# Intuition for Volume of Frustum vs. Trapezoid

The area of a trapezoid with height $$h$$ and base lengths $$b_1$$ and $$b_2$$ is given by

$$\frac{b_1 + b_2}{2} h.$$

I like to think of this as saying that the area of a trapezoid is the same as the area of the rectangle with the same height and average width of the trapezoid.

It seems to me that a frustum is the 3D analog of a trapezoid. If we reason analogously to the trapezoid, we might expect the volume of a frustum with height $$h$$ and base areas $$A_1$$ and $$A_2$$ to be

$$\frac{A_1 + A_2}{2} h,$$

which is the volume of a prism with the same height and average base area. But that's not what we get; instead, we get

$$\frac{A_1 + A_2 + \sqrt{A_1A_2}}{3} h,$$

which we can derive by extrapolating the frustum to a pyramid and removing the part outside the frustum.

My question is: is there any intuitive interpretation for this extra $$\sqrt{A_1A_2}$$ term? What does it represent?

While the cross-section varies linearly in the trapezoid, the cross-section of a frustum varies quadratically. Consider $$A=ay^2$$, the average function value on the interval $$[y_1,y_2]$$ is $$\frac{a\times (y_2^3-y_1^3)}{3(y_2-y_1)}=\frac{a(y_2^2+y_1^2+y_2y_1)}{3}$$. Let $$a=\pi$$ and $$y=r$$ for example. $$A_\text{avg}=\frac{\pi(r_2^2+r_1^2+r_2r_1)}{3}$$

Per your formula, $$r_1=\frac{\sqrt{A_1}}{\sqrt {\pi}}$$ and $$r_2=\frac{\sqrt{A_2}}{\sqrt {\pi}}$$, or, generally, $$r_1=\frac{\sqrt{A_1}}{\sqrt {a}}$$ and $$r_2=\frac{\sqrt{A_2}}{\sqrt {a}}$$

We could look back at one of the ways we obtain the area of the trapezoid to see if it provides insight into the volume of the conical frustum. If we start with a pair of similar right triangles of heights $$\ H \$$ and $$\ H + h \ \ , \$$ which have bases $$\ b_1 \$$ and $$\ b_2 \ \ , \$$ respectively, the area of the trapezoid is given by $$\mathcal{A} \ \ = \ \ \frac12·(H \ + \ h)·b_2 \ - \ \frac12·H·b_1 \ \ = \ \ \frac12·H·(b_2 \ - \ b_1) \ + \frac12·h·b_2 \ \ . \$$

We can form the similarity proportion $$\frac{H}{b_1} \ \ = \ \ \frac{H \ + \ h}{b_2} \ \ \Rightarrow \ \ H·b_2 \ \ = \ \ H·b_1 \ + \ h·b_1 \ \ \Rightarrow \ \ h·b_1 \ \ = \ \ H·(b_2 \ - \ b_1) \ \ .$$

Inserting this into the area equation then yields $$\ \mathcal{A} \ = \ \frac12·h· b_1 \ + \ \frac12·h·b_2 \ \ . \$$

We will now imagine revolving these triangles about the vertical leg and apply Pappus' (second) centroid theorem, which states that the volume of a solid of revolution is given by the area of the figure revolved times the circumference of the circle "traveled" by the figure's centroid. The centroid of one of the right triangles is located one-third of the way from its base toward its apex and one-third of the way from the vertical leg parallel to its base (a special case of the theorem concerning the centroid of a triangle).

The centroid of the smaller triangle (green dot) is then at a distance $$\ \frac13·b_1 \$$ from its vertical leg and that of the larger triangle (red dot) is at a distance $$\ \frac13·b_2 \$$ from its vertical leg. The volume of the smaller cone is given by Pappus as $$\mathcal{V}_1 \ \ = \ \ \left(\frac12·H·b_1 \right)·\left(2 \pi · \frac13·b_1 \right) \ \ = \ \ \frac13·H·(\pi·b_1^2) \ \ ,$$

the volume of the larger cone as $$\mathcal{V}_2 \ \ = \ \ \left(\frac12·[H \ + h]·b_2 \right)·\left(2 \pi · \frac13·b_2 \right) \ \ = \ \ \frac13·[H \ + h]·(\pi·b_2^2) \ \ ,$$

and hence the volume of the frustum is $$\mathcal{V}_{fr} \ \ = \ \ \frac13·H ·( \ \pi·b_2^2 \ - \ \pi·b_1^2 \ ) \ + \ \frac13·h·(\pi·b_2^2)$$ $$= \ \ \frac13·H · \ \pi·(b_2 \ - \ b_1 )·(b_2 \ + \ b_1 ) \ + \ \frac13·h·(\pi·b_2^2) \ \ .$$

Again using the result from the similarity proportion, this becomes $$\mathcal{V}_{fr} \ \ = \ \ \frac13·h · \ \pi·b_1 ·(b_2 \ + \ b_1 ) \ + \ \frac13·h·(\pi·b_2^2)$$ $$= \ \ \frac13·h · ( \ \pi·b_1 · b_2 \ + \ \pi·b_1^2 \ + \ \pi·b_2^2 \ ) \ \ .$$

The geometric mean of the two circular conical bases is $$\ \sqrt{\pi·b_1^2 \ · \ \pi·b_2^2} \ = \ \pi·b_1 · b_2 \ \ , \$$ which is the first term in the parentheses above. I'm not sure there's an intuitive way to see that this geometric-mean-of-areas term should appear in the frustum volume formula, but perhaps this gives another way to see how it arises.

ADDENDUM -- It might be mentioned that the frustum volume formula $$\ \mathcal{V}_{fr} \ = \ \frac{A_1 + A_2 + \sqrt{A_1A_2}}{3} · h \$$ is completely general for any frustum with similar top and bottom surfaces. For example, the volume of a square-pyramidal frustum would have $$\ A_1 \ = \ s_1^2 \ \ , \ \ A_2 \ = \ s_2^2 \ \ , \$$ a frustum with equilateral triangular "bases" (tetrahedronal frustum?) would use $$\ A_1 \ = \ \frac{\sqrt3}{4}s_1^2 \ \ , \ \ A_2 \ = \ \frac{\sqrt3}{4}s_2^2 \ \ , \$$ and so forth (see the "Volume" section of Formulas here).