As ane fule kno the volume of a ball (ie., the interior of a sphere) is .
But I have just had one of those “why’s that then” moments and sought to prove to myself, using integration, why this would be so.
But my logic is flawed and I get the wrong result – so risking looking stupid – I am asking for someone to correct the error in my logic.
Starting from the start and a familiar (I hope!) lemma:
is the ratio of a circle’s circumference () to its diameter (), hence or, more familiarly, where is the radius of the circle.
Now the area (a) of a circle can then be found by integration:
, giving the familiar
So, my reasoning runs, the volume of a ball would then be:
or , which is precisely half the figure it should be – so where have I gone wrong?
Update: with thanks to Hugh in the comments – of course what I have described is a (double) cone with a height equal to the radius of the base. Cones and circles are not the same, obviously.
- Equivalence between weak convergence and uniform tightness. The Helly’s lemma and the Prohorov’s theorem (maikolsolis.wordpress.com)
- Giant watery balls (howtospotapsychopath.com)
- Trend Alert: Volumized Updos at the 2009 Grammys (bellasugar.com)
2 responses to “Volume of a ball”
Picture what you are integrating: the area of a disc as it grows to the full radius. But that isn’t a set of slices that makes up a ball. It’s just a set of overlapping slices.
You want to integrate the surface area of a sphere that grows from 0 to the radius. Or you want to integrate a set of slices along a diameter, with radius appropriate (not linear).
Thanks – I’ve updated the article to reflect this!