Printable View
An easter egg manufacturer has an idea for a new product called an Easter Ring.
Take a sphere, drill a vertical hole down through the centre of the sphere to remove the core. The vertical height of the cylindrical hole is 10cm.
What volume of chocolate is required to make the remaining 'D' ring?
Is it possible to answer this question without knowing the width of the hole? :confused:
The width of the hole depends on the size of the drill.
So you want the volumne expressed in terms of the width of the drill (call it W)?
((Pi * (100 + w^2)^(3/2)) / 6) - (10 * Pi * w)
Where w = Width of drill bit.
It might be able to be simplified further but my algebra is so rusty, I can't think of how.
The first bit is the volume of the sphere and the second bit is the volumn of the cylinder.
It's supposed to be a puzzle so I don't want to give too much away.
You do have enough information to find the volume of the ring.
Just to help me along, could you give me the formula for the volume of a dome? I could probably look it up (if I had my Maths book) but I'm sure it is something to do with the height and the base area.
Please...It's not very often I do these sorts of problems attall. :(
some formulae:
volume of a cube:
x * y * z
volume of a cone:
1/3 * Pi * r^2 * (vertical)h
frustrum of a cone:
1/3 * Pi * r(total)^2 * (vertical)h - 1/3 * Pi * r(top)^2 * (top)h
volume of a cylinder:
Pi* r^2 * height
volume of sphere:
4/3 * Pi * r^3
If this is any help at all please don't post your results directly on here as someone else may like to try the puzzle - you will know when you have the answer - just post 'I've got it' or something.
Can you tell me what a 'frustrum' is?
Did you ommit the formula for a dome on purpose?:confused:
As I understand it, a frustrum of a cone is the bit thats left after you cut the pointy end off.
So the formula is for whole cone minus the bit that was removed.
Formula for a dome:
Sorry I thought this was a typo for cone - I'm not sure of a formula for this - I suppose you have to do a similar thing as for frustrum of a cone. Maybe a bit of integral calculus would help? :7)
Actually the formula for a pyramid is very similar to the one for a cone - 1/3 base area * height. This may not be relevant though.
Now I'm totally confused... :confused:
If you don't know the formula for a dome then I must be going down a 'blind alley'.
I can't for the life of me think why I would need the formulas for cones (or even the frustrum of a cone)!
Please, somone else, say something usefull!
I'm sorry I can't reply for a few days now as I am away for Easter. I will stress that YOU DO HAVE ALL THE INFORMATION YOU NEED!
A bit of lateral thinking is always useful.
See you after Easter.
I have seen this puzzle before.
If you assume there is a solution with the information given, you can give a direct formula for the answer. It just takes a clever point of view. otherwise, do some work.
Hint: Choose two possible drill diameters and calculate the two resulting volumes. Given drill diameter and height, you can easily calculate the volume of the cylindrical hole in the final object. It takes a little more work to calculate the volume of the ring.
Do not worry about the domes removed. Just calculate the volume of the circumscribing sphere and subtract the volume of the cylindrical hole.
Sorry folks. To do it the hard way, you really need to know a formula for the volume of the missing domes.
If you trust the poser of the problem, there is an intuitive way to solve this problem.
Trust me :7)
Just found a formula for the dome (end cap) of a sphere.
Vol = ((Pi * h^2)/3) * (3R – h) , where h is the height of the cap and R is the radius of the sphere.
I'm not really here, just visiting - remembered my password on about the fifteenth attempt - hope the above helps and Happy Easter all.
I don't really know what you mean by 'D' ring.
'D' ring - after drilling out the center of the sphere you are left with a ring, the cross-section of this ring has one flat side 10 cm high (where the drill went) and one curved side which is/was the outside of the sphere. Roughly a capital 'D' in shape.
The question is 'What is the volume of the ring that's left after drilling out the center of the sphere?'
If you trust the poser of the problem, the drill diameter does not matter. If it mattered, you would need to know it to solve the problem. he claimed that you had enough data to solve the problem.
If drill diameter does not matter, an infinitesimal or zero diameter should give the same answer as other diameters.
