Step 1

We are given the height of the frustrum of the right-circular cone as h and the radius of the top and bottom portion as r and R respectively.

We are asked to locate the centroid $\bar z$ of the frustum of the right-circular cone.

 
Step 2

We can consider the differential element of the circular disk parallel to the x-axis with radius y and thickness of $dz$.

To calculate the value of y, we need to apply the property of the similar triangle as:

\begin{array}{c} \left( {\frac{{y - r}}{{h - z}}} \right) = \left( {\frac{{R - r}}{h}} \right)\\ y = \frac{{\left( {R - r} \right)\left( {h - z} \right)}}{h} + r\\ y = \frac{{Rh - Rz - rh + rz}}{h} + r\\ y = \frac{{Rh + \left( {r - R} \right)z}}{h} \end{array}

The volume of the differential element can be given as:

\[dV = \pi {y^2}dz\]

Substitute the values in the above expression, we get:

\begin{array}{c} dV = \pi \times {\left( {\frac{{Rh + \left( {r - R} \right)z}}{h}} \right)^2}dz\\ dV = \frac{\pi }{{{h^2}}}\left[ {{R^2}{h^2} + {{\left( {r - R} \right)}^2}{z^2} + 2Rh\left( {r - R} \right)z} \right]dz \end{array} … (1)
 
Step 3

To calculate the centroid $\bar z$ of the frustum of the right-circular cone, we have:

\[\bar z = \frac{{\int\limits_V {z'dV} }}{{\int\limits_V {dV} }}\] …. (2)

Here, $z'$ is the z-coordinate of centroid of the differential element and consider $z' = z$.

To calculate the integral of the differential element, we need to integrate equation (1) with the limits from $0$ to h as:

\[\begin{array}{l} \int\limits_V {dV} = \int\limits_0^h {\frac{\pi }{{{h^2}}}\left[ {{R^2}{h^2} + {{\left( {r - R} \right)}^2}{z^2} + 2Rh\left( {r - R} \right)z} \right]} \cdot dz\\ \int\limits_V {dV} = \frac{\pi }{{{h^2}}}\left[ {{R^2}{h^2}z + {{\left( {r - R} \right)}^2} \times \left( {\frac{{{z^3}}}{3}} \right) + 2Rh\left( {r - R} \right)\left( {\frac{{{z^2}}}{2}} \right)} \right]_0^h \end{array}\]

Substitute the limits in z, we get:

\[\begin{array}{c} \int\limits_V {dV} = \frac{\pi }{{{h^2}}}\left[ {{R^2}{h^2} \times \left( {h - 0} \right) + {{\left( {r - R} \right)}^2} \times \left( {\frac{{{h^3}}}{3} - 0} \right) + 2Rh\left( {r - R} \right)\left( {\frac{{{h^2}}}{2} - 0} \right)} \right]\\ \int\limits_V {dV} = \frac{\pi }{{{h^2}}}\left[ {{R^2}{h^3} + \left( {{r^2} + {R^2} - 2rR} \right)\left( {\frac{{{h^3}}}{3}} \right) + R{h^3}\left( {r - R} \right)} \right]\\ \int\limits_V {dV} = \frac{\pi }{{{h^2}}} \times \frac{{{h^3}}}{3}\left[ {3{R^2} + {r^2} + {R^2} - 2rR + 3rR - 3{R^2}} \right]\\ \int\limits_V {dV} = \frac{{\pi h}}{3}\left[ {{R^2} + {r^2} + rR} \right] \end{array}\]

Similarly, to calculate the integral of the z-coordinate of centroid of the differential element, we need to integrate \[z'dV\] with the limit $0$ to h as:

\[\begin{array}{c} \int\limits_V {z'dV} = \int\limits_0^h {\left[ {z \times \frac{\pi }{{{h^2}}}\left[ {{R^2}{h^2} + {{\left( {r - R} \right)}^2}{z^2} + 2Rh\left( {r - R} \right)z} \right]dz} \right]} \\ \int\limits_V {z'dV} = \frac{\pi }{{{h^2}}}\left[ {{R^2}{h^2}\left( {\frac{{{z^2}}}{2}} \right) + {{\left( {r - R} \right)}^2} \times \left( {\frac{{{z^4}}}{4}} \right) + 2Rh\left( {r - R} \right)\left( {\frac{{{z^3}}}{3}} \right)} \right]_0^h\\ \int\limits_V {z'dV} = \frac{\pi }{{{h^2}}}\left[ {{R^2}{h^2}\left( {\frac{{{h^2}}}{2} - 0} \right) + {{\left( {r - R} \right)}^2} \times \left( {\frac{{{h^4}}}{4} - 0} \right) + 2Rh\left( {r - R} \right)\left( {\frac{{{h^3}}}{3} - 0} \right)} \right] \end{array}\]

On further solving, we get:

\[\begin{array}{c} \int\limits_V {z'dV} = \frac{\pi }{{{h^2}}} \times \frac{{{h^4}}}{{12}}\left[ {6{R^2} + 3{r^2} + 3{R^2} - 6rR + 8rR - 8{R^2}} \right]\\ \int\limits_V {z'dV} = \frac{{\pi {h^2}}}{{12}}\left[ {{R^2} + 3{r^2} + 2rR} \right] \end{array}\]

Substitute the values in equation (2), we get:

\begin{array}{c} \bar z = \left( {\frac{{\frac{{\pi {h^2}}}{{12}}\left[ {{R^2} + 3{r^2} + 2rR} \right]}}{{\frac{{\pi h}}{3}\left[ {{R^2} + {r^2} + rR} \right]}}} \right)\\ \bar z = \frac{{h\left( {{R^2} + 3{r^2} + 2rR} \right)}}{{4\left( {{R^2} + {r^2} + rR} \right)}} \end{array}