After the interesting Puzzle about the Lawn at Quiz Master Shop Towers, with the paths now sorted, we posted a puzzle about the lawn ornaments, as follows.

Now it's quite a big lawn, so we want some big ornaments, and to our own bespoke design. Each ornament is a sphere with a cylindrical hole through it, and to fit into the design the cylindrical hole must be exactly one metre long.

We are trying to work out which metal to use - bronze or copper would look brilliant, but steel would be cheaper. So we need to know how much metal is needed for each ornament, to see how much it would cost to cast them.

Do you have enough information to work out the volume of one of the ornaments, and if you do what is the volume?

Well, clearly you must have enough information to solve this, otherwise it would not have been very interesting. Which means, somewhat surprisingly, that the diameter of the sphere is irrelevant.

This seems a bit counter intuitive - what about a sphere the size of the earth. To get a band one metre high with the same diameter as the earth, the band would be very, very thin, so perhaps . . .

If this is the case then the volume of the ornament is the same as a sphere with a very narrow (actually zero width) cylinder through it. That is a sphere with no hole at all. The volume of a sphere is

$$\frac{4}{3} \pi r^3$$

And as the radius is half a metre, this drops out to

$$\frac{\pi}{6}$$

The fun comes trying to show that any sphere with a one metre high cylinder through it has the same volume.

We already have the formula for the volume of a sphere, above.

The volume of a cylinder is

$$\pi a^2 h$$

where a is the radius of the cylinder and h is its height.

h = 1 and from Pythagoras

$$a = \sqrt{r^2 - (1/4)}$$

The volume of a cap on a sphere (and we looked up this one!) is

$$\frac{\pi c^2}{3} (3r - c)$$

where r is the radius of the sphere and c the height of the cap. And c = r - (1/2).

The volume of the ornament is the volume of the sphere, minus the volume of the cylinder, minus the volume of two caps (top and bottom).

$$\frac{4}{3} \pi r^3 - \pi a^2 h - \frac{2 \pi c^2}{3} (3r - c)$$

And then substituting values for a, h and c from above:

$$\frac{4}{3} \pi r^3 - \pi (r^2 - \frac{1}{4}) - \frac{2 \pi (r - (1/2))^2}{3} (3r - (r - (1/2))$$

Expanding this out, and the cancelling out terms, amazingly, this reduces all the way back to

$$\frac{\pi}{6}$$