Beady Eyed
If you drill a diametrical hole of final length L through a sphere, the volume of the remainder is the same as the volume of a sphere of diameter L. How come?
Labels: mathemagic
posted by Chris at
5:30 PM
Wednesday, September 9, 2009
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5 Comments:
As pi is going to crop up again, just for amusement follow the link:
http://en.wikipedia.org/wiki/Feynman_point
What does "final length" mean? If it means the length of hole from one side of the sphere to the other then the hole's diameter becomes relevant.
Since this diameter is not specified then one can assume that the question applies to ANY diameter of the hole.
If the diameter of the hole is infinitely small then what is left is the original sphere of diameter L.
If, on the other hand, the diameter of the hole is equal to that of the sphere then it has zero length as it consumes all of the sphere, and the remaining volume of the sphere is also zero.
I claim to have answered the question as it stands. I didn't need pi for this. Heh, heh!
Hi Wiz. You're doing very good so far. I really mean the measured length of the hole (after you have actually made the hole). You seem to have understood it correctly.
All. For consistency and convenience, I suggest that you use h=L/2 (h=half-length) when you get to the details. It'll save a lot of (L/2) tedium.
Wizard of Oz, got it right (As far as he went).
The solution: Read for a bit then make yourself a 2D sketch (I'll use 3D terminology though). Imagine a planet similar to the Earth. It has two identical circular icecaps. Drill a hole from North to South, exactly cutting out the icecaps. The length of this hole is measured from the outer edge of the ex-Northern icecap to the outer edge of the ex-Southern icecap and its length is L. Let h= L/2. Draw the axis from N to S. Draw the equatorial plane. Let the radius of the hole be a, and the radius of the sphere be r. As the distance between the centre of the sphere/circle and an edge of either ex-icecap is r we have a^2+h^2=r^2 (Pythagoras).
Now draw a plane parallel to the equatorial plane, but about half-way to the North pole. Let the distance between this plane and the equatorial plane be y, and the radius of the new plane be x. Note that x^2+y^2=r^2, and the area of the annulus (the washer shape) is pi(x^2-a^2) = pi((r^2-y^2)-(r^2-h^2)) = pi(h^2-y^2)
Now consider a similar situation, but with a whole sphere of radius h. The radius of the non-equatorial plane is x^2+y^2 = h^2 and the area of the non-equatorial disc is pi*x^2, where the plane is at the same distance y from the equatorial plane as before. So area of disc = pi(h^2-y^2).
Comparing the two equations of area immediately shows that the areas of the corresponding annulus and disc are identical. So the volumes of the holed and whole sphere are the same, as one can be constructed from the other by deformation. Note it was important to convert the equations into functions of y for the comparison.
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