cartesian product … stuff about computing mostly

Volume of a ball

As ane fule kno the volume of a ball (ie., the interior of a sphere) is $\frac{4}{3}\pi r^3$ .

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:

$\pi$ is the ratio of a circle’s circumference ( $c$ ) to its diameter ( $d$ ), hence $c = \pi d$ or, more familiarly, $c = 2 \pi r$ where $r$ is the radius of the circle.

Now the area (a) of a circle can then be found by integration:

$a = \int_0^r 2\pi x dx$ , giving the familiar $\pi r^2$

So, my reasoning runs, the volume of a ball would then be:

$2 \int_0^r \pi x^2 dx$ or $\frac{2}{3}\pi r^3$ , 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.

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Adrian McMenamin

May 12, 2012

Uncategorized

ball, integration, mathematics, maths, sphere, volume

2 responses to “Volume of a ball”

Hugh

May 12, 2012 at 3:40 pm

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).
1. Adrian McMenamin
  
  May 12, 2012 at 4:50 pm
  
  Thanks – I’ve updated the article to reflect this!

Volume of a ball

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