# Video: Deriving the Formula for the Volume of a Cone Without Calculus

Tim Burnham

Calculate the volume of a functionally infinite number of disks of decreasing radii, approximating a cone, to produce a formula for the volume of a cone.

14:53

### Video Transcript

To come up with the formula for the volume of a cone, you can use integration to calculate the volume of revolution of the line representing the slope side of the cone. But in this video, we’re gonna look at another way of deriving the formula by representing the cone with a series of thin discs placed on top of each other. The thinner we make the discs, the closer we’ll get to the real volume of the cone. But we’ll try just a few discs first to see how it works and then we’ll build out to a general formula.

First, we’ll represent our cone using just two discs. So first of all, we’re gonna split the cone into two heightwise. And then we’re gonna take the mid-slice of each of the discs halfway between each of those dividing lines like this. So we’re looking side-on of the cone here and we can see that the disc at the bottom isn’t quite as wide as the bottom of the cone. But then it sticks out a little bit above that. And we hope that this sort of the extra bits here sort of balance the bits that we’ve missed here. And that’s not gonna be a hundred percent accurate, but in fact the more discs that we have, the more accurate that will be.

Now we’ve split this into two discs. But in doing so, we’ve actually divided the height of the cone into four equal chunks. And because the slopy side is a straight line, we can see we’ve got some equal triangles or some similar triangles. So this triangle here is similar to this triangle here, is similar to this triangle here, is similar to this triangle here. So this triangle — the top — has a height which is a quarter of the large triangle. So the radius of the bottom — the length of the bottom side — will also be a quarter of that length. This triangle here has a height which is three quarters the size of the big triangle. So the base would also be three quarters of the size of the base of the big triangle. And the height of each of the discs is half the height of the overall cone.

So now we have all the information that we need to work out the volume of those two discs. Remember the discs are basically cylinders, so the volume of a cylinder is 𝜋 times the radius squared times the height of the cylinder. So the total volume is the volume of the bottom disk 𝜋𝑟 one squared times the height of that disc ℎ over two plus the height of the top desk 𝜋 times 𝑟 two squared times ℎ over two, the height of the disc.

So now substituting the values for radius one and radius two in terms of 𝑟, the radius of the base of the cone, we can see that the volume is 𝜋 times three quarters of 𝑟 all squared times ℎ over two plus 𝜋 times a quarter of 𝑟 all squared times ℎ over two. Now we’ve got some numbers in common. So in common- so we could factorize this. So they both got a 𝜋, they both got an 𝑟 here, they both got an ℎ over two, and in fact they both got a denominator of four squared because they both got squared too as well. So we can rearrange these a little bit. Let’s start to do some factorization.

So we factored out the 𝜋 and the ℎ and the two and I’ve just split up. So three quarters of 𝑟 all squared is three quarters squared times 𝑟 squared and a quarter of 𝑟 all squared is a quarter squared times 𝑟 squared. So now you can see that we’ve got the 𝑟 squared terms, which we can also factor out. Then of course, three quarters squared is the same as three squared over four squared and one quarter squared is the same as one squared over four squared. So now we’ve got the four squared on the denominator, which we can also factor out.

So that leaves us with 𝜋ℎ𝑟 squared all over two times four squared times the whole of three squared plus one squared. Well, four squared is sixteen and two lots of sixteen is thirty-two, three squared is nine, and one squared is one. Add those two together and you get ten, so it’s ten over thirty-two times 𝜋𝑟 squared ℎ. I’ve just swapped around the 𝑟 squared and the ℎ there because it doesn’t matter what order we multiply them together in. So the formula that we’ve come up with is ten thirty-twoths of 𝜋𝑟 squared ℎ. So that’s pretty close to a third times 𝜋𝑟 squared times ℎ. But as we said with two discs, there were quite a lot of inaccuracies. And because we had sort of spare bits up here and then we had to- we hoped that they balanced out these bits down here, but maybe they wouldn’t quite accurately do that. So let’s move on and try with three discs and see if we get a slightly more accurate picture.

Now with three discs, each disc is a third of the height of the cone. And we can see that the radius one, radius two, and radius three, we’ve had to split the cone up into six, so the cone height up into six. So we can see that the first radius is gonna be five-sixths of the radius of the 𝑟. The second radius is going to be three-sixths of the radius of 𝑟; yes I know that’s a half, but we’ll leave it at six for now. And then the third radius is going to be one-sixth of the radius of 𝑟.

So let’s put all that information together and work out the volume of those three discs added together. Well, the volume of the first disc at the bottom is 𝜋 times its radius squared, which is 𝑟 one squared, times its height, which is ℎ over three.

And we can add the volume of the second disc which has a radius 𝑟 two and the third disc which has a radius of 𝑟 three. So I’m just gonna factor out the 𝜋 and the ℎ over three before I substitute in the other values for 𝑟 one, 𝑟 two, and 𝑟 three in terms of 𝑟. And those are the values for 𝑟 one, 𝑟 two, and 𝑟 three. And then I’m just gonna square each of those individually. And when I do that, we can see that we’ve got some more common factors; they’ve all got six squared as a denominator; they’ve all got 𝑟 squared up here. So factor those out.

And we can see a bit of a pattern emerging here because what we were left with with the premier looking at the volume of two discs in the brackets over on the right-hand side there was one squared plus three squared. But now we’ve got one squared plus three squared plus five squared. And in fact, that’s a pattern that continues, so it’s gonna be the sum of the squares of the odd numbers. The first however many if we split the-the cone into three discs, then we’re gonna have the first three odd numbers. And if we split it into ten discs, it would be the first ten odd numbers, and so on. So let’s just evaluate those numbers. And three times six squared is a hundred and eight and one squared plus three squared plus five squared is thirty-five. So if we use this approximation to a cone, our formula for the volume would be thirty-five over a hundred and eight times 𝜋𝑟 squared ℎ. And thirty-five over a hundred and eight is- well, it’s a little bit closer than the last one to a third.

