Answer
The coordinates of the centroid is $\left( {0,0,\frac{{3h}}{4}} \right)$.
Work Step by Step
Recall from Exercise 19 in Section 15.8, the volume of a right circular cone of height $h$ and radius $R$ is $V = \frac{1}{3}\pi h{R^2}$. Let ${\cal W}$ denote the solid region of the right circular cone and the $z$-axis be the central axis. It is illustrated in Figure 22 (Exercise 37 in Section 16.4), and also in the figure attached.
From the figure attached, we see that the right circular cone has axial symmetry with respect to the $z$-axis. Since the right circular cone is the surface consisting of all rays passing through the origin and a point on the cone, we see that the angle the rays make with the $z$-axis is given by
$\tan \theta = \frac{{\sqrt {{x^2} + {y^2}} }}{z} = \frac{r}{z}$
$z = \frac{r}{{\tan \theta }}$
But we also have $\tan \theta = \frac{R}{h}$. So, $z = \frac{{rh}}{R}$.
Thus, the equation of the right-circular cone in Figure 22 in cylindrical coordinates is $z = \frac{{rh}}{R}$.
The description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le R,0 \le \theta \le 2\pi ,\frac{{rh}}{R} \le z \le h} \right\}$
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}{{\pi h{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = rh/R}^h {r^2}\cos \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{3}{{\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 = rh/R}^h {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}{{\pi h{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = rh/R}^h {r^2}\sin \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{3}{{\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 = rh/R}^h {r^2}{\rm{d}}z{\rm{d}}r} \right)$
Since $\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta = - \cos \theta |_0^{2\pi } = 0$, 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{3}{{\pi h{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \mathop \smallint \limits_{z = rh/R}^h zr{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{3}{{2\pi h{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \left( {{z^2}|_{rh/R}^h} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{3}{{2\pi h{R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R \left( {{h^2} - \frac{{{r^2}{h^2}}}{{{R^2}}}} \right)r{\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 \left( {r{h^2} - \frac{{{r^3}{h^2}}}{{{R^2}}}} \right){\rm{d}}r} \right)$
$ = \frac{3}{{h{R^2}}}\left( {\left( {\frac{{{r^2}{h^2}}}{2} - \frac{{{r^4}{h^2}}}{{4{R^2}}}} \right)|_0^R} \right)$
$ = \frac{3}{{h{R^2}}}\left( {\frac{{{R^2}{h^2}}}{2} - \frac{{{R^2}{h^2}}}{4}} \right) = \frac{{3h}}{4}$
So, the coordinates of the centroid is $\left( {0,0,\frac{{3h}}{4}} \right)$.
