# Filling Cone, Hemisphere and Cylinder: Easy as 1:2:3

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Filling a cone, a hemisphere and a cylinder is as easy as 1:2:3, the ratios of their volumes.

Contributed by: Kenneth E. Caviness and R. Lewis Caviness (June 2020)

Open content licensed under CC BY-NC-SA

## Snapshots

## Details

A display at an interactive science center (the best kind!) consists of three water containers: a cone, hemisphere and cylinder. Visitors can use the cone twice to fill the hemisphere with water, then combine that with a third cone's worth to completely fill the cylinder. This is especially intriguing because of the various-shaped containers. This Demonstration allows the user to interactively fill these containers and gain insight about why their volumes conform to the 1:2:3 ratios. Two related visualizations show that the volume of the cone is 1/2 the volume of the hemisphere and 1/3 that of the cylinder, assuming that their heights are all equal to the radius of the hemisphere. A comparison of the formulas shows this:

In the "fill all" view, experiment with simultaneously filling (stacked) cones, hemispheres and cylinders with water. Stop the animation at integer multiples of to see how many cones, hemispheres and cylinders are needed for the same volume. Notice that when the cone is not full, there is no simple relationship between the water height in the different containers. (For example, the water height rises steadily in the cylinder, but initially changes very rapidly in the cone and hemisphere, slowing as it fills.)

The " versus graph" view shows graphically the dependence of water volume on height for the different cases. Note the 1:2:3 ratios when full height is reached.

The "submerged hemisphere" and "submerged cone" views allow either a solid hemisphere or a solid cone to be submerged in the cylinder, showing that and . Interestingly enough, now the water heights in the containers shown together remain exactly in sync.

The "geometric explanation" view shows graphically that the volume of the cylinder minus that of the cone equals the volume of the hemisphere, and moreover shows that for every water height the surface area is the same in the two cases. This neatly explains not only the equal surface areas, but also the equal volumes for each height up to the total container height, since the volume can be found by integrating slices of area and thickness : .

Snapshot 1: equal amounts of water fill 2 cones, 1 hemisphere and 2/3 of the cylinder

Snapshot 2: equal amounts of water fill a cylinder with a submerged cone and a hemisphere to equal heights

Snapshot 3: geometric explanation for Snapshot 2

Snapshot 4: equal amounts of water fill a cylinder with submerged hemisphere and a cone to equal heights

This Demonstration was inspired by an interactive display at Leoni Meadows Camp and Nature Center.

## Permanent Citation