# Video: Volume of Cylinders, Cones and Spheres - Problems in Context

Learn to calculate volumes of cylinders, cones, and spheres through a series of problems in context. For example, calculating the percentage of the volume of a cylindrical can is occupied by tennis balls or the volume of a chocolate layer on a sphere.

14:08

### Video Transcript

In this video we are going to look at calculating the volume of three different types of 3D solids: cylinders, cones, and spheres. I’m going to look specifically at problems in context, so real-world problems, where these skills might be required.

So firstly just a very brief reminder of the three formulae that we’re going to need. So the cylinder first of all. The cylinder has two dimensions that we’re interested in: the radius of the circle at the base and then the height of the cylinder. And the volume of the cylinder is calculated using this formula here: 𝜋𝑟 squared ℎ.

Secondly the cone; so it has the same two pieces of information that we’re interested in: the radius of the circle at the base and the height of the cone. And the formula for its volume is very similar, but it has an extra factor of a third. So the volume is one-third times 𝜋𝑟 squared ℎ. And you can imagine it as a cylinder around this cone. But some of that volume — in fact two-thirds of that volume — has been cut away. So it’s just one-third of the formula for the volume of a cylinder.

Finally then the sphere, and there’s only one piece of information this time, which is the radius of the sphere. And the formula for the volume: four-thirds 𝜋𝑟 cubed. So those are the three formulae that we’ll need and we will refer to them at different points during this video.

So let’s look at our first example. It says “Tennis balls are spheres of diameter seventy milliammetres. They’re packed in threes into cylindrical cans which of diameter eighty milliammetres and height two-hundred and twenty milliammetres. What percentage of the space inside the can is filled by the three tennis balls?” So it may be helpful just to sketch a quick diagram first of all to make sure entirely clear what the question is saying. So what we have then just is a rough sketch; we have a cylindrical can and then we have three spherical tennis balls positioned inside it.

So let’s put some of the key information onto our diagram. We told that the tennis balls are spheres of diameter seventy milliammetres. Now that means the radius of those spheres is half of that, so the radius is thirty-five milliammetres. So all three of the tennis balls have a radius of thirty-five milliammetres. Now the can has a diameter of eighty milliammetres. So the radius is again half of that, so the radius of the can is forty milliammetres. And then the height of the can is two hundred and twenty milliammetres.

So there we have all the information in the question drawn onto our diagram. So the question ask what percentage of the space inside the can is filled by the three tennis balls. So we need to work out a couple of different things. We need to work out the volume of the tennis balls first of all. We need the volume of the can and then we can look at this percentage. So let’s start off with the tennis balls.

Now the tennis balls are spheres. So we need the formula for the volume of a sphere. And if we recall that from the previous slide, it was four-thirds 𝜋𝑟 cubed. So we have three of them. so our volume is going to be three lots of and then four-thirds multiplied by 𝜋 multiplied by 𝑟 which is thirty-five cubed. So there’s our calculation for the volume of these three tennis balls. Now actually these two threes is just going to cancel out directly. And if I evaluate the remainder of that constant on my calculator, so four times thirty-five cubed, I will be able to calculate the volume of the three tennis balls as a hundred and seventy-one thousand five hundred 𝜋 milliammetres cubed. And I’ve kept it in terms of 𝜋 at this point in order to make sure it’s an exact value that I’m carrying through to the next stage of the calculation to make sure I don’t introduce any rounding errors later on.

Now we need to think about the volume of this can, which is of course a cylinder. So we need to recall the formula for the volume of a cylinder, which was that the volume is equal to 𝜋𝑟 squared ℎ. So we can use the value of 𝑟 for the cylinder. So the radius there is forty and the height of the cylinder is two hundred and twenty. And so that will give us a calculation of 𝜋 multiplied by forty squared multiplied by two hundred and twenty. And again if I just evaluate the constant — so forty squared times two hundred and twenty — I get three hundred and fifty-two thousand 𝜋 as the volume of the can.

Final step is to work out what percentage of the can is taken up by these three tennis balls. So I need to divide the volume of the tennis balls by the volume of the can that will give me a decimal. So then I’ll need to multiply three by a hundred at the end. So my calculation the volume of the tennis balls divided by the volume of the can multiplied by a hundred. Now these two factors of 𝜋 here will again just cancel out. And that’s why I kept the values exact because I knew that was going to happen. And it means that I haven’t introduced any rounding errors into my calculation. Now I can actually evaluate this using my calculator. And it comes out to be forty-eight point seven percent to three significant figures.

So in this question we’ve used the volume of a sphere to calculate the volume of the tennis balls. We’ve used the volume of a cylinder to work out the volume of the can and then we’ve combined them into a percentage, in order to answer the question that was being asked.

Okay let’s look at the second question. It says “A chocolate sweet has a truffle centre which is a sphere of radius twelve milliammetres. It is dipped in white chocolate so that the coating is two milliammetres thick all around. Find the volume of white chocolate used in the coating.”

So as usual a diagram would be helpful here so we can visualize the situation. So we start off with a spherical centre of radius twelve milliammetres and then we’re adding this coating around the outside. So we have an extra layer of white chocolate, where the thickness is two milliammetres all the way around. And we want the volume of white chocolate used, so we want to calculate this volume here. Now it looks two-dimensional, but remember is of course a three-dimensional sphere that we’re interested in.

So we’re going to need the formula for the volume of a sphere first of all. So we’ll recall that formula; it’s the volume is equal to four-thirds times 𝜋 times the radius of the sphere cubed. Now let’s think about how we’re gonna calculate this volume because the volume we’re interested in isn’t actually a sphere itself.

It says coating; that’s going around the outside. But what we can do is we can work out the volume of the spherical centre. We can then work out the volume of the overall sweet when it’s been coated in chocolate. Both of those are spheres. And then we can look at the difference between those two values in order to work out how much the volume of the chocolate is that’s been added.

So let’s start off with the volume of the centre. So the centre is a sphere and it has a radius of twelve. So we can substitute directly into the formula that we’ve got. And we will have that the volume of the centre is equal to four-thirds times 𝜋 times twelve cubed. And so if I just evaluate the constant — the four-thirds times twelve cubed — I get two thousand, three hundred and four. So my overall answer for the volume of the centre of this sweet two thousand, three hundred and four 𝜋 milliammetres cubed. And I’ll keep it in terms of 𝜋 for now.

Next we want to work out the total volume of this sweet once it’s been covered in this white chocolate coating. So the total volume, well it’s still a sphere, but the radius this time is going to be the sum of that twelve milliammetres and the two milliammetres.

So we have a radius for this larger sphere of fourteen milliammetres. Now we can substitute directly into the volume formula again. So we will have that the total volume is four-thirds multiplied by 𝜋 multiplied by fourteen cubed and again will evaluate the constant. and I’ll leave it as an exact fraction for now ten thousand, nine hundred and seventy-six over three 𝜋. So there is our volume for the total of this sweet.

Final step then to calculate the volume of the white chocolate is we need to look at the difference between these two values. So we need to subtract the volume of the centre from the total volume. So we have ten thousand, nine hundred and seventy-six 𝜋 over three subtract two thousand, three hundred and four 𝜋. And we can evaluate that on a calculator, which will give an answer of four thousand, two hundred and fifty-five point eight one something. So if I round that value to perhaps the nearest integer this time, then we will have an answer of four thousand, two hundred and fifty-six milliammetres cubed to the nearest integer. So we have been able to work out the volume of the white chocolate by doing the volume of a larger sphere and then subtracting the volume of the smaller sphere.

Okay let’s look at our final question. So it says a circus plinth is in the shape of a frustum and this is a particular type of shape that is formed when you take a cone and then you chop off the top of it. So you start with the large cone and you chop off a smaller cone. And what you’re left with is called “a frustum”; that’s the shape at the bottom here in blue. And the question says calculate the volume of plastic used in order to make the circus plinth. So calculate the volume of this frustum shape here.

So we can use the description to help. It’s formed by cutting a small cone off the top of a larger cone. So what we need to do is work out the volume of the large cone that we started with and then subtract the volume of the small cone that we’ve cut off. So in order to do that, we’re going to need our formula for the volume of a cone. So if you recall that formula, it said that the volume is equal to one-third 𝜋𝑟 squared ℎ, where 𝑟 is the radius of the cone and ℎ is the height.

So we’ve got two cones to work with: we’ve got a large cone and we’ve got a small cone. And let’s start with the large cone first of all. So the volume of the large cone is gonna be one-third times 𝜋. Now we need the radius. So we have the diameter of the large cone, which is thirty centimetres. So we need to halve that the radius of the large cone is fifteen, so 𝑟 squared fifteen squared. And then the height of the large cone is twelve. It’s the full measurement here.

So we have one-third multiplied by 𝜋 multiplied by fifteen squared multiplied by twelve for the large cone. And we just evaluate the constant. So we’re keeping our answer in terms of 𝜋 for now. So fifteen squared multiplied by twelve divided by three and we have an answer of nine hundred 𝜋 centimetres cubed for the volume of the larger cone.

Now we can do the same thing. We can calculate the volume of the smaller cone. So for the small cone, we have one-third multiplied by 𝜋; now the radius of the smaller cone, well we can see that the diameter is ten. So therefore the radius is half of that; it’s five. And the height of the small cone is four centimetres. So we have one-third times 𝜋 times five squared times four.

And again if we just evaluate the constant for now. We have one hundred over three 𝜋 centimetres cubed for the volume of the smaller cone. Final step then is to find the difference between these two because this will give us the remaining volume of the plastic. So the volume of this plinth is nine hundred 𝜋 and then we need to subtract one hundred 𝜋 over three. And if I evaluate this using a calculator, I get a value of two thousand and seven hundred and twenty-two point seven one three and so on. And if I round this perhaps to the nearest integer in this case, then I have a value of two thousand, seven hundred and twenty-three centimetres cubed to the nearest integer. So there we have an example of using the volume of two separate cones to work out the volume of a frustum shape.

Now this may be presented slightly differently. We were given the height of each of the cones: the four centimetres and the twelve centimetres. But what may happen is that you’re given perhaps the height of the smaller cone, so the four centimetres. And you’re given instead the height of this part here, so you’re told that this is eight centimetres. And you would just need to combine those two together in order to give the total height of the large cone.

So you just need to think carefully if you’re given the information in a slightly different way, you just need to calculate the overall height. Today you have three examples of real-life problems that require you to calculate the volumes of cylinders, cones, and spheres.