## 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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