In this video, we’re going to look at how to apply your knowledge of the volume and surface area of cylinders to some more complex problems. First of all, a quick reminder of the formulae that we’re going to need. So we have here a diagram of a cylinder and we have its two key dimensions marked on it. So we have the height of the cylinder represented by ℎ and then the radius of the base of the cylinder represented by 𝑟.
So the volume formula first of all, the formula for calculating the volume is this: 𝜋𝑟 squared multiplied by ℎ. 𝜋𝑟 squared remember gives the area of the base of this cylinder and then multiply it by ℎ, the height. Next the formula for the surface area of the cylinder, and there are two types of surface area that we need to consider. The first is the lateral surface area, which is just the curved surface, and this is given by two 𝜋𝑟ℎ. Remember that came about because the curved surface if you were to flatten it out is actually a rectangle with one dimension equal to ℎ and the other dimension two 𝜋𝑟, which is the circumference of the circular base. The other formula we need is for the total surface area of the cylinder, so the curved part and the top and the base. And so this formula has a two 𝜋𝑟ℎ as before, but it also has a contribution of two 𝜋𝑟 squared as I mentioned for the circle on the top and the circle on the base of the cylinder. So those the formulae that we’re going to need and now we will see how to apply them to a couple of different questions.
So here is our first question. It says a cylinder has a height of five centimetres and its lateral surface area is thirty-two 𝜋 centimetre squared. We are asked to calculate the volume of this cylinder. So in order to calculate the volume, we’re going to need that volume formula, which is that the volume is equal to 𝜋𝑟 squared ℎ. But we haven’t actually been given the radius in the information in the question. We’ve been given something else instead; we’ve been given the lateral surface area. So this question is going to involve using that known surface area to calculate the radius so that we can then use that in our calculation of the volume. So let’s recall what the formula for the lateral surface area was, and it was this that is equal to two 𝜋𝑟ℎ.
So what we can do is use the information that we know to form an equation. So two 𝜋𝑟ℎ, that is two multiplied by 𝜋 multiplied by 𝑟, which we don’t know, multiplied by ℎ, which from the information given in the question is five centimetres. And we’re also told in the question that this is equal to a lateral surface area of thirty-two 𝜋. So I’ve used the information in the question to set up this equation here. Now I can simplify this equation slightly. So on the left-hand side, I have two times five, which is ten. So I’ve got ten 𝜋𝑟. So I can write it out without those multiplication signs. So I want to go ahead and solve this equation to work out the value of 𝑟. So I have a factor of 𝜋 on both sides of the equation; I can divide three by 𝜋, which will cancel those out. And so now what I have is ten 𝑟 is equal to thirty-two. Well in order to work out the radius, I need to divide by ten. And so this gives me the 𝑟 is equal to three point two, and of course it’s three point two centimetres because those are the units in the question.
Now I have all the information that I need in order to calculate the volume of this cylinder. So I can return to my volume formula. And now I know that 𝑟 is three point two and ℎ is five, so I can substitute both of those values into this formula. So I have volume equals 𝜋 multiplied by three point two squared multiplied by five; this gives me an answer in terms of 𝜋 of two hundred and fifty-six 𝜋 over five. And I could leave my answer at this stage here if I didn’t have a calculator or if I was asked to leave my answer in terms of 𝜋. But if I evaluate this as a decimal, then this gives me an answer of one hundred and sixty point eight centimetres cubed for the volume, and that has been rounded to one decimal place. So in this question, we needed to record two relevant formulae, and we needed to work backwards from knowing the lateral surface area to calculate the radius so that we can then go on to calculate the volume of this cylinder.
This is the next question that we’re going to look at. The volume of a cylinder of height seven centimetres is one hundred and ninety-six centimetres cubed. We are asked to calculate to the nearest tenth the circumference of the base of the cylinder. So let’s think about the key piece of information that we’re going to need during this question. We’re told the volume of the cylinder is a hundred and ninety-six centimetres cubed; so we’re going to need to use our volume formula at some point, which remember is this formula here volume is equal to 𝜋𝑟 squared ℎ. Now we’re told that the height of the cylinder is seven centimetres, but we’re not told explicitly what the radius of the cylinder is. So we’re going to have to work backwards from knowing the volume and the height to working out the radius because we need the radius in order to calculate the circumference.
