The volumes and areas of solids like spheres or cones are a bit more complicated to work out than say for a cube.

### The sphere

A sphere is a perfectly round object like a ball. It has only one surface, which is curved, and no corners or edges.

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The formula for the volume of a sphere is:

_{}

The formula for the surface area of a sphere is:

_{}

Handy Hint #1 - Dimensionality of formulas

Say you’re in an exam, and you can’t
quite remember whether the formula for the volume of a sphere is _{} or _{}. You’re a bit
stuck, because you’ve only got a 50-50 chance of picking the correct formula.
However, you can help yourself out by thinking about the *dimensionality*
of the formula.

The volume of a cube is calculated by
simply *cubing* the side length:

_{}

Because a cube is a *3-dimensional*
object, there are *three* dimensions multiplied by each other – in this case
the three dimensions are all the same – ‘L’, ‘L’ and ‘L’. What about for a
rectangular based prism? Well, then the formula for the volume is:

_{}

Once again, because we’re working out
volume, which is a 3-dimensional quantity, we have *three* dimensions
multiplying each other.

Area on the other hand is a *2-dimensional*
quantity. The area of a rectangle for instance is:

_{}

Because it’s only 2-dimensional, we
only have *two* dimensions multiplying each other – ‘L’ and ‘W’. So in
general, formulas for volumes need to have **3*** *dimensions multiply
each other, while area formulas need only 2 dimensions multiplying each other.
So if we’re trying to pick between _{} or _{} for the volume of a sphere – we
know that it is probably going to be _{}, because it has *three* lots
of dimensions (r, r and r) multiplying each other. We can guess that the other
formula – _{} is
probably the formula for the *area* of something, since it only has two
dimensions multiplying each other (r and r).

### The pyramid

Pyramids actually have *two* different heights. The *slant
height* is the distance along one of the edges from the base to the pointy
tip – it’s ‘slant’ because it’s at an angle to the vertical. The *vertical
height*, usually just called the *height*, is the vertical distance
from the centre of the base to the tip, and is shorter than the slant height.
Usually ‘h’ is used to represent the vertical height, and ‘s’ to represent the slant
height.

If you make a prism which has the exact same base as the
pyramid and the same height, then it has a volume *three* times as great.
This is where the formula for the volume of a pyramid comes from:

_{}

Remembering the one third volume fact about the pyramid is a good way of remembering the formula – it’s just one third the volume of a prism with the same base and height.

### Proving the volume of a pyramid formula

Here’s a way to show how to get to this formula using geometry. What we want to do is represent a pyramid by a series of levels. Say we’re trying to represent a square based pyramid, we can build up one by using thin square based prisms:

Now, the volume of this ‘pyramid’ is made up of the volumes of each layer. The volume of each layer can be calculated by multiplying its height (h / 4) by its cross sectional area:

_{}

There are *four* levels in our diagram, and the side
length of each level decreases by a regular amount as you go up each level.
The second bottom level’s side is *three quarters* the length of the
bottom level. The third level’s side is only *half* as long as the bottom
level. The top level has a side only *one quarter *the length of the
bottom level’s side. The side lengths decrease in *quarter* amounts
because we have *four* levels.

_{}

How does the *area* change as we go up levels? Well,
area is a *two-dimensional* quantity, so it will decrease at a faster rate
than the side length. Here’s the calculation for the area of the 2^{nd}
bottom level:

_{}

So in general, the side length and area of any level depends on how high up in the pyramid it is and on how many levels there are in total.

Now, this four level pyramid doesn’t really look much like a real pyramid, it’s too chunky. However, if we use lots and lots of levels, it starts to look just like a real pyramid:

If we use ‘N’ number of levels, then the height of each
level is going to be _{}. Now, we can number the levels
from 1 to N, starting at the *top* of the pyramid. For a pyramid with N
levels, the side length of level number ‘i’ is (if the top level counts as level
number 1):

_{}

So the area of level ‘i’ is going to be:

_{}

We can get the volume of this pyramid by starting at the top and adding up the volumes of each layer:

_{}

I have factorised out the _{} part from each term. Now, what
about those areas? Well, we can substitute our formula for the area of layer
number ‘i’ into this equation:

Because we have an ‘A_{base}’ in every term inside
the square brackets, we can *factorise* it outside as well:

_{}

Now we know that our final formula for the volume of a pyramid has ‘A’s in it and ‘h’s in it, so that’s fine, but it doesn’t have any ‘N’s in it. Let’s group these together and try and get rid of them. First we can change how the inside of the bracket looks, then we can bring the ‘one on N’ in from outside the brackets:

_{}

Now, what does 1^{2} + 2^{2} + … N^{2}
equal? Well, there is a formula which tells you the sum of any series of
square numbers from 1 to N:

_{}

There is a way to prove this formula as well, but I won’t go into it here. You can satisfy yourself that it works by trying it out.

