Yesterday (as I type this) Fermat’s Library on Twitter Tweet this:

Click here to see the original

Then there was an image with the actual formula: $ V= frac {4} {3} { pi} r ^ 3 $.

Over thousand people “Liked” the tweet, but someone said:

Volume of a cone

Firstly, think about a cube. We can think of the cube as six pyramids put together with their apexes meeting in the middle. The apex of each pyramid is in the exact center of the cube, so if the cube has a side length of 1, then the pyramids will have to have “height” $ frac {1} {2} $. Then because there are six identical pyramids we can conclude that each pyramid is one sixth of the volume of the cube. So a square-based pyramid of side length 1 and height $ frac {1} {2} $ has a volume of $ frac {1} {6} $.

and the height is $ h $, then the volume is $ frac {1} { 3} Ah $.

Using that as our inspiration we introduce “Cavalieri’s principle” which says this:

So in some cases we can compare volumes as follows:

Now pick some value, $ R $, which we are going to use as a radius for a cone, and hemisphere, and a cylinder.

Arrange our cone, hemisphere, and cylinder as follows:

We have a “side on” view here, seeing the hemisphere, inverted cone, and cylinder, all from the side, and we have taken a slice through them all at some arbitrary height $ h $. The three objects here are all of height $ R $ and radius $ R $, so the sides of the cone are at 90 degrees to the table on which they rest (even though in my rough, hand-drawn diagram they don’t look like it).

We know that taking a slice through the cylinder will result in a circle of radius $ R $, and hence area $ { pi} R ^ 2 $. The slice through the cone will also give a circle, and when we realise that the red triangle is (despite the drawing!) a – – 728 triangle, we can see that the radius of the circle will equal the height of the slice, and so the radius of the slice in the cone will be $ h $, the height of the slice. So its area will be $ { pi} h ^ 2 $.

In the case of the slice through the hemisphere, that will also be a circle, and we can compute its radius (and hence area) by using our old friend, Pythagoras’ Theorem. We have a right-angled triangle with hypotenuse the radius of the hemisphere, hence $ R $. The height, and hence one short side, is $ h $, so we have the equation:

## Combining the results

Now for the pay-off. Looking at the slices through the hemisphere and cone, their total area is:

The cylinder has volume $ { pi} R ^ 3 $, because that is the height times the area of the base. The cone has volume $ frac {1} {3} { pi} R ^ 3 $, being one third of the base area times the height. So the volume of the hemisphere is the difference, which is $ frac {2} {3} { pi} R ^ 3 $.

Therefore the volume of the full sphere is twice that.

Thus the volume of a sphere of radius $ R $ is $ frac {4} {3} { pi} R ^ 3 $.

Further reading …

A previous blog post gave a different derivation of this result:

If there’s anything on this page that you think is unclear or needs expanding, please let me know and I’d be happy to revisit anything that I think should be enhanced or updated.

Good luck!

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