So let’s start then by setting up an equation involving the known volume and the known height and this unknown radius 𝑟. So we know that 𝜋 multiplied by 𝑟 squared multiplied by the height which is seven we know that this is equal to one hundred and ninety-six centimetres cubed. So here is our equation. Now I can simplify the left-hand side slightly; I can write it as seven 𝜋𝑟 squared. And that’s just as that you need two ways of writing it because we don’t need the multiplication signs in the algebra. Now I’m looking to solve this equation to work out the value of 𝑟. So the first step would be to divide both sides of this equation by seven 𝜋. So I will have 𝑟 squared is equal to one hundred and ninety-six over seven 𝜋. Now I could evaluate it at this point. But in order to keep it exactly I want, the next step is I need to take the square root of both sides. So I have 𝑟 is equal to the square root of a hundred and ninety-six over seven 𝜋. And if I then evaluate this using my calculator, then I’ll have 𝑟 is equal to two point nine-eight five four.
Now I’m not going to round this value of 𝑟 at this point because if I do, I could introduce the rounding error to the next stage of the calculation. So I’m just gonna keep that value on my calculator screen. Now remember the question asks us to calculate the circumference of the base of the cylinder. So I need to think back to my other work on circles and will call that the formula for calculating the circumference of a circle is either 𝜋 times 𝑑, the diameter, or two 𝜋𝑟. So I’ll use this formula with 𝑟 in as that’s what I’ve just calculated. So now I just need to substitute that value of 𝑟 into this formula. Remember I’ve kept it on my calculator. So I’m just gonna multiply the value on my calculator display by two 𝜋. And when I do that, I get a value of eighteen point seven five seven and so on. Now remember the question asked me to calculate this to the nearest tenth. So I’m going to round my answer and to the nearest tenth, it will be eighteen point eight. Remember a circumference is a length, so the units for this in this case are centimeters because those were the units of the other measurements in the question.
Right, the final question that we’re going to look at in this video, it says the volume of the cylinder shown is one thousand two hundred and seventy-eight 𝜋 over five cubic inches. We are asked to calculate the height of this cylinder. So looking at the diagram, we can see that we are given the radius of the cylinder — it’s six inches — but we’re missing this measurement of the height. So let’s think about the key formulae that we’re going to need. We’re told the volume of the cylinder. So we’re going to need our formula for the volume. And here it is; it is 𝜋𝑟 squared ℎ that we’re now quite familiar with.
So as in previous questions, we can use the information we do know to set up an equation that we can then solve to work out the information we don’t. So 𝜋𝑟 squared ℎ, that is going to be 𝜋 multiplied by six squared multiplied by ℎ and we’re told that this is equal to one thousand two hundred and seventy-eight 𝜋 all over five. So here is our equation involving that unknown height. Right, next let’s just tidy this equation up a bit. So six square is thirty-six. So on the left-hand side, I have thirty-six 𝜋ℎ and the right-hand side is exactly as it was before. Now I have a factor of 𝜋 on both sides, so dividing the equation through by 𝜋 will just cancel out those two factors of 𝜋. And now on the left-hand side, what I’ve got now is thirty-six ℎ. So if I want to know just what ℎ is, I need to divide both sides of this equation by thirty-six. That has the effect of the thirty-six appearing as another factor on the denominator of the right-hand side. So now I have an explicit expression for ℎ, so I can evaluate this. And doing so, I get that ℎ is equal to seven point one. Units remember this time are inches. So my answer to the question is that the height of the cylinder is equal to seven point one inches.
So to summarise then, we’ve looked at how you can use your knowledge of volume and surface area of a cylinder to solve some more complex problems, most of which have involved working backwards from knowing either the volume or the surface area to then calculating a missing dimension in the cylinder.