| 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?
| 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
| Posted on Saturday, 01 February, 2003 - 07:30 pm: |
Arthur,
I make the volume
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
| Posted on Saturday, 01 February, 2003 - 09:03 pm: |
Just for fun, this is what it looks like:
David
| 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.
| 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
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
| 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
| 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 =q, and the radius of the circle r. The area of the sector OEF with the curved boundary is 1/2 r2q; that of the triangle OEF is 1/2 r2 sinq. So the area of the sliver is 1/2 r2(q-sinq). Now what is q? 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 cosq/2=x/2r. We can then express sinq in terms of this. Now if we substitute in
| r= | _____ Öx(h-x) |
, we obtain the formula I gave above for the area of this slice. David
| Posted on Sunday, 02 February, 2003 - 12:38 pm: |
ah!! understood !!!
love arun
| 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?