Answer
The coordinates of the centroid:
$\left( {\bar x,\bar y,\bar z} \right) = \left( { - \frac{R}{4},0,\frac{{5H}}{8}} \right)$
Work Step by Step
Using $\frac{x}{R} + \frac{z}{H} = 1$, we obtain $z = H - \frac{H}{R}x$.
We can consider the region ${\cal W}$ as a $z$-simple region bounded below by $z=0$ and bounded above by $z = H - \frac{H}{R}x$. The projection of ${\cal W}$ onto the $xy$-plane is a disk ${\cal D}$ of radius $R$ defined by ${x^2} + {y^2} \le {R^2}$.
In cylindrical coordinates, $z = H - \frac{H}{R}r\cos \theta $. Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le R,0 \le \theta \le 2\pi ,0 \le z \le H - \frac{H}{R}r\cos \theta } \right\}$
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}^{H - \frac{H}{R}r\cos \theta } r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R r\left( {z|_0^{H - \frac{H}{R}r\cos \theta }} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R rH\left( {1 - \frac{r}{R}\cos \theta } \right){\rm{d}}r{\rm{d}}\theta $
$ = H\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \left( {r - \frac{{{r^2}}}{R}\cos \theta } \right){\rm{d}}r{\rm{d}}\theta $
$ = H\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {\frac{1}{2}{r^2} - \frac{1}{{3R}}{r^3}\cos \theta } \right)|_0^R} \right){\rm{d}}\theta $
$ = H\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{1}{2}{R^2} - \frac{1}{3}{R^2}\cos \theta } \right){\rm{d}}\theta $
$ = \frac{1}{2}H{R^2}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta - \frac{1}{3}H{R^2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta $
$ = \pi H{R^2} - \frac{1}{3}H{R^2}\left( {\sin \theta |_0^{2\pi }} \right)$
So, $V = \pi H{R^2}$.
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{1}{{\pi H{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^{H - \frac{H}{R}r\cos \theta } {r^2}\cos \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\pi H{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R {r^2}\cos \theta \left( {z|_0^{H - \frac{H}{R}r\cos \theta }} \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\pi H{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R {r^2}\cos \theta \left( {H - \frac{H}{R}r\cos \theta } \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \left( {{r^2}\cos \theta - \frac{{{r^3}}}{R}{{\cos }^2}\theta } \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R {r^2}\cos \theta {\rm{d}}r{\rm{d}}\theta - \frac{1}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \frac{{{r^3}}}{R}{\cos ^2}\theta {\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{1}{3}{r^3}|_0^R} \right)\cos \theta {\rm{d}}\theta - \frac{1}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{{{r^4}}}{{4R}}|_0^R} \right){\cos ^2}\theta {\rm{d}}\theta $
$ = \frac{R}{{3\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta - \frac{R}{{4\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\cos ^2}\theta {\rm{d}}\theta $
Using the Double-angle formulas in Section 1.4:
${\cos ^2}x = \frac{1}{2}\left( {1 + \cos 2x} \right)$
we get
$\bar x = \frac{R}{{3\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta - \frac{R}{{8\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta - \frac{R}{{8\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos 2\theta {\rm{d}}\theta $
$ = \frac{R}{{3\pi }}\left( {\sin \theta |_0^{2\pi }} \right) - \frac{R}{{8\pi }}\left( {\theta |_0^{2\pi }} \right) - \frac{R}{{16\pi }}\left( {\sin 2\theta |_0^{2\pi }} \right)$
$ = - \frac{R}{{8\pi }}\left( {2\pi } \right) = - \frac{R}{4}$
So, $\bar x = - \frac{R}{4}$.
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}{{\pi H{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^{H - \frac{H}{R}r\cos \theta } {r^2}\sin \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\pi H{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R {r^2}\sin \theta \left( {z|_0^{H - \frac{H}{R}r\cos \theta }} \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\pi H{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R {r^2}\sin \theta \left( {H - \frac{H}{R}r\cos \theta } \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \left( {{r^2}\sin \theta - \frac{{{r^3}}}{R}\cos \theta \sin \theta } \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R {r^2}\sin \theta {\rm{d}}r{\rm{d}}\theta - \frac{1}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \frac{{{r^3}}}{R}\cos \theta \sin \theta {\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{1}{3}{r^3}|_0^R} \right)\sin \theta {\rm{d}}\theta - \frac{1}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{{{r^4}}}{{4R}}|_0^R} \right)\cos \theta \sin \theta {\rm{d}}\theta $
$ = \frac{R}{{3\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta - \frac{R}{{4\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta \sin \theta {\rm{d}}\theta $
$ = - \frac{R}{{3\pi }}\left( {\cos \theta |_0^{2\pi }} \right) - \frac{R}{{8\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin 2\theta {\rm{d}}\theta $
$ = \frac{R}{{16\pi }}\left( {\cos 2\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{1}{{\pi H{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = 0}^{H - \frac{H}{R}r\cos \theta } zr{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{\pi H{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R r\left( {\frac{1}{2}{z^2}|_0^{H - \frac{H}{R}r\cos \theta }} \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{H}{{2\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R r\left( {1 - 2\frac{r}{R}\cos \theta + \frac{{{r^2}}}{{{R^2}}}{{\cos }^2}\theta } \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{H}{{2\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \left( {r - 2\frac{{{r^2}}}{R}\cos \theta + \frac{{{r^3}}}{{{R^2}}}{{\cos }^2}\theta } \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{H}{{2\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{1}{2}\left( {{r^2}|_0^R} \right) - \frac{2}{{3R}}\left( {{r^3}|_0^R} \right)\cos \theta + \frac{1}{{4{R^2}}}\left( {{r^4}|_0^R} \right){{\cos }^2}\theta } \right){\rm{d}}\theta $
$ = \frac{H}{{2\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{{{R^2}}}{2} - \frac{{2{R^2}}}{3}\cos \theta + \frac{{{R^2}}}{4}{{\cos }^2}\theta } \right){\rm{d}}\theta $
$ = \frac{H}{{4\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta - \frac{H}{{3\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta + \frac{H}{{8\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\cos ^2}\theta {\rm{d}}\theta $
Using the Double-angle formulas in Section 1.4:
${\cos ^2}x = \frac{1}{2}\left( {1 + \cos 2x} \right)$
we get
$\bar z = \frac{H}{{4\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta - \frac{H}{{3\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta + \frac{H}{{16\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta + \frac{H}{{16\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos 2\theta {\rm{d}}\theta $
$ = \frac{H}{2} - \frac{H}{{3\pi }}\left( {\sin \theta |_0^{2\pi }} \right) + \frac{H}{8} + \frac{H}{{32\pi }}\left( {\sin 2\theta |_0^{2\pi }} \right)$
$ = \frac{H}{2} + \frac{H}{8} = \frac{{5H}}{8}$
So, the coordinates of the centroid: $\left( {\bar x,\bar y,\bar z} \right) = \left( { - \frac{R}{4},0,\frac{{5H}}{8}} \right)$.
