Can't Dig to China: Man Finds Out The Hard Way

[Gokul43201]

I hope this is not old too...

A man decides the find out for himself whether or not "if you dig a hole straight down from America, you get to China". So he grabs his beautiful rosewood globe and drills a cylindrical tunnel diametrically through it, starting from America. He finds the other end is far from China... in fact it's in the southern hemispere.

So here's the question : If the length of the tunnel measured along its wall is 'L', what the volume of the material left in the globe after drilling out the tunnel ?

 

[jcsd]

I hope this is not old too...

A man decides the find out for himself whether or not "if you dig a hole straight down from America, you get to China". So he grabs his beautiful rosewood globe and drills a cylindrical tunnel diametrically through it, starting from America. He finds the other end is far from China... in fact it's in the southern hemispere.

So here's the question : If the length of the tunnel measured along its wall is 'L', what the volume of the material left in the globe after drilling out the tunnel ?

Surely the radius of the tunnel is important too?

As r tends to zero the volume tends to:

\pi(\frac{L^3}{6} - Lr^2)

where r is the radius of the tunnel.

 

[Gokul43201]

Surely the radius of the tunnel is important too?

No, the question is complete !

 

[jcsd]

Of course! The length of the tunnel is dependent on it's radius, so it's going to disappear in the final equation.

 

[jcsd]

Okay the missing parts of the globe are made up of the volume of the tunnel plus extra missing volume from the ends due to curved surface of the globe.

If we define R as the radius of the globe and r as a radius of a section of the circle of the tunnel we can define a length a:

R^2 - a^2 = r^2

As when a = L/2, r = the radius of the tunnel, the volume of the tunnel is given by:

\pi LR^2 - \frac{\pi L^3}{4}


To find the area of the 'missing volume' at the ends we can us tehe fact that a radius of a cross section is

\pi R^2 - \pi a^2

So the volume of the 'missing ends' is given by (the factor of 2 appears as there are 2 of them):
2\int^{R}_{L/2} (\pi R^2 - \pi a^2) da = 2\left[\pi R^2 a - \frac{\pi a^3}{3}\right]^{R}_{L/2}
= \frac{4R^3\pi}{3} - \pi L R^2+ \frac{\pi L^3}{12}

therfore the volume of left is (the volume of the sphere - the volume of the cynclinder and the 'ends':

\frac{4R^3\pi}{3} - \frac{4R^3\pi}{3} + L\pi R^2 - \frac{\pi L^3}{12} - \pi LR^2 + \frac{\pi L^3}{4}

=\frac{\pi L^3}{6}

(barring any silly mistakes).

 

[Njorl]

Neat problem. My initial reaction also was that we would need the radius of the tunnel. Tricky.



Njorl

 

[jcsd]

Infact I've just relaized looking at my first post, if you know that the voulem is dependt only on L you can guess from the fact that as r tends to zero the volume tends to:

\frac{\pi L^3}{6}

without haveing to integrate.

 

[Gokul43201]

Infact I've just relaized looking at my first post, if you know that the voulem is dependt only on L you can guess from the fact that as r tends to zero the volume tends to:

\frac{\pi L^3}{6}

without haveing to integrate.

Exactly !
jcsd with the point (I'm feeling generous, so I'll ignore the spellings).

 

[jcsd]

Exactly !
jcsd with the point (I'm feeling generous, so I'll ignore the spellings).

oi! there called typos

 

[jcsd]

And don't say: they're called typos.

 

[Gokul43201]

Okay ! Guess it's hard to type correctly when you're drowning.