Hello everyone,

I have come across this following question which has bothered me for a whole day!

Two circular cylinders each have a radius r. The cylinders intersect so that their axes of symmetry intersect at right angles. Find, in terms of r, the volume common to both cylinders.

I cannot seem to be able to imagine the common volume!
[Picture here - you can even rotate it! - The Editor]

I hope you can help me!

Thank you

I'm afraid I can't visualise this either, so I'm going to have to resort to brute force.

Now I think that points inside the first cylinder satisfy:

$x^2+y^2<r^2$

And the second:

$x^2+z^2<r^2$

So we want to find the volume enclosed by these equations.

Now consider a slice given by:

$a<x<a+\delta a$

We find that:

$-\sqrt{r^2-a^2}<y<\sqrt{r^2-a^2}$

$-\sqrt{r^2-a^2}<z<\sqrt{r^2-a^2}$

In other words we are considering a rectangular slice. The volume of it is:

$\delta a\times 2\sqrt{r^2-a^2}\times 2\sqrt{r^2-a^2}$

$=4(r^2-a^2)\delta a$

Now integrate this from $a=-r$ to $a=r$ (to cover the entire range of $x$ values):

${\int }_{-r}^r[4(r^2-a^2)\mathrm{da}]$

$=4r^{2}a-4/3a^3$

$=8r^3-8/3r^3=16r^3/3$.
Hope this is right,

Michael

Michael; you are correct. However, Archimedes provided a more elegant method of solution. In the volume common to both cylinders place a sphere of radius r having its centre at the intersection of the two axes of the cylinders.

Now take any plane section through the cylinders. This will give you a square cross section of the area common to both cylinders which circumscribes a circle as the cross section of the sphere. Now, clearly the volume of the sphere is the sum of all circular cross sections and the volume of the solid common to both cylinders is the sum of all square cross sections. The ratio of the volume of the sphere to the volume of the solid is therefore the same as the ratio of the area of a circle to the area of a circumscribed square. So

$x=16r^3/3$
Tom

An Egyptian method using no calculus is described in this article . In this case, you don't even need to know the volume of a sphere before you start, and so you can then derive the volume of the sphere using Tom's method.

What about the case when you have three cylinders intersecting orthogonally? I think that the common volume is 8(2-sqrt(2))r³ , but my derivation of this is really nasty and involves setting up integrals for the areas of slices which are themselves the intersections of circles and squares. Did Archimedes have a clever trick for this one too?

David

Thanks Tom, that's very nice. I had to spend about 15 minutes trying to visualise where the rectangle came from (of course I could calculate its existence - but I don't have very good visual intuition). Actually the two methods are more similar than you would think. Both ways are considering a horizontal slice, and if you look at my formula, $4(r^2-a^2)\delta a$... this is proportional to the area of the circle you were talking about. So actually pretty similar.

But how would Archemedies have known the volume of a sphere, without knowing integration?

I agree with David's answer for the extension question, but cannot see an elegant method of approach. I thought I had a good way using a higher-dimensional generalisation, but it didn't work.

Yours,

Michael

As to how Archimedes would have known the volume of a sphere, he could have possibly found it by displacement of water. I think this is how the volume of a cone was known a long time ago. And, who knows, he could have estimated a formula using a bunch of pyramids put together, although this would be a good deal of work.

Brad

Thanks Brad. Of course you are correct - but I would be surprised if Archimedes knew that the constant of proportionality between $V_{\mathrm{sphere}}$ and $r^3$ was $4\pi /3$. Perhaps he had a good numerical estimate. The volume of a cone is a good deal easier - you can prove it without calculus.

Hi, everyone!

Did Archimedes know about the formula:

inverted right circular cone + hemisphere of same radius = cylinder?

If so, then he could have derived the constant in a (very) simple manner.

Er... he might have done but I certainly didn't know that till now! Thank you for informing me.

Michael