Center of gravity of a cone

The cone has a cross-sectional area $A(z) = \pi r^2 = \pi \left(R - \frac{1}{2} R z \right)^2 = \pi \left(\frac{z^2}{4} - z + 1 \right)$ The area $A(z)$ is a quadratic function in $z$ . The volume of the truncated cone results to $V = \int_0^1 A(z) dz = \pi \left[\frac{z^3}{12} - \frac{z^2}{2} + z \right]_0^1 = \frac{7 \pi}{12}$ Therefore, the center of gravity results $\begin{aligned} z_0 &= \frac{1}{V} \int_0^1 z A(z) dz = \frac{\pi}{V} \int_0^1 \left(\frac{z^3}{4} - z^2 + z \right) dz \\ &= \frac{\pi}{V} \left[\frac{z^4}{16} - \frac{z^3}{3} + \frac{z^2}{2} \right]_0^1 = \frac{\pi}{V} \frac{11}{48} \\ &= \frac{11}{28} \approx 0.3929 \end{aligned}$