Arthur Redsell

Posted on Friday, 31 January, 2003 - 10:53 pm:

I've had this problem for many years and only ever achieved a result by computer spreadsheet. Integration by calculus defeated me and I have never found an elegant solution. Maybe one doesn't exist but can you think of one?

A trophy is to be cast in the combined shape of a pyramid and sphere. The pyramid's base is a square and has a length of side equal to its height. The pyramid coincides with a sphere which is of diameter equal to the pyramid's height and whose vertical diameter touches the vertex of the pyramid and the mid- point of its base.
If the pyramid sides, its height and the diameter of the sphere are all 6 inches what is the volume of metal required to cast it?

Arun Iyer

Posted on Saturday, 01 February, 2003 - 05:35 am:

well,
here is an idea.

your trophy here can be thought of as a fusion of
1> entire sphere with altered pyramid or
2> entire pyramid with altered sphere.

i will consider the first one here.

now if you try to visualize this trophy,you would see that the entire sphere remains intact and the pyramid is altered to 4 identical pieces which look like this ....

(Sorry for the bad-looking diagram)

now this thing is actually an intersection of three planes.Out of which two are z=2x and y=2x(i think!!!i derived this from the front view of the trophy).The third plane is the surface of the sphere..(sorry but i am not aware of its equation)

So with this information at hand,i think we can find the volume with a triple integral. However,the equation of the third plane is what is required in this case.

hope this was of some help!

love arun

David Loeffler

Posted on Saturday, 01 February, 2003 - 07:30 pm:

Arthur,

I make the volume

LaTeX Image

where h is the side-length; this works out at 0.63695690384789601629 h³ . So in your example 137.58269123114553952 cubic inches of metal are needed.

Is this anything like your numerical approximations? It should clearly be something larger than either of pi /6 h³ (the volume of the pyramid) and 1/3 h³ (the volume of the cone) but smaller than the sum of the two, which fits; but I may still have made some errors in my working.

David

David Loeffler

Posted on Saturday, 01 February, 2003 - 09:03 pm:

Just for fun, this is what it looks like:
trophy

David

Arthur Redsell

Posted on Saturday, 01 February, 2003 - 09:08 pm:

Arun
Yes, you have outlined the difficult bit but I could not solve the triple integral!

David
My computed answer achieved by slicing and adding up the awkward corners, outlined by Arun, into 200 slices of .004" thickness was
Volume = 137.58268 cu.ins.
so I reckon we agree.

David Loeffler

Posted on Saturday, 01 February, 2003 - 09:25 pm:

My calculation above is rather uninspired: I considered a slice taken through the cone at height x measured downwards from the vertex. If x < = 2/3 h then the sphere lies wholly outside the cone at that level, so we need only work out the area of the spherical slice, which is pi x(h-x).

If x > = 4/5 h then we need only consider the square slice of pyramid, as the sphere now lies inside this; the area of this is simply x² .

In the remaining interval we can decompose the strange shape we end up with into a square plus some 'slivers' from the circle; it isn't difficult to calculate the area, which turns out to be
LaTeX Image

Integrating this from x=2/3 h to x = 4/5 h and adding the remaining contributions gives the result. As for how you actually do the integral, I just fed it to Mathematica and it spat out the result I gave above; don't ask me how it did it!

David

Arun Iyer

Posted on Sunday, 02 February, 2003 - 06:57 am:

David,
could you please elaborate on that area result (and if possible even the volume result).... the expressions are not something that i would call "obvious" .

love arun

David Loeffler

Posted on Sunday, 02 February, 2003 - 10:55 am:

Well, we have a shape like this:

We'd like to add up the area as a sum of the square ABCD and the four slivers, of which EF is one.

Now let the angle EOF $= θ$ , and the radius of the circle $r$ . The area of the sector OEF with the curved boundary is $1 / 2 r^{2} θ$ ; that of the triangle OEF is $1 / 2 r^{2} \sin θ$ .

So the area of the sliver is $1 / 2 r^{2} (θ - \sin θ)$ .

Now what is $θ$ ? The perpendicular distance from O to the sides is $x / 2$ . Hence one may draw a right-angled triangle OEG where G = midpoint of EF, and thus obtain $\cos θ / 2 = x / 2 r$ . We can then express $\sin θ$ in terms of this.

Now if we substitute in $r = \sqrt{x (h - x)}$ , we obtain the formula I gave above for the area of this slice.

David

Arun Iyer

Posted on Sunday, 02 February, 2003 - 12:38 pm:

ah!! understood !!!

love arun

Arthur Redsell

Posted on Saturday, 08 March, 2003 - 09:49 pm:

Thanks folks. It looks as though we might have gone as far as we can. Sometimes there is a flash of insight which enables a short cut solution.
It now seems that this one that has to be done the hard way.
David. I would like to know a bit more about Mathematica. Is it black magic?