Volume of sphere with hole in it

Cifrocco · #1 $Old$ 11-02-2008, 09:26 AM

A cylindrical hole is bored centrally through a ball, you are told the length of the hole, now find the volume of the remaining material. The surprising thing is that the data is sufficient - you do not need to know the radius of the ball.

Can this problem be solved, if so, what is the solution?

Mathstud28 · #2 $Old$ 11-02-2008, 10:06 PM

Quote:

Originally Posted by Cifrocco $View Post$

A cylindrical hole is bored centrally through a ball, you are told the length of the hole, now find the volume of the remaining material. The surprising thing is that the data is sufficient - you do not need to know the radius of the ball.

Can this problem be solved, if so, what is the solution?

If the whole is of length l then the volume 4/3pi(l/2)^3-pi(l/2)^2h

CaptainBlack · #3 $Old$ 11-03-2008, 01:07 AM

Quote:

Originally Posted by Cifrocco $View Post$

A cylindrical hole is bored centrally through a ball, you are told the length of the hole, now find the volume of the remaining material. The surprising thing is that the data is sufficient - you do not need to know the radius of the ball.

Can this problem be solved, if so, what is the solution?

You know the volume of the hole, and the intact sphere.

You have enough information to compute the diameter of the hole, and so the volumes of the two spherical caps that are needed in addition to a cylinder the same size as the hole to complete the sphere.

So yes, you have enough information to solve this problem.

CB

Cifrocco · #4 $Old$ 11-03-2008, 05:15 AM

I tried to solve it but I don't succeed.

Mathstud, what is the "h" in the exponent "2h" at the end of your formula?

CaptainBlack, how do you know the volume of the hole and the intact sphere at the outset?

Can you give examples of solutions?

The problem comes from the website of W. W. Sawyer.

CaptainBlack · #5 $Old$ 11-03-2008, 05:39 AM

Quote:

Originally Posted by Cifrocco $View Post$

I tried to solve it but I don't succeed.

Mathstud, what is the "h" in the exponent "2h" at the end of your formula?

CaptainBlack, how do you know the volume of the hole and the intact sphere at the outset?

Can you give examples of solutions?

The problem comes from the website of W. W. Sawyer.

Volume of sphere:

$V_{sphere}=\frac{4}{3}\pi r^3$

where r is the radius of the sphere

$V_{cylinder}=\pi \rho^2 h$

where h is the height of the hole and $\rho$ the radius of the cylinder, and we also know:

$\rho^2=r^2-(h/2)^2$

Then the total volume of the caps is:

$V_{2 \times cap}=\frac{2}{3}\pi (r-h/2)^2(2r+h/2)$

$V_{remains}=V_{sphere}-V_{cylinder}-V_{2 \times cap}$

(you will need to check the volume of the caps as I changed variables on the fly so there could easily be an error)

If this has all gone right when you simplify that it should be independent of $r$ .

CB

CaptainBlack · #6 $Old$ 11-03-2008, 05:49 AM

Quote:

Originally Posted by Cifrocco $View Post$

A cylindrical hole is bored centrally through a ball, you are told the length of the hole, now find the volume of the remaining material. The surprising thing is that the data is sufficient - you do not need to know the radius of the ball.

Can this problem be solved, if so, what is the solution?

Well if you know that the data is sufficient the volume in question must be equal to that of a sphere of radius h/2.

There is probably a simple trick that will solve this.

Here is a method using calculus

The Math Forum's Ask Dr Math also has a discussion of this.

CB

Cifrocco · #7 $Old$ 11-06-2008, 04:24 PM

Fascinating.

HallsofIvy · #8 $Old$ 11-09-2008, 12:40 PM

Quote:

Originally Posted by Cifrocco $View Post$

A cylindrical hole is bored centrally through a ball, you are told the length of the hole, now find the volume of the remaining material. The surprising thing is that the data is sufficient - you do not need to know the radius of the ball.

Can this problem be solved, if so, what is the solution?

The simplest way to solve that problem is to ASSUME that it has a solution. If the volume of the remaining material is independent of the actual radius of the ball, assume the radius is l/2 where l is the given length of the hole. In order that a "hole" of length l fit in a sphere of radius l/2 (and diameter l) the hole must have radius 0 and so NO material is removed. The "material remaining" is just the volume of a sphere of radius l/2: $\frac{4}{3}\pi \frac{l^3}{8}= \frac{1}{6}\pi l^3$ .

It can be shown, by carefully integrating the volume of the "caps" on either end of such a hole, that this is, in fact, independent of the radius of the sphere, as claimed in the problem.