So let’s try five. So for five discs, each of their heights is gonna be a fifth of the height of the cone. And their radiuses are gonna be one-tenth, three-tenths, five-tenths, seven-tenths, and nine-tenths of the radius of the base of the original cone. So let’s add together the volume of each of those discs. So remember the volume of a cylinder is 𝜋 times its radius squared times its height. So we’ve put all those down here; let’s do some factoring. And then let’s substitute in our values for 𝑟 one, 𝑟 two, 𝑟 three, 𝑟 four, and 𝑟 five in terms of the original radius of the base of the cone. And then we can sort of split those up individually. And we can factor out the ten squared and the 𝑟 squared from each of these expressions. And then we can evaluate those sets of numbers. So that gives us a hundred and sixty-five 𝜋𝑟 squared ℎ over five hundred. So a hundred and sixty-five over five hundred simplifies down to thirty-three over a hundred.

So it looks like the more discs that we use, the closer this coefficient here is getting to a third. But what we’re gonna do is a little bit of mathematical magic and we’re going to sort of do this not for a specific number of discs, but for 𝑛 discs. And we’re gonna try come up with a general formula. And then this sort of the magic trick at the end is to say what if 𝑛 was infinitely big. What would that do to our formula? So here we’ve split the cone up into 𝑛 discs. Obviously, I can’t draw them all because I don’t know what 𝑛 is. But we can see that each of the discs has a height, which is one 𝑛th of the height of the total cone. So that’s ℎ over 𝑛. So now we need to look at working out the radii of each of these discs.

So when we split this into two discs, in order to work out the denominator for working out the radii of each individual disc, we had to double the number of discs because each disc was gonna split into a top half and a bottom half. And the radius came halfway up that disk. So the first radius there was three quarters of the big radius — the base radius of the cone. And the second radius was only one quarter of that. So when we had three discs, we needed to double three to make six to get the denominator for the radii of the discs. So what we can see is if we double the number of discs we’ve got, that gives us the denominator for our radius. So doubling 𝑛 gives us two 𝑛 for that denominator.

And to work out the numerators, it was just the denominator minus one for the first one, the denominator minus three for the second one, and the denominator minus five for the next one. So we can do the same again for 𝑛 disc situation. So two 𝑛 take away one is gonna be the first numerator, we gonna have two 𝑛 minus three over two 𝑛𝑟 for our second radius, and then so on, and so on, and so on. And eventually, we’ll work down to the last two discs at the top. They will have radii of three over two 𝑛𝑟 and one over two 𝑛𝑟.

So let’s work out the volume of those discs. Then now we’ve got a lot of information. It’s 𝜋 times the first radius squared times the height of that disc which is ℎ over 𝑛 plus 𝜋 times the radius of the second disc squared times its height which is ℎ over 𝑛, and so on, and so on for all of the discs. Okay, let’s factor out the 𝜋 and the ℎ and the 𝑛, which gives us this — much the same as we saw before a similar sort of pattern. And now let’s substitute in the values that we’ve got for 𝑟 one, 𝑟 two, 𝑟 three, and so on in terms of just 𝑟 the base of the- the base radius of the cone. So this time, I’ve actually squared all the individual terms out. So we’ve got two 𝑛 minus one squared times 𝑟 squared all over two 𝑛 all squared, and so on, and so on. So now I can factor out the common denominators that I’ve got here of two 𝑛 all squared and the 𝑟 squareds that are in each of those terms.

So this is what we end up with. And the contents of these parentheses over here, don’t be put off by those horrible two 𝑛 minus one all squared, and so on. This is just one squared plus three squared plus five squared plus seven squared plus nine squared if we saw it from the other end and so on up to the first 𝑛 odd numbers all squared. And it turns out there’s a nice formula for adding all of those up. Now we’re not gonna go into how that’s derived in this video. But basically that whole set of parentheses there can be replaced by this formula 𝑛 lots of two 𝑛 minus one lots of two 𝑛 plus one all over three.

Right, so now I’ve got this. I can cancel out these two 𝑛s: divide the top by 𝑛, divide the bottom by 𝑛, so they’ll cancel out. And now I’m going to multiply out these brackets over here. So two 𝑛 times two 𝑛 is four 𝑛 squared, two 𝑛 times positive one is positive two 𝑛, negative one times two 𝑛 is negative two 𝑛, so when I add those two together, they’re gonna cancel out, and then negative one times one is one. So this pair of parentheses here multiplied out equals this over here, four 𝑛 squared minus one. If I expand up two 𝑛 times two 𝑛, that’s four 𝑛 squared. So that’s where this bit came from up here and then I’ve just swapped around the ℎ and the 𝑟 squared because it doesn’t matter what order we multiply those two together in. So I’ve got this formula right over here. Now what I can do now is multiply those together. I’m gonna multiply the denominators together first.

So going back up to the top of the page, I’ve got that whole thing over twelve 𝑛 squared now. So what this means is I’ve got 𝜋𝑟 squared ℎ times four 𝑛 squared; I could separate these out into two separate fractions. And 𝜋𝑟 squared ℎ times negative one. So let’s do that. And we can simplify each of those terms a little bit. So let’s look at the left-hand term. First, I’ve got an 𝑛 squared on the top