# Digging to China

1. Jun 4, 2004

### Gokul43201

Staff Emeritus
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 ?

2. Jun 4, 2004

### jcsd

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.

3. Jun 4, 2004

### Gokul43201

Staff Emeritus
No, the question is complete !

4. Jun 4, 2004

### jcsd

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

5. Jun 4, 2004

### 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 voulme 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).

Last edited: Jun 4, 2004
6. Jun 4, 2004

### Njorl

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

Njorl

7. Jun 4, 2004

### 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.

Last edited: Jun 4, 2004
8. Jun 4, 2004

### Gokul43201

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

9. Jun 4, 2004

### jcsd

oi! there called typos

10. Jun 4, 2004

### jcsd

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

11. Jun 4, 2004

### Gokul43201

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