Answer: The cone has height \(h\) and base radius \(r\), so at distance \(x\) from the origin its radius is \(y=\frac{r}{h}x\).

The volume is

\(V=\frac{1}{3}\pi r^{2}h\).

To find the distance of the centroid from the origin, let \(\bar{x}\) be the x-coordinate of the centroid. Using moments about the origin,

\(V\bar{x}=\int_{0}^{h} \pi y^{2} x \, dx\).

Substitute \(y=\frac{r}{h}x\):

\(V\bar{x}=\int_{0}^{h} \pi \left(\frac{r}{h}x\right)^{2}x \, dx=\frac{\pi r^{2}}{h^{2}}\int_{0}^{h} x^{3}\,dx\).

So

\(V\bar{x}=\frac{\pi r^{2}}{h^{2}}\left[\frac{x^{4}}{4}\right]_{0}^{h}=\frac{\pi r^{2}h^{2}}{4}\).

Hence

\(\bar{x}=\frac{\pi r^{2}h^{2}/4}{(1/3)\pi r^{2}h}=\frac{3h}{4}.\)

Therefore the centroid is at a distance \(\frac{3}{4}h\) from the origin \(O\).

Consider the cone formed by rotating the region under the line \(y=kx\) from \(x=0\) to \(x=h\) about the x-axis. The centroid must lie on the axis of symmetry, so we only need its x-coordinate.

At position \(x\), the cross-section perpendicular to the x-axis is a disc of radius \(y\), so its area is \(\pi y^{2}\). The moment of this thin slice about the origin is therefore \(x\cdot \pi y^{2}\,dx\).

Since the line passes through \((h,r)\), we have \(y=\frac{r}{h}x\). Hence

\(\text{moment about }O=\int_{0}^{h} x\pi y^{2}\,dx=\int_{0}^{h} x\pi \left(\frac{r}{h}x\right)^{2}dx\).

So

\(\text{moment about }O=\frac{\pi r^{2}}{h^{2}}\int_{0}^{h} x^{3}\,dx=\frac{\pi r^{2}}{h^{2}}\left[\frac{x^{4}}{4}\right]_{0}^{h}=\frac{\pi r^{2}h^{2}}{4}.\)

The volume of the cone is

\(V=\frac{1}{3}\pi r^{2}h.\)

Therefore the x-coordinate of the centroid is

\(\bar{x}=\frac{\text{moment about }O}{V}=\frac{\pi r^{2}h^{2}/4}{(1/3)\pi r^{2}h}=\frac{3h}{4}.\)

So the centroid is a distance \(\frac{3}{4}h\) from the origin \(O\).