Sharing a Pint

Four friends, exhausted after a long hike, stagger into a pub to slake their thirst. But, pooling their funds, they have enough money for only one pint.

Annie drinks first, until the surface of the beer is half way down the side (Fig. 1(A)). Then Barry drinks until the surface touches the bottom corner (Fig. 1(B)). Cathy then takes a sup, leaving the level as in Fig. 1(C), with the surface through the centre of the bottom. Finally, Danny empties the glass.

Question: Do all four friends drink the same amount? If not, who gets most and who gets least?

Fig. !. Beer remaining after (A) Annie, (B) Barry and (C) Cathy.

Fig. 1. Beer remaining after (A) Annie, (B) Barry and (C) Cathy.


By symmetry, Annie has drunk half-of-the-top-half of the glass. So she has consumed 25% of the beer. Again, by symmetry, Barry has left exactly 50% of the beer in the glass, so he has swallowed 25%. So far so good.

But Cathy has left beer forming a less regular shape: the liquid remaining in Fig. 1(C) is in the shape of an ungula, the volume formed by a plane slicing a cylinder and passing through the centre of the base. We have to calculate the volume of the ungula to see how much beer is left for Danny.

“Ungula” is Latin for “hoof”, and a section of a cylinder or cone cut off by a plane oblique to the base is so-called because it resembles a horse’s hoof. The shape is shown in Figure 2 below. Its volume can be calculated by a straightforward integration.

Figure 2: Cross-sections of an ungula perpendicular to (A) the x axis, (B) y axis and (C) z axis.

Figure 2: Cross-sections of an ungula perpendicular to (A) the x axis, (B) the y axis and (C) the z axis.

The three panels in Figure 2 show cross-sections of the ungula perpendicular to the x, y and z axes. They are respectively a rectangle, a triangle and a segment, and the volume is obtained by integration along the relevant axis. So, schematically, we can write the volume as

V = \int (Rectangle) dx = \int (Triangle) dy = \int (Segment) dz

Naturally, all three yield the same result, V = (2/3)r2h where r is the radius and h the height.

Now the volume of the cylinder is πr2h, so the fraction left for Danny is 2/(3π), or about 21%. Thus, while Annie and Barry drank 25% of the beer, Cathy must have drunk about 29%, leaving Danny short.

There is something remarkable about the volume of the ungula: it does not involve π, even though one of the surfaces is curved. Where has π gone?

More remarkable still is that Archimedes showed that the volume of the ungula is one sixth that of the surrounding cube or block. The volume is (2/3)r2h, and the volume of the rectangular box containing the cylinder is 2r x 2r x h = 4r2h, so indeed the ratio is 1/6.

Archimedes used the triangular cross-section (Fig 2(B)), integrating in the y direction. His reasoning was not some crude approximation, but a true application of the method of integral calculus!


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