 1 + \frac{1}{2}\sqrt 3 } \right)\left( {\frac{1}{3}{R^3}} \right)H = \frac{{2\left( {\sqrt 3 - 2} \right)R}}{{11\pi }}$
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{{12}}{{11\pi H{R^2}}}\mathop \smallint \limits_{\theta = \pi /6}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^H rz{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{{12}}{{11\pi H{R^2}}}\left( {\mathop \smallint \limits_{\theta = \pi /6}^{2\pi } {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R r{\rm{d}}r} \right)\left( {\mathop \smallint \limits_{z = 0}^H z{\rm{d}}z} \right)$
$ = \frac{{12}}{{11\pi H{R^2}}}\left( {\frac{{11}}{6}\pi } \right)\left( {\frac{1}{2}{r^2}|_0^R} \right)\left( {\frac{1}{2}{z^2}|_0^H} \right)$
$ = \frac{{12}}{{11\pi H{R^2}}}\left( {\frac{{11}}{6}\pi } \right)\left( {\frac{1}{2}{R^2}} \right)\left( {\frac{1}{2}{H^2}} \right) = \frac{H}{2}$
So, the coordinates of the centroid is $\left( { - \frac{{2R}}{{11\pi }},\frac{{2\left( {\sqrt 3 - 2} \right)R}}{{11\pi }},\frac{H}{2}} \right)$.
