Monty Hall

[JohnW2]

No, I didn't read it all. sorry.

I'm getting confused. Are you now saying that you agree that:

Therefore the residue must equal the volume of a sphere with a diameter of six inches, namely 36π.

is correct (which it is!).

Note that it says the volume of a sphere with a diameter of six inches, NOT the volume of the sphere in this puzzle (whose diameter would be greater than 6 inches).

Kind Regards, John

 

[TonyW2]

Edit ...missed quote

 

[JohnW2]

I think I see what you`re getting at John. The length of the hole in the remaining metal of what was originally a true sphere is 10cm ... But the diameter of the sphere to start with would vary from this 10cm figure by an amount dependant upon the diameter of the hole.

Exactly. The length of the hole will always be less than the diameter of the original sphere, increasingly so as the the diameter of the hole increases.

If one progressively increases the diameter of the hole, the length of that hole will progressively decrease, approaching the ultimate limit when the diameter of the hole approaches the diameter of the sphere, when the length of the hole will approach zero (i.e the drill will have completely destroyed the sphere!).

Edit: phonetic typo ('whole' corrected
Kind Regards, John

 

[JohnW2]

To be honest, I'm still trying to find a link which explains how to derive the volume of a spherical cap from first principles. I suppose it's a double integration but I haven't done that for some time.

It's simple enough, and only requires a one, pretty simple, integration. See the very last example in this link . That proof results in the familiar formula in terms of the height of the cap (h, analagous to 'A' in your link) and the radius of the sphere (r). I can but presume that the 'r' in your link is something different (hence the unfamiliar formula), possibly the diameter of the base of the cap, or something like that. I'll try doing some sums and see if I can work out what their 'r' actually is!

Kind Regards, John

 

[TonyW2]

Does a point have an area or a line a width?

 

[DetlefSchmitz]

It's simple enough, and only requires a one, pretty simple, integration. See...

Thanks John. That is much easier than the approach I was trying. Just shows that it's best to start from the right place.

I'm on holiday at the moment and trying to avoid thinking about work, so this was good fun.

 

[JohnW2]

Does a point have an area or a line a width?

By conventional definition, no.

If they have an area, then the point is some sort of 'shape' (circle or whatever), rather than a point, and the line is a rectangle, rather than a line.

Kind Regards, John

 

[DetlefSchmitz]

I should add that I've always had a dread of these 'part of circle' question ever since I was about 14. A local builder gave my dad a question to give to me which I spent many hours on without any success.

A goat is tethered by a rope to the edge of a circular field 100m in diameter, such that the goat is able to eat exactly half the grass in the field. How long is the rope?

Edit: I'm beginning to think it's not as hard as I used to think.

 

[JohnW2]

Thanks John. That is much easier than the approach I was trying. Just shows that it's best to start from the right place. I'm on holiday at the moment and trying to avoid thinking about work, so this was good fun.

You're welcome. I'm still struggling to work out what the 'r'in the formula per your link represents - unless I've gone wrong, it doesn't seem to be anything particularly obvious!

Kind Regards, John

 

[EFLImpudence]

No, I didn't read it all. sorry.

I'm getting confused. Are you now saying that you agree that:

Therefore the residue must equal the volume of a sphere with a diameter of six inches, namely 36π.

is correct (which it is!).

No, I just glanced at it and thought it said the volume of a sphere was 36π, i.e. πr².

 

[JohnW2]

I should add that I've always had a dread of these 'part of circle' question ever since I was about 14. A local builder gave my dad a question to give to me which I spent many hours on without any success.

A goat is tethered by a rope to the edge of a circular field 100m in diameter, such that the goat is able to eat exactly half the grass in the field. How long is the rope?

At least that's only 2-dimensional. It was 3-D geometry and trig which used to get my brain tangled (and fortunately I very rarely have reason to think about such things these days ).

Kind Regards, John

 

[DetlefSchmitz]

John: In your derivation paper the r used is the sphere's radius. In all the formula I have seen including in my link, the r is the radius of the bottom circle of the spherical cap. Hence the difference in formulae. I'll work through your paper using the different r and see what I get.

 

[JohnW2]

No, I didn't read it all. sorry.

I'm getting confused. Are you now saying that you agree that:

Therefore the residue must equal the volume of a sphere with a diameter of six inches, namely 36π.

is correct (which it is!).

No, I just glanced at it and thought it said the volume of a sphere was 36π, i.e. πr².

Fair enough. I take it thatthe 'No' there actually means 'Yes'?

It's obvioulsy just a 'coincidence' that πr², if r were equal to 6 inches, which it wasn't!), and 4πr³/3, when r=3, both evaluate to 36π !

Kind Regards, John

 

[DetlefSchmitz]

OK John. I substituted an equation for the different use of radius in the equation from the paper you linked to, and got the result in the puzzle I linked to, so I'm happy they're the same.

 

[JohnW2]

John: In your derivation paper the r used is the sphere's radius.

Indeed, that's what I said, and that resulting formula is the one which I'm familiar with (and which you'll find all over the place if you Google for 'volume of cap of sphere').

In all the formula I have seen including in my link, the r is the radius of the bottom circle of the spherical cap. Hence the difference in formulae. I'll work through your paper using the different r and see what I get.

That's what I suggested, but (maybe just through errors), I can't (yet) get that to work. The 'radius of the bottom circle of the spherical cap' is a bit messy when expressed in terms of radius of sphere (r) and height of cap (h), namely (if I've got it right) ... √(2rh -h²) ... and substituting that into the formulae doesn't yet work for me. I'll check my working and try again!

Edit: typo corrected (in red)
Kind Regards, John