Hence Volume = 4*PI*5^3 / 3
or volume = 500*Pi/3
That's it Guv.
I originally had two methods for finding this volume, the intuitive way which you have described and some calculus, finding the volume of part of a sphere by rotating a circle about the x-axis and integrating up to the depth of the hole. I have now added a third method - the hard way which cuts out the calculus and involves the formula for the end caps.
Please excuse the length of this post and any repetitions of your answer - here is one I prepared earlier.
The three methods:
The Intuitive Way
If all the required information is here, then the volume of the ring must be constant for all sizes of the sphere (The drill size will vary accordingly) – so if you take a sphere of radius 5cm, drill 10cm through the centre with a drill bit of zero diameter, you are left with a sphere of diameter 10cm, radius 5cm.
Formula for the volume of a sphere is 4/3*Pi*R^3.
Volume of the remaining ring (The whole sphere in this case) is 4/3*Pi*5*5*5 = 523.599cc (roughly).
The ‘hard’ way
Volume of the ring = volume of sphere – volume of cylinder(height 10cm) – 2 * volume of end cap.
Vol of Sphere = 4/3 * Pi * R^3 where R is the radius of the sphere
Vol of Cylinder = 2 * Pi * r^2 * L (usually h but L will save confusion later)
Vol of End Cap = ((Pi * h^2)/3)*(3R–h) where h is the height of the cap from the outside of the sphere.
In the End Cap equation h = R – L.
In the cylinder equation, r^2 = R^2 – L^2
RingVol = (4/3 * Pi * R^3) – (2*Pi(R^2 – L^2)*L) – ((Pi * (R – L )^2)/3)*(3R–( R – L))
Multiplying this little lot out you get:
= 4/3*Pi*R^3 – 2*Pi*R^2*L + 2*Pi*L^3 – 4/3*Pi*R^3 + 8/3*Pi*R^2*L – 4/3*Pi*R*L^2 – 2/3*Pi*R^2*L + 4/3*Pi*R*L^2 – 2/3*Pi*L^3 (whew!)
Grouping the terms:
= [4/3*Pi*R^3 – 4/3*Pi*R^3] + [8/3*Pi*R^2*L – 2/3*Pi*R^2*L – 2*Pi*R^2*L] + [4/3*Pi*R*L^2 – 4/3*Pi*R*L^2] + 2*Pi*L^3 – 2/3*Pi*L^3
The bits in the square brackets cancel out to leave:
RingV = 2*Pi*L^3 – 2/3*Pi*L^3
Simplify further:
RingV = 4/3*Pi*L^3 – The formula for a sphere where R is equal to L, in this case equal to 5cm.
The Calculus Way
Volume of a hemisphere is found by integrating (R^2 – (x^2)) between zero and R and multiplying by Pi.
So Vol = Pi * integral [ R^2*x – (x^3)/3 ]
The section we are interested in is only 5cm high from the centre so just integrate from zero to 5.
This finds the volume of half the ring so double it to get the whole volume.
Vol = 2*Pi*R^2*h – 2*(Pi*h^3)/3
Where h=5cm this is the volume of the sphere without the end caps.
To subtract the cylinder (Pi*y^2*h) calculate y^2 in terms of the sphere’s radius, y^2=(R^2 – h^2)
So cylinder volume = Pi*R^2*h – Pi*h^3, h is only half the height of the whole cylinder so double it:
Cylinder volume = 2*Pi*R^2*h – 2*Pi*h^3
To get the volume of the ring:
RingV = (2*Pi*R^2*h – 2*(Pi*h^3)/3) – (2*Pi*R^2*h – 2*Pi*h^3)
RingV = 2*Pi*R^2*h – 2*(Pi*h^3)/3 – 2*Pi*R^2*h + 2*Pi*h^3
RingV = 2*Pi*h^3 – 2*(Pi*h^3)/3
RingV = 4/3*Pi*h^3 – The formula for a sphere where R is equal to h
When h=5, Ring Volume = 523.599cc (roughly).
I find it amazing to try to visualise the ring left by drilling out a sphere the size of this planet - a sort of ribbon 10cm wide that still has the same volume as a carefully selected grapefruit.