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I have a solid sphere. I drill a whole right through the middle so that the middle of the cylinder that is cut out is also a diameter of the sphere and the whole is 4 inches long. The big question is, what is the volume of the object that is left over?

There are two ways of getting the answer – can you give them both (or a third)?

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I don’t understand your question:
What part of the cylinder’s dimension is equal to the diameter of the sphere.
The “hole” is 4 inches long – does that mean the cylindrical hole has a ht of 4 inches?

Steve

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Ok, a little bit of clarity might help. After you have drilled the hole in the sphere, the resulting solid is 4 inches high – the cut in the sides of the sphere is 4 inches long.

Does that help? If you want the volume of the “cap”, then it is …
v=pi/2 h r^2 + pi/6 h^3 where r is the radius of the bottom of the cap and h is the height of the cap

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By my reckoning, the remaining solid is equal the the volume of a 4″ diameter sphere.
Tak a 4″ diameter sphere and drill an infinitely thin hole through it. The length (or height) of the hole by definition is 4″ long, the remaining volume of the sphere is as it was at the beginning (as the hole is infinetely thin); the volume of a 4″ sphere.

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One answer is based on the hole being infinitely small in diameter, the other based on the sphere being infinitely large in diameter . . .

And no, I don’t have a third

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My method used Geometry to get volumes, Pyth theorem for relative variables, algebra to solve equations

The solution of the 4″ cylinder is ca 33.5 inches ^3

The general number is: pi / 6 * h^3 where h is the cylinder ht

Steve

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Here is the answer that I like (already shown above) …

Assuming that there is a solution and given that I’m not given the radius of the sphere, then, the radius of the sphere must be immaterial. Hence, take a sphere of radius 1″ and drill a hole with length 2″ through it. Hence, the resultant cylinder has zero volume and the remaining shape’s volume is the volume of a sphere with radius 1″.

There is a more technical answer in the attachment.

There is also a third way to get the answer that involves integration – I might work on that on the week end and post it next week (if I get really bored).

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I don’t neccessarily agree with the statement:
[indent]

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Assuming that there is a solution and given that I’m not given the radius of the sphere, then, the radius of the sphere must be immaterial.

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[though it is true in this case]

The answer could have involved finding a volume as a function of the radius or even as a percentage. This one happens to be immaterial, but I have seen questions where it is NOT and just because something is NOT given, does NOT make it immaterial.

Steve

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As promised, here is a file that contains three solutions. One based on geometry, one on calculus and one on general principles.

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