Lesson video

Hello, I'm Mrs. Lashley, and I'm gonna be working with you as we go through the lesson today.

I really hope you're looking forward to it and you're ready to try your best.

So the lesson outcome today is to be able to calculate the volume of a sphere.

The definition of a sphere is on the screen.

You will have met that before, but you may wish to pause the video here so that you can re-familiarize yourself before we make a start.

So when we're looking at finding the volume of a sphere, we're gonna break it into two learning cycles.

The first learning cycle is to look at finding the volume using displacement.

And the second learning cycle is where we're going to look at using a formula instead.

So let's make a start of thinking about how we can use displacement to find the volume.

So here we have a large jug, and it's filled with 2.

5 litres of water, which occupies 2,500 cubic centimetres.

If we remind ourselves that 2.

5 litres is equivalent to 2,500 millilitres, then we can see that one millilitre is equivalent to one cubic centimetre.

One millilitre of liquid will occupy the same amount of space as one cubic centimetre.

So Sam is going to use displacement to calculate the volume of various objects.

Sam is telling us what that actually means.

So displacement is going to be looking at the change in the water level, and that change will be the volume of the object.

So the change in the water level is the displacement of water.

In this case, that's the liquid that's being used, and that will represent and be equivalent to the volume.

So firstly, Sam has put a strawberry into this jug.

Remember, the jug already had 2.

5 litres of water in it.

And Sam has noted that the strawberry has increased the water level, so it's displaced the water, it has increased by 50 millilitres, so its volume is 50 cubic centimetres because once again, one millilitre occupies the amount of space as one cubic centimetre.

How does she get the 50 millilitres? Well, that's where we need to think about the scale that's on this jug.

We can see that the litres are marked one litre, two litre, three litre, four litre, five litre, and then there's some larger marks, larger increment marks, where we can see the halves and the quarters, but then there are the smaller ones.

And within one litre there are 20 increments, or 20 jumps, which tells us that the scale is going up in 50 millilitres because one litre is 1,000 millilitres and 1,000 split into 20 parts would give you 50 millilitres per part, so each increment on the scale is 50 millilitres.

And hopefully you can see that that water level is just one increment above the 2.

5, so it is gone up by 50 millilitres, and that's how we know that the volume of this strawberry is 50 cubic centimetres.

So this time Sam puts a golf ball into the jug, and they notice that the ball has increased the water by 100 millilitres this time 'cause it went back to 2.

5 litres.

Having removed the strawberry, the pre-level was 2.

5 litres.

It now reads as 2,600 millilitres, or 2.

6 litres.

So that increase, that displacement of 100 millilitres, means that the golf ball has a volume of 100 cubic centimetres.

Here's a check for you.

What is the volume of this ornament if there had been 2.

5 litres in the jug before the object was placed in it? So pause the video, look at the scale, and when you are ready to check your answer, press play.

The change was 1.

5 litres, or 1,500 millilitres, so it has a volume of 1,500 cubic centimetres.

This ornament has a volume of 1,500 cubic centimetres.

Sam is still using this displacement method, and this time they've placed a cricket ball in.

Sam comments that the cricket ball has a volume somewhere between 150 cubic centimetres and 200 cubic centimetres.

Can you see why Sam has said "somewhere between"? The water level is not on, perfectly on one of those scale increments, and because it's going up in 50 millilitres, that's as accurate as we can be in terms of our measurement.

So it's rather challenging to be accurate with this method.

So a check, which of these is a fair approximation for the volume of this object if the jug started with 2,500 millilitres of water? So pause the video, and when you're ready to check, press play.

So B is the fair approximation for the volume of this object.

So let's go through why A and C were not.

So C is the reading from the scale, and so that is not the volume of the object because the volume comes from the displacement, the change in the water level.

So just taking that reading is not giving us the volume.

We need to subtract the pre-level from that to get the change.

A is close to the volume, yes, but because we were told that it started with 2.

