#3**+10 **

Archimedes provided a rather ingenious proof of this by noting that the volume of a hemisphere is equal to the sum of the differences in the cross-sections of a cylinder and an associated cone both having the same radius as the hemisphere, i.e....... pi*r^2h - (1/3)pi*r^2h = (2/3)pi*r^2h. But, the cylinder and the cone also have a height equal to this radius, so we have (2/3)pi*r^2h = (2/3)pi*r^2*r = (2/3)pi*r^3

So....twice this equals the volume of a whole sphere = 4/3pi*r^3

CPhill
Dec 8, 2014

#1**+10 **

Volume of the sphere = 4/3ㅠr^3

Where, r = radius of the sphere

Derivation for Volume of the Sphere

The differential element shown in the figure is cylindrical with radius x and altitude dy. The volume of cylindrical element is...

The sum of the cylindrical elements from 0 to r is a hemisphere, twice the hemisphere will give the volume of the sphere. Thus,

From the equation of the circle x^{2} + y^{2} = r^{2}; x^{2} = r^{2} - y^{2}.

flflvm97
Dec 8, 2014

#3**+10 **

Best Answer

Archimedes provided a rather ingenious proof of this by noting that the volume of a hemisphere is equal to the sum of the differences in the cross-sections of a cylinder and an associated cone both having the same radius as the hemisphere, i.e....... pi*r^2h - (1/3)pi*r^2h = (2/3)pi*r^2h. But, the cylinder and the cone also have a height equal to this radius, so we have (2/3)pi*r^2h = (2/3)pi*r^2*r = (2/3)pi*r^3

So....twice this equals the volume of a whole sphere = 4/3pi*r^3

CPhill
Dec 8, 2014

#4**+5 **

Welcome to the forum fiflvm97. I hope that ou like it here. I can see that you will be a great asset to us if you choose to stay around :)

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I have been looking at these answers.

Geno, yours is concise but exactly what the asker would have wanted.

flflvm, I loved your analysis of this answer. Adding the diagram made all the difference too.

I hope that some of our senior students really think about what you said - as I did. It would be of great benefit to them.

Chris, I had not thought before about the volume of a cone being one third of a hemisphere. It's an obvious conclusion now that you point it out.

As for the rest of what archimedes said that will require a more concentrated effort from me. Thank you for discussing it.

Melody
Dec 8, 2014

#5**+5 **

Melody......there used to be a very good illustration of this on the net, but I'm unable to find it, now. However, this one is pretty good......http://mathcentral.uregina.ca/QQ/database/QQ.09.01/rahul1.html

CPhill
Dec 9, 2014