[Guest]

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Please forgive my rather crude drawing above. I am trying to represent a sphere with a hole drilled through the centre of it. As you might be able to make out, it would resemble a ring shaped object afterwards. If we say that the depth of the "ring" is 150mm after the hole has been drilled, can you give an expression for the volume of the ring?

Edited October 15, 2009 by Forcedhand

[Guest]

Actually I can't get the drawing to resemble anything useful. Try and picture a sphere with a hole drilled through he centrer. What you will be left with is a ring shaped object. If the depth of the ring is 150mm, what is the volume?

[Guest]

longest length of drill-nose xPIEx150mm?

idk how to do anything to do with pie

Edited October 15, 2009 by Jonn

[Guest]

*cough* Find the volume of a "ring"? I believe you need to reword this entire question, as for the depression of 150mm into a sphere, in-order to find the volume of the depression, the radius, diameter, or even circumfrence must be given or possibly found in some way, and then the volume of the amount of the sphere shaven at the edge of the sphere (as if looking from the side) the size of the sphere itself would be needed to find out. THIS QUESTION SIMPLY DOES NOT GIVE ENOUGH INFORMATION TO PROVIDE AN ANSWER... and your drawing is misshapen btw

[plainglazed]

There is indeed enough information to calculate an answer. Knowing this should give you a hint into how it can be done.

there is no limit to how big the diameter of the sphere can be

EDIT: I think this is very similar to a Bonanova topic awhile back.

Edited October 16, 2009 by plainglazed

[Prof. Templeton]

There are two ways to solve this. Calculus or logic. Logic is quicker...

Assume the drilled out portion of the "ring" has zero volume. What are we left with?

bn's topic One of my favorites.

You can use a codebox for crude drawings and such.

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[/codebox]

[Guest]

So then, it's simply Pi x 150 as the drilled out portions will still leave end caps that complete the perfection of the sphere thus not changing its mass.

er, displacement mass.

Edited October 16, 2009 by alphajonny

[Guest]

If this question had appeared on a math exam the first thing that would likely pop into your mind is that the solution didn't depend on the diameter of the circle (since it isn't given) meaning you could pick any diameter and get the same answer. So by picking a diameter of 150mm the solution can be computed very quickly (the logical solution). But if you are like me you'd probably want to do the math anyways just to be sure your reasoning, or maybe the question, wasn't flawed. Anyways, the solution uses some pretty easy calculus so here it is:

R = Radius of the Original Sphere

r = Radius of the Cylinder

r = sqrt (R²-75²) ==> See Fig 1.

If we fill in the cylinder but leave the top and bottom of the sphere cut off we get something that resembles a 150mm thick wheel of cheese. If we find this volume and subtract the volume of the cylinder (150piR²-843750pi) we have our solution. Fig 2 is a cross section of the wheel. If we take out a tiny slice of the wheel at x with thickness dx, then as dx approaches 0 the volume of the slice will approach pi(R²-x²)dx. So to get the volume we integrate from -75 to 75:

pi[integral(R²-x²)dx][-75 to 75] (R is constant)

pi(xR²-1/3x³)[-75 to 75]

pi[75R²-140625-(-75R²+140625)]

150piR²-281250pi

Now subtract the volume of the cylinder:

150piR²-281250-(150piR²-843750pi) = 562500pi or 4/3(75)³pi

Sphere.bmp

[Guest]

If this question had appeared on a math exam the first thing that would likely pop into your mind is that the solution didn't depend on the diameter of the circle (since it isn't given) meaning you could pick any diameter and get the same answer. So by picking a diameter of 150mm the solution can be computed very quickly (the logical solution). But if you are like me you'd probably want to do the math anyways just to be sure your reasoning, or maybe the question, wasn't flawed. Anyways, the solution uses some pretty easy calculus so here it is:

R = Radius of the Original Sphere

r = Radius of the Cylinder

r = sqrt (R²-75²) ==> See Fig 1.

If we fill in the cylinder but leave the top and bottom of the sphere cut off we get something that resembles a 150mm thick wheel of cheese. If we find this volume and subtract the volume of the cylinder (150piR²-843750pi) we have our solution. Fig 2 is a cross section of the wheel. If we take out a tiny slice of the wheel at x with thickness dx, then as dx approaches 0 the volume of the slice will approach pi(R²-x²)dx. So to get the volume we integrate from -75 to 75:

pi[integral(R²-x²)dx][-75 to 75] (R is constant)

pi(xR²-1/3x³)[-75 to 75]

pi[75R²-140625-(-75R²+140625)]

150piR²-281250pi

Now subtract the volume of the cylinder:

150piR²-281250-(150piR²-843750pi) = 562500pi or 4/3(75)³pi

Very good. It is (as you've established) the same volume as a 150mm diameter sphere, regardless of the size of the sphere before the hole was drilled.

[roolstar]

I think your ring might be a Hole in a Sphere already posted on this forum!

Check it out:

Edited October 16, 2009 by roolstar

[Guest]

*cough* Find the volume of a "ring"? I believe you need to reword this entire question, as for the depression of 150mm into a sphere, in-order to find the volume of the depression, the radius, diameter, or even circumfrence must be given or possibly found in some way, and then the volume of the amount of the sphere shaven at the edge of the sphere (as if looking from the side) the size of the sphere itself would be needed to find out. THIS QUESTION SIMPLY DOES NOT GIVE ENOUGH INFORMATION TO PROVIDE AN ANSWER... and your drawing is misshapen btw

You should try and be more studious in your analysis of the problem. Your use of capitals is unwarranted and rather rude. Other, more tolerant readers of this board have solved this problem with the information provided. I regret that it was beyond your capabilities.