5 litres of water, or 2,500 millilitres of water, a change of 500 would get us to the three-liter mark.

And we can quite clearly see that the water level is above the three litres, so we know that the volume of that object is more than 500.

So we're up to the first task of the lesson for you with using displacement to find volume.

So in question one, you need to match the object, the objects are at the bottom, to the correct jug.

So each jug has got a question mark 'cause we're hiding which object was in there.

You were told the pre-level, so that was the water level before an object was placed in.

And then you can read off of the diagram off the jug the current level with the object inside.

The objects are a piece of stone, an ornament, a duck, and an egg.

Pause the video whilst you're doing the matching, and when you're ready for the next question, press play.

So the next question is, what is the volume of this crystal? So the jug on the left-hand side is without the crystal, and the jug on the right-hand side is with the crystal in the water.

So what is the volume of this crystal? Press pause, and then, when you're ready for question three, press play.

Here's the last question of this task.

So, what is the volume of this solid sphere? So similar to the crystal question, the left-hand side is without this object, which is a sphere, and the right-hand jug is the one with the sphere.

So what is the volume? Press pause, and then when you're ready for the answers to task A, press play, and we'll go through them.

Question one, you needed to match up.

So you need to look at the change in the level from the pre-level, which was written in text, to the scale on the jug.

So if we look at the first jug, it was 2.

75 litres before the object was placed in it, and now it reads at three litres.

So the displacement, the change in the water level, is 250 millilitres, which is equivalent to 250 cubic centimetres, which is why it matches with the dragon ornament.

The second jug started at three litres and now reads at just about 3.

5 litres.

So there's a change there of half a litre.

Half a litre is equivalent to 500 millilitres, and for every one millilitre, that's equivalent to one cubic centimetre, so it matches with the stone.

This third jug matches with the duck.

It started with 1.

75 litres and it now reads at 2.

5 litres, so that is a change of 0.

75 litres, or 750 millilitres, which is equivalent to a volume of 750 cubic centimetres, which leads to the last jug with the egg.

And there is a 2.

25 is the reading on the jug, so it's a change of 10 millilitres, and so that's 10 cubic centimetres.

Question two, you needed to work out the volume of the crystal.

So the left-hand jug gave you the pre-level, the right-hand jug gave you the level with the crystal in there, so the displacement is the change.

Using the scales, there is a change of 150 millilitres, which means that the volume is 150 cubic centimetres.

Once again, remembering that the scale was going up in 50-milliliter increments.

So we can see once the crystal has been put inside the jug that the reading is 2.

9 litres.

So that change from 2.

75 to 2.

9 was what was giving you the volume.

And the last question, the sphere would have a volume approximately 825.

So if you've got a value close to 825, then you can say that you are correct.

This one, the water levels were not on in the pre-level, and after the sphere had been dropped into the water, they weren't exactly, the water level was not clearly on an incurrent, so you were doing a bit of guesswork with in terms of the approximation.

So we're now gonna go to the second learning cycle, where we are gonna look at the formula to find the volume.

So using displacement to find the volume of an object can have a fair amount of inaccuracy, and we saw that if you had a better or more accurate scale, then your accuracy would improve.

But there is going to have quite a lot of inaccuracy.

However, it is a really useful way of finding the volume of irregular objects such as the ornament, or the bottle, or that ornament there.

So some objects are not easy to find the volume, however, with this idea of using displacement and the change in the water level is a good method for that, but remembering that there is an element of inaccuracy.

However, some 3D shapes have other ways to calculate the volume.

So there's some 3D shapes on the screen here.

We've got the cylinder, we've got hexagonal prism, our sphere, the focus of our lesson today, and a square-based pyramid.

And so for these 3D objects, there is other ways to get the volume.

We're gonna focus on how to find the volume of a sphere.

So for any sphere, the volume can be calculated by using this formula.

So the volume is 4/3 pi r cubed, where r is the radius of the sphere.