We can use this formula to replace the 1^{2} + 2^{2}
+ … N^{2} line in our equation:

_{}

Now we can expand out the top line of the fraction and simplify it:

Now, the more layers we have in our pyramid, the closer it looks to an actual pyramid. So as we make pyramids with more and more layers, N becomes a very large number (since N is how many layers there are). When N becomes a very, very large number, two of the three fractions inside the brackets become very small numbers, since N is in their denominators only. As N becomes even larger these two fractions eventually become zero, and we’re just left with the one third:

_{}

There’s our proof! Although we proved it for a *square
based* pyramid, it applies to any pyramid, because the cross sectional areas
would change as you moved through the pyramid in exactly the same way,
regardless of what shape the area was. You’d never be expected to actually
come up with a proof like this yourself in an exam, but it is good to at least
understand how it works.

The surface area of a pyramid is calculated by just adding up the areas of all the sides and the base:

_{}

There’s always one base, and then as many triangular sides as there are sides on the base. Calculating the base area can be a little tricky, especially if it’s something like a decagon (10 sided shape). To work out the area of the triangular sides, you need to use the triangle area formula:

_{}

Be careful! The ‘h’ in this equation is *not* the
slant height *or* the vertical pyramid height. It’s the height shown in
the diagram:

So if you’re only given the slant height or the vertical pyramid
height, you’ll need to use Pythagoras’ Theorem to get the actual *triangle
height*. For instance, if you had a square based pyramid, and you only knew
the square side length and the pyramid’s vertical height, you could work out
the triangle height like this:

Notice how the triangle height line, the pyramid height line and the ‘half base side length’ line together form a right-angled triangle? This means we can use Pythagoras’ Theorem:

_{}

If the base of your pyramid isn’t a regular shape like a square or hexagon, then you’re going to have different size triangular sides. This means you’ll have to work out the area individually for each triangle. For symmetrical pyramids like a square based pyramid though, you only need to work out the area of one triangular side, then multiply it by however many there are.

### The cone

The cone is like a pyramid except with a *circular*
base. It has *two* heights like a pyramid – a *slant height* and a *vertical
height*, normally just called the ‘height’.

A pyramid’s volume is one third the volume of a prism having
the same base and height. Well, the volume of a cone is *one third *the
volume of a *cylinder* which has the same base and height:

_{}

You can actually work out the formula for the surface area
of a cone by imagining *unwrapping* the cone to form its net (the flat
shape that you fold together to get the solid shape):

So the surface area of a cone is made up of two areas – the
area of the circular base, which is easy to calculate, and the area of the
curved side. The area of the circle is just _{}. The other area is just a sector
of a circle. The arc length of the sector is just _{}, since we need it to join
up with every point on our base circle. So the fraction that the sector is of
the entire dotted circle is its arc length divided by the circumference of the
dotted circle:

_{}

This means that the *area* of the sector is just this
fraction of the area of the dotted circle:

_{}

The total surface area of the cone is:

_{}

Now, if you don’t have the slant height of the cone, you can calculate it from the vertical height and the base radius using Pythagoras’ Theorem:

_{}

Cone nets look different depending on how ‘tall’ the cone
is compared to the radius of its base. If we have a cone with a base radius
‘r’, and a slant height ‘s’, then we can calculate the angle θ of the
sector because we know the length of the sector arc has to be _{} long.

_{}

Now here when we’re talking about a circle we mean the
large dotted one in the diagram. The small circle that is the bottom of the
cone is referred to as the ‘base’. So we have two radii – r_{base} and
r_{circle}. Now r_{circle} is just the slant height of the
cone, so we can replace it and rearrange the equation like this:

_{}

So say I wanted to draw the net for a cone that had a base radius of 1 and a slant height of 8. I could work out the angle of the sector using this formula:

_{}

So the net would look something like this:

Notice how the sector or wedge is a lot narrower than the
first cone net we looked at, this is because the *slant height* to *base
radius* ratio is a lot larger – the cone is a lot taller and skinnier.

A martini is a very famous alcoholic drink which Mr James Bond is in the habit of ordering – “shaken, not stirred” being a classic quote. It’s served in a conical glass, with an olive in it. Mr James Bond is arguing with Mrs Bond over how much liquid there is in his Martini glass. Mr Bond guesses 40 mL. Mrs Bond thinks it’s more like 55 mL. The liquid is filled to within a centimetre of the top of the glass, which has a diameter of 8 cm. The height of the conical part of the glass is 6 cm (assume the top of the glass is like an upside down cone). The olive is spherical and has a diameter of 2 cm. How much liquid is there in the glass? |

Solution |

This is quite a tricky question, because you aren’t given
the exact information you need. We want to work out how much liquid there is
in the glass So, the height of the liquid is easy to work out – it’s just one centimetre less than the overall height of the conical part: We also need to know the radius of the top of the liquid part. This is a bit trickier to work out, we need to use ratios to help us. What we’re doing is comparing the conical part of the glass with the conical section of liquid inside it: We can write a ratio comparing the two diameters and the two heights: Now we can work out the volume of the liquid inside the
glass, Now 1 cubic centimetre of fluid is the same as 1 millilitre (mL) of fluid: We however want the volume of liquid inside the glass, So we just subtract this from our total liquid volume: Looks like Mrs Bond is much better at estimating volumes than Mr Bond! Oh well, he can’t be good at everything. |