Use cylinder's formula for frustum (conical frustum)

Question

I know that frustum(conical) has a formula for its volume,i.e. $\frac1 3\pi h(r^2+R^2+rR)$, but why can't we place the average of two radii into cylinder's formula: $\pi(\frac{r+R}2)^2(h)$?

I need the reason why I get wrong answer on doing this.

Answer 1

Suppose the conical frustum ("cone", for short) is standing upright, with the larger radius ($R$) at the bottom, and the smaller one ($r$) at the top.

Suppose you replace the cone by an "average" cylinder of radius $\tfrac12(r+R)$.

At the base, we will be removing a volume $V_{\text{base}}$, since the cylinder is smaller than the cone in this region. At the top, we will be adding a volume $V_{\text{top}}$. The cylinder will give us the same volume as the cone if $V_{\text{base}} = V_{\text{top}}$.

The added and subtracted volumes, $V_{\text{base}}$ and $V_{\text{top}}$, are both solids of revolution, formed by rotating triangles around the cylinder/cone axis. The two triangles have identical shapes, which is what leads you to believe that they produce the same volumes, I guess. But, in $V_{\text{base}}$, the triangle is swept along a larger circular path, so it has a larger volume than $V_{\text{top}}$. In short, $V_{\text{base}} > V_{\text{top}}$, which means we're removing more than we're adding. So, the "average" cylinder has a smaller volume than the cone.

There's a picture below. The yellow and blue triangles are the same size, but, when revolved, the yellow ones give a larger volume.

Answer 2

Essentially, what you'd need is the average of the areas of the horizontal slices into which the frustrum is cut by planes paralell to its base, not their diameters.