And there's a diagram there of where the radius, but remember the radius could be pointing straight up, vertically upwards, or vertically down, or at any angle in this 3D space.

The definition of a sphere is that all points on the surface of the sphere are equidistant from its centre.

So this is the formula to find the volume for a perfect sphere.

So let's have a look at using it.

A sphere has a radius of nine centimetres.

What is its volume? So the volume formula is in the box, 4/3 pi r cubed.

The r, the radius, is the variable.

The 4/3 pi is a constant.

The reason this is volume is because we are cubing the radius, so that means that we're going to end up with a cubic unit.

Andeep says, "The volume is 972 pi cubic centimetres." So that's being calculated by doing 4/3 times pi times 9 cubed, and that simplifies to 972 pi.

Andeep tells us that this is the most exact answer.

So just have a think.

Why is this the most exact answer that Andeep could have given? I'm hoping you've thought about what pi is.

Pi is an irrational number.

If we use pi to any other degree of accuracy, 3.

14 or three, quite a crude approximation of pi there, then you've included these inaccuracies in the rounding.

So because pi is irrational, leaving it in terms of pi is the most exact answer.

You have not rounded any values.

But let's use our calculator.

What is the volume to three significant figures? So our calculator has pi programmed onto it.

So let's type 4/3 times pi times 9 cubed and find this value of the volume to three significant figures.

So pause the video whilst you grab your calculator and you do that, and then press play to check it.

So to three significant figures, it's 3050 cubic centimetres.

So by rounding it, we've now decreased the accuracy.

A check for you.

A sphere has a radius of six centimetres.

What is the exact volume? So remember, that means you should be leaving it in terms of pi.

And I'd like you to do this without a calculator.

Press pause, and then, when you're ready to check, press play.

288 to pi.

So you needed to cube the six, six cubed, and you can do that with written methods without a calculator.

And then you need to times it by 4/3.

It might be that you divide it by three and then times by four as a way of times by 4/3, and that would be 288, and then we are leaving it in terms of pi, 288 times pi.

We've got formula for a volume of a sphere, and we can calculate that.

We can give it as an exact answer in terms of pi, and we can also round it using our calculator.

A hemisphere is exactly half of a sphere with the same radius.

So I'm guessing you've used that word previously.

You may have used it in more geographical context.

Hemisphere, it splits the sphere into two equal parts.

Hemisphere, half of a sphere.

What proportion of the volume would a hemisphere have? It would have half the volume.

So it split the sphere into two equal parts into halves, and therefore the volume of each would be half of the original sphere.

So if we have a formula for the volume of a sphere and we would like to get the volume of a hemisphere, then we can do that by calculating the volume of the sphere with the same radius and dividing it by two, or halving it.

So for this one here, the radius is nine.

So 4/3 times pi times 9 cubed.

The part in brackets there is the volume of a sphere with a radius of nine, and then I'm dividing it into two equal parts to find the volume of the hemisphere.

But we can do that a different way.

Rather than dividing by two, we know that that is equivalent to multiplying by a half, a half of the volume of the sphere.

Well, this actually has some common factors that can cancel down.

And so if we look at those two fractions, 4/3 times a 1/2, well, that would be 2/3 times 1.

So 2/3 times pi times 9 cubed is the volume of the hemisphere with a radius of nine.

When we evaluate that, that's 486 pi cubic centimetres.

So halving the 4/3 gives you 2/3.

So 2/3 times pi times 9 cubed is the volume of the hemisphere.

So here's a check for you.

Which of these calculations finds the volume of a hemisphere with a radius of 10 centimetres? Pause the video, and then, when you're ready to check, press play.

There were actually three that would get you the volume of a hemisphere with a radius of 10.

A, which is finding the volume of the whole sphere and then dividing it by two to get half, or B, which is writing it as half of the volume of the whole sphere.

Or D, where the half of the 4/3 has been cancelled down to 2/3.

