If a hole of radius *r* is drilled through the center of a sphere of radius *R*, we refer to theremaining portion of the sphere as a bead with inner radius *r* and outer radius *R*. The *height* of the bead is *h*. It can be shown (using calculus) that the volume of a bead depends only on its height. Since the volume is the same whatever the value of *r*, you can find the answer to parts (a) and (b) by thinking of a special case where *r*=0 and the bead is actually a sphere.

**a.** What is the volume of a bead with height 12cm.

**b.** Give a formula for the volume V of a bead in terms of the height *h*.

notsomathpro
Jun 5, 2014

#1**+8 **

I'm going to buck the trend of this forum and ask you why I should answer this question ?

It looks to me like a homework question or (part of) an assignment. Why should I do it for you ?

Have you actually tried to answer it ? If you have, show us some working and one of us may be able (and willing) to help.

Bertie
Jun 5, 2014

#1**+8 **

Best Answer

I'm going to buck the trend of this forum and ask you why I should answer this question ?

It looks to me like a homework question or (part of) an assignment. Why should I do it for you ?

Have you actually tried to answer it ? If you have, show us some working and one of us may be able (and willing) to help.

Bertie
Jun 5, 2014

#2**0 **

Hi notsomathpro and Bertie,

Firstly it is fine for you to answer this way Bertie.

notsomathpro, you did not do anything wrong either.

We want to work with you to help you nut out your problems. You are older than many of the question askers and it is reasonable for Bertie to ask you to interact with him. So try and talk about the question. What have you tried. Why won't it work etc.

On other forums this is always demanded. On this forum we are often more relaxed but sometimes we are more relaxed than we should be.

Melody
Jun 5, 2014

#4**+5 **

I actually don't know how to answer this, but I think this is what you're looking for:

http://en.wikipedia.org/wiki/Napkin_ring_problem

Hey.....maybe we'll both read it together and learn something, huh ....???

CPhill
Jun 5, 2014

#6**+5 **

This is actually quite a well-known recreational maths problem. I first came across it in a Martin Gardner book xxx years ago! However, in the usual statement of the problem it doesn't tell you the answer.

In the statement of the question here, the hint in the last sentence gives the game away. In the limit when r is zero, the height, h, becomes the diameter of the sphere. (In this limit the sphere has an infinitely thin hole of zero volume, so the bead becomes the whole sphere of diameter h).

Alan
Jun 5, 2014

#7**0 **

I don't get it.

Certainly I agree with what Alan has said. The question tells you to work it out like as if there is no hole. That is fine. BUT

Isn't this bead just a sphere with a hole in it? If you put a hole through the middle of the bead then the volume if the bead has to decrease. Certainly if the diameter of the hole is 0 then there really is no hole and the volume is as it was to start with. But if the diamer of the whole is equal to the diameter of the bead then there is no bead left and the volume of the bead is 0. I obviously do not understand the wording of this problem.

I think the penny just fell! The volume is the constant, the radius is the variable, now I've got it!!

The answer was obvious from the wording, but I haven't even thought about why that would be so yet. - But finally I think I understand the problem!

I do not know what level you are studying at nosomathpro but the answer is given in the question. You do not even have to understand the question to get the answer (I did not). Is this question just trying to teach you to pull out the relevant details and ignore the rest? Not a difficult problem, just a word exercise??

WOW - I am going to be thinking about this for days. Thanks for sharing it with me notsomathpro and thank you Bertie for drawing my attention to this question. Otherwise i probably would not have even thought about it at all.

-------------------------

Everything i have written is real but I have also demonstrated how a question asker can 'discussed' their question.

Melody
Jun 5, 2014

#8**0 **

So....from my reading, the answer to "a" would be in this set-up??

pi *∫ (h/2)^{2} - y^{2}) dy = pi * ∫ (12/2)^{2} - y^{2}) dy = pi * ∫ (6)^{2} - y^{2}) dy ..... where the limits on the integral are -6 and 6 ......???

CPhill
Jun 5, 2014

#9**0 **

The answer is just $$Volume = \frac{4}{3}\pi (\frac{h}{2})^3$$

For part a) just plug in 12 cm for h.

It is possible to do this the hard way, by calculating the bead volume for a non-zero radius hole, and, strictly one should do this in order to prove that the bead volume is independent of the hole radius. However, since the question here *specifies* that the volume is independent of hole radius, why make life difficult?!!

Alan
Jun 6, 2014