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Video: Volume of Cylinders, Cones and Spheres

Lauren McNaughten

Learn about the formulae for calculating volumes of cylinders, cones, and spheres. We also apply them to compound shapes made up of more than one of the three shapes, or parts of them, such as hemispheres.

17:58

Video Transcript

In this video, we’re going to look at the formulae for calculating the volume of three different types of 3D shapes, so looking at cylinders, cones, and spheres, and then we’ll look at some questions where we need to apply these skills in order to calculate some volumes.

So let’s start off by looking at the formula for calculating the volume of a cylinder first of all. Now a cylinder is defined by two particular dimensions; it’s defined by the radius of the cylinder, so this dimension here, and it’s defined by the height of the cylinder, this dimension here. So we have ℎ and we have 𝑟.

Now a cylinder is really just a special type of prism; it’s a 3D shape which has a constant cross section, so we can calculate its volume in exactly the same way as we do for any prism.

And if you recall the way we calculate the volume of a prism is we first calculate the area of its cross section. So for the cylinder, its cross section will be this circle shape here.

And then we multiply the area of that cross section by the depth of the prism. So in that- this case, that’s going to be the height. So the formula for the volume of a cylinder, the volume is equal to the area of the cross section; well it’s a circle, so the area of a circle is 𝜋𝑟 squared, and then we’re gonna multiply it by the depth of the prism or the height which is ℎ, so the volume of a cylinder is equal to 𝜋𝑟 squared ℎ.

And there we have the first of the three formulae that we’re going to use in this video. Now let’s look at the formula for the volume of a cone. Now again a cone has two key measurements that we’re interested in. Again, it has the radius of the circle on the base of that cone and then it has the vertical height of the cone here. And that height is perpendicular to the radius, so those two lengths meet at a right angle.

Now the best way to understand the formula for the volume of a cone is, if you understand that it’s actually just one-third of the cylinder that surrounds it, so if I sketch in a cylinder around this cone. So if you imagine it like that, then the volume of the cone is just one-third of the total volume of that cylinder.

So we’ve seen already that the volume of the cylinder would be 𝜋𝑟 squared ℎ. Therefore, the volume of the cone is just one-third of that, so its one-third 𝜋𝑟 squared ℎ. And there we have the second of the three formulae that we’re going to be using.

Finally, let’s look at the volume of a sphere. Now a sphere just has one particular measurement that we’re interested in, and it is the radius of the sphere, which again we’ll denote by 𝑟. And the formula for calculating the volume of a sphere, so this one is four-thirds multiplied by 𝜋 multiplied by 𝑟 cubed, so volume is equal to four-thirds 𝜋𝑟 cubed.

Now just something to note about this one, when we calculate the circumference of a circle, so in one dimension we’re calculating a length, the formula for that is 2𝜋𝑟, so it involves 𝑟 with a power of one. When we calculate the area of a two-dimensional circle, we use the formula 𝜋𝑟 squared, so it’s got an 𝑟 squared component to it. And then when we’re calculating the volume, we now have this 𝑟 cubed component because we’re calculating something three-dimensional. So that’s one way to remember that it involves 𝑟 cubed.

So there we have three formulae that we’re interested in. Obviously, they all involve a 𝜋 because they’re all linked to circles. Now you need to know these three formulae to learn them off by heart. We’ve talked a little bit about where they come from there, so you can remember some of the logic behind them, but you do need to learn these three formulae. So now we’ve got the three formulae; we’ll look at some questions on how we can apply these.

So here’s our first question. It says a cone has a base radius of five centimeters and a slant height of thirteen centimeters. Calculate its volume. Now we need to make sure we read the information in the question very carefully and we understand exactly what is meant by this phrase slant height here.

So let’s start off with a little sketch of a cone. So here we are, and now let’s put the information in the question onto this diagram. So we’re told the cone has a base radius of five centimeters. So this measurement here is five, and we’re told it has a slant height of thirteen.

Now the slant height it’s not the same as the perpendicular height of the cone; the slant height is actually the diagonal height of the cone. So the slant height of thirteen is actually this measurement here. Now let’s recall our formula for the volume of a cone.

And we had that the volume was one-third 𝜋𝑟 squared ℎ, so 𝑟 was the base radius; that’s five. But ℎ is the vertical height of the cone. So ℎ is this measurement here that I’ve marked in purple. Now we don’t know that, but we can calculate it. And the way to calculate it is to understand that the relationship between these three lengths between the radius, the perpendicular height, and the slant height is that they form a right-angled triangle.

So if I sketch that triangle out, here it is, we’ve got the length of five on the base of the triangle, the ℎ is it’s vertical height, and then thirteen, the hypotenuse of that right-angled triangle. So we don’t know ℎ immediately, but if we recall the Pythagorean theorem, we can use that in order to calculate the value of ℎ.

So the Pythagorean theorem remember tells us that the sum of the squares of the two shorter sides in a right-angled triangle is equal to the square of the hypotenuse. So that tells me that this triangle ℎ squared plus five squared is equal to thirteen squared.

Now I can just work through the remainder of this method using the Pythagorean theorem to work out the value of ℎ. So if ℎ squared plus twenty-five is equal to a hundred and sixty-nine, subtracting twenty-five from both sides, ℎ squared is equal to a hundred and forty-four. Square rooting both sides, ℎ is equal to the square root of a hundred and forty-four.

And therefore ℎ is equal to twelve centimeters. Now you may actually have been able to spot that if you’re familiar with Pythagorean triples, you may recall that five, twelve, thirteen is in fact the Pythagorean triple, and you could perhaps have got to the value of ℎ without going through the formal method using the Pythagorean theorem.

In any case, we now know that ℎ is equal to twelve centimeters. So to work out the volume, we just need to substitute into the formula. So the formula was volume is equal to one-third 𝜋𝑟 squared ℎ.

So if I substitute 𝑟 is equal to five and ℎ is equal to twelve, I get the volume is one-third times 𝜋 times five squared times twelve. So if I first of all work out the constant involved here, five squared multiplied by twelve and divided by three, I have a hundred, so the volume is equal to a hundred 𝜋, that would be an exact answer in terms of 𝜋, or I can go further and I can then evaluate that using my calculator.

And it will tell me that the volume is equal to three hundred and fourteen point one five nine something. And if I made that a three significant figure values, so if I rounded to three significant figures, I will have a volume of this cone is equal to three hundred and fourteen centimeters cubed because I’m calculating a volume, so my units need to be cubic units.

So key points about this question, we had to sketch a diagram first to make sure we could visualize the situation. We weren’t given the perpendicular height, so we had to use the Pythagorean theorem in order to calculate it from knowing the slant height and the radius of the cone, so an extra step in this question that you may not always see. You just need to make sure that you read the question very carefully and make sure you’re clear up front have you been given the perpendicular height, in which case you’re okay to proceed, or have you been given the slant height in which case you need an extra step in your method.

Okay let’s now look at a second example. So this question says calculate the volume of a hemisphere of diameter twenty centimeters. So there are two things that we need to watch out for in this question. First of all, it’ not a full sphere; it’s a hemisphere. So we’re only looking for half of a sphere. Secondly, our formula for the volume of a hemisphere, of a volume of a sphere, uses the radius. And we haven’t been given the radius; we’ve been given the diameter. So we need to make sure that we halve it when we’re using this value.

So a quick diagram first of all, there we have a hemisphere, the diameter, so the full length here is twenty centimeters, which means that the radius of this hemisphere must be half of that, so the radius is ten centimeters.

Let’s recall our formula for the volume of a sphere. So we had volume is equal to four-thirds 𝜋𝑟 cubed, but we have a hemisphere, so we only want half of that. So for our volume, we need to multiply through by half. So we will have the volume of this hemisphere is equal to half times four-thirds times 𝜋𝑟 cubed.

Now that will actually simplify directly to two-thirds 𝜋𝑟 cubed, and what we’ve done there is just cancel this two and this four, so leading to a two in the numerator here by dividing both of those through by two. So now we just need to substitute the values from this question, so the radius of this hemisphere is ten centimeters. So we have two-thirds times 𝜋 times ten cubed.

And again if I evaluate this constant, I will have two thousand over three lots of 𝜋, so that would again be an exact answer, or I can then evaluate it on my calculator. And this gives me an answer of two thousand ninety-four point three nine five one something something. So again if I round this value to three significant figures, I now have an answer of two thousand and ninety centimeters cubed for the volume of this hemisphere.

So key points here, read the question carefully; it’s a hemisphere not a full sphere, so we needed the factor of a half in order to account for that, and make sure you’re clear up front about whether you’ve been given the diameter or the radius.

Okay the third question, find the volume of the composite shape shown, leaving your answer in terms of 𝜋. So we have a composite shape which means it’s made up of more than one thing. So we have a cone and it’s sitting on top of a cylinder. And the question also says leave your answer in terms of 𝜋, so our final answer needs to be given as a multiple of 𝜋.

So we’ve got two different things that we need to work out here. We need to work out the volume of the cone first of all and then the volume of the cylinder that it’s sitting on top of.

So for the cone, the volume of the cone is given by one-third 𝜋𝑟 squared ℎ, and we have both of those pieces of information. We can see that the radius, it’s down the bottom of the diagram, that the radius is equal to eight centimeters and the height of the cone is this fifteen centimeters here, so we have all the information we need to work out the volume of the cone directly.

So substituting the values of 𝑟 and ℎ into the formula, we have the volume is equal to one-third times 𝜋 times eight squared times fifteen. And if I evaluate that constant so if I work out eight squared which is sixty-four and then multiply it by fifteen and divide by three, then I obtain the value three hundred and twenty 𝜋 for the volume of the cone. And I’m gonna leave my answer there because remember the question said give your answer in terms of 𝜋. So I don’t want to go any further than that in the working out.

Now let’s think about the cylinder. So the formula for the volume of our cylinder was 𝜋𝑟 squared ℎ, so essentially the same formula but without this factor of a third, and we have the radius again; it’s eight centimeters. We just need to be a little bit careful looking at the height of this cylinder because we have the overall height of the figure which is forty centimeters and we have the height of the cone at the top which is fifteen, so the height of the cylinder is going to be the difference between those two values.

So the height of our cylinder is forty subtract fifteen, which is twenty-five centimeters. So now I can just substitute the values of 𝑟 and ℎ into the formula for the volume of the cylinder, so I have the volume is equal to 𝜋 multiplied by eight squared multiplied by twenty-five.

And again if I just work out what that constant comes out as, so eight squared which is sixty-four and then multiplied by twenty-five is sixteen hundred. So it gives me an answer of sixteen hundred 𝜋, and again I’m leaving my answer like that for the cylinder because I want the overall answer to be in terms of 𝜋.

The final step will then be to add these two volumes together. So the total volume three hundred and twenty 𝜋 plus sixteen hundred 𝜋, and that will give me a final answer then of one thousand nine hundred and twenty 𝜋; unit’s centimeters cubed because I’m calculating a volume. And I’ve left my answer in terms of 𝜋 as the question asked.

Okay the final thing we’re going to look at is how to convert volume units. So we have a question, a sphere has a volume of fourteen thousand three hundred centimeters cubed and we’d like to know its volume first of all in millimeters cubed and then secondly in meters cubed.

Now your first thought may be that we need to work backwards through the volume formula to work out what the radius of the sphere is in centimeters and then convert it to millimeters and meters and then go back through again. But actually we don’t need to do that. We can just look at how we calculate volume units. So let’s think about the millimeters cubed first of all, and we need some key conversions.

So first of all, we know that one centimeter is equal to ten millimeters; that doesn’t mean though that one centimeter cubed is equal to ten millimeters cubed, because this first scale factor here, this one-to-ten ratio, this is telling us about a ratio of lengths, and what we’re interested in is the ratio of volumes. So let’s think about areas secondly.

If you want to know about what one centimeter squared is equal to in millimeters squared, well it’s not just ten millimeters squared; it’s actually ten squared so ten times ten, which is a hundred millimeters squared. And this is because if we’re thinking about an area, then we’re thinking in two dimensions, so we have two values we need to convert, which means we need to multiply by ten twice.

What that means when we come to volume then and converting volume units is because we’re thinking in three dimensions we have to multiply by ten three times. And multiplying by ten three times is equivalent to multiplying by a thousand. So it gives us this conversion factor here that one centimeter cubed is equal to one thousand millimeters cubed, and that’s a really important fact to remember.

So now I can think about answering the first part of this question. If the volume is fourteen thousand three hundred centimeters cubed, we want to convert that to millimeters cubed. So we need to take that value and we need to multiply it by a thousand.

So we’ll start with fourteen thousand three hundred and multiply that by a thousand which is our conversion factor for volume. And I get an answer of fourteen million three hundred thousand millimeters cubed. So it’s not just a case of multiplying by ten. If you’re working with volume, we have to multiply by ten three times.

So there’s our answer for part a. Part b then, we can approach in a very similar way. So I’ve just written the conversion factors out. If we’re converting from centimeters cubed to meters cubed, we need to divide by a million because that’s dividing by a hundred three times. So we have a final answer of nought point nought one four three meters cubed.