So 2/3 times pi times 10 cubed.

We're gonna continue working now with the volume of a sphere because we have a formula to find the volume of a sphere, we have a formula to find the volume of a hemisphere.

So here we're told that 269 cubic millimetres of metal is used to create a spherical marble.

So spherical means it's the shape of a sphere.

What is the radius of the marble to the nearest millimetre? So here we are looking at going from volume 'cause that's the amount of metal that is going to occupy the space of the sphere.

And so what is the radius of that said marble? So an equation can be set up and solved to find the radius.

The volume of the sphere is found by doing 4/3 times pi times radius cubed.

It's the radius that we're trying to calculate.

We know the volume of our marble is 269 millimetres cubed because that is the amount of metal being used.

So now we've set the equation up, we can use all of our algebraic skills to solve this to find the radius.

So I've times by three, divided by four, divided by pi.

Dividing my pi because pi being irrational, I end up with another irrational number.

So r cubed equals 64.

219.

I'm gonna be using the most exact answer I can.

So the dot dot dot indicates that this is not rounded, there are further digits.

I'm gonna cube root, and r is 4.

004.

Again, there are some more digits after that.

So to the nearest millimetre, that's what the question asked, the radius of this marble is four millimetres.

Let's continue with working with spheres in context.

So the Earth's equator is 24,901 miles long.

What is the volume of the Earth? So we are gonna model the Earth as a sphere.

You might think, well it is a sphere, it isn't a perfect sphere, but we are gonna model it to be a perfect sphere.

So let's think about what we're asked for.

We're asked for the volume.

How do we calculate the volume of a sphere? Well, we use the formula 4/3 times pi times radius cubed.

Radius, it's the radius of the Earth that we need.

We don't have the radius of the Earth, but instead we have the length of the Earth's equator.

The Earth's equator is that green line that goes around the central middle part of our globe.

It's where we split into the Northern Hemisphere and the Southern Hemisphere.

What is the equator in terms of geometry? Well, the equator is the circumference of a circle.

If you think about when we have our two hemispheres, we have that circular face.

And so the equator is the maximum length around the sphere.

A circumference of a circle.

So circumference of a circle is two pi r, or pi times diameter.

So again, I can set up an equation.

We know the length of that circumference, and we have a formula for the length of a circumference in terms of radius.

It's the radius that we need.

So solving this by dividing by two pi, we get the radius as 3,963 roughly.

So what did we want? We want the volume.

What did we need? The radius.

We have the radius.

So now we're gonna use our formula, substituting that in.

Again, I'm gonna use the most exact answer on my calculator, and I end up in standard form.

Why do I end up in standard form? Well, think how large the Earth is.

So standard form is being used.

In normal form, we can see the volume of the Earth in cubic miles.

Here's a check for you.

The Earth's equator is 40,075 kilometres long.

What is the volume of the Earth in cubic kilometres? So once again, model the Earth as a sphere.

Press pause, and when you're ready to check, press play.

So the question is the same, but the units is different.

So the equator is the circumference of a circle.

So set up the equation, solve it to find the radius.

Now we can use the radius in our volume formula, and we again, we're gonna have it in standard form, and in cubic kilometres, that is the volume of the Earth.

Well done if you manage to get that.

So onto the last task of the lesson.

So on question one, there are six parts, A to F.

Read them carefully.

Is it a sphere, is it a radius, is it a diameter? The volume of a sphere formula is at the bottom of the slide, so you can make use of that.

I'd like the exact volume, so that means you're leaving it in terms of pi, so you should be doing this without a calculator.

Press pause, and when you're ready for the next question, press play.

Question two, you are going to use a calculator on this question.

By modelling each as a sphere or a hemisphere as appropriate, work out the volume of the following items, giving your answer to an appropriate degree of accuracy.

So there's a football, an orange, a candlepin bowling ball, which is basically a perfect sphere.

It doesn't have any holes, finger holes in it, or anything, so it's just perfect sphere.

A marble and the amount of soup held in a ladle of radius 4.

5 centimetres.

Pause the video, and when you're ready to go to the next questions, press play.

So questions three, four, and five are on the screen now.

So I'd encourage you to read them carefully, think about what it's asking for, and therefore what you need and how you'll get it.

Press pause whilst you work through those three questions.

When you press play, we're gonna go through the answers.

Question one, all the answers are on the screen.

So exact value, so we needed it in terms of pi.

The first two was a sphere with a radius given.

So that was just substituting and evaluating the formula.

C was a sphere, but it was given with the diameter.

So the first step would be to half the diameter to give you the radius.

D, E, and F were all hemispheres, so you needed to half the volume of a full sphere.

D and E was the radius, whereas F was the diameter.

Maybe you noticed that if you half the diameter on F, you get 15.

And part A was a radius of 15 centimetres of a sphere and part F was the hemisphere, so you could half your answer to A and that would give you the answer to F.

Question two, we were modelling each as either a sphere or a hemisphere, depending on what the context was.

And using our calculator, rounding to an appropriate degree of accuracy.

So a football with a diameter of 22 centimetres.

So for that one, we're going to half the diameter to get our radius of 11.

It's gonna be modelled as a sphere because it's spherical, and that would be 5575.

3 cubic centimetres to one decimal place.

You may have decided a different degree of accuracy, so just check that your calculation was correct and that your answer rounds to one decimal place this value.

An orange of a radius of 38 millimetres.

Well, an orange, again, we're gonna model to be a perfect sphere, and I've used centimetres rather than millimetres.

When we use the millimetre, the value will be just much larger.

So I've changed it to centimetres, and therefore the volume of the orange to the nearest cubic centimetre is 230 cubic centimetres.

C and D are on the screen now.

So C is modelled as a sphere.

The diameter was given, so you did need to half that to get your radius of 5.

5.

And D was also going to be a sphere.

The radius was given.

So substituting that in, I've given one to one decimal place and one to three significant figures.

Lastly, E was the amount of soup held in a ladle of radius of 4.

5 centimetres.

So we are modelling the spoon part of the ladle as a hemisphere.

We're not including the handle.

And so we're gonna do 2/3 because half of 4/3 is 2/3.

2/3 times pi times 4.

5 cubed, which gives you 190.

9 cubic centimetres.

So in terms of the amount of soup, you may have written 191 millilitres or 190.

9 millilitres.

Question three, the volume was given, and you needed the radius.

So set up the equation and solve it.

So I've divided by pi, first of all, and then divided by 4/3, and then cube-rooted it, the radius would be 18 metres.

The unit was metres cubed, so the radius would be in metres.

Question four, the volume of a sphere was given, and you needed to work out the diameter.

So it was like question three, there was just one more step, which is once you had the radius, which was 15, to double it to give you the diameter.

So the diameter of the sphere is 30 centimetres.

Lastly, the volume of a hemisphere was given, work out the circumference of the circular face.

So the volume of the hemisphere is equivalent to 2/3 pi r cubed.

Solve it to find the radius.

The radius was 15.

So a hemisphere, when you think of a hemisphere, it has the curved surface, and there is also a circular face.

So the circular face, if we're trying to get the circumference, and when we had equator in the context, our equator was the circumference, we'd need two times pi times r.

r is 15, we just found that out, so 30 pi centimetres would be the circumference in its exact form.

So to summarise today's lesson on the volume of a sphere, one method is displacement of liquid, and that can be used to find the volume of shapes where a formula is not known.

However, our sphere is a shape, a 3D object, where we do have a formula, so we can be very exact.

So if the radius of the sphere is known, then the volume can be found by using 4/3 pi r cubed.

The radius can be found from a known diameter, circumference, or volume.

So there are times where we need to make use of that formula and work backwards to get the radius.

Really well done today, and I look forward to working with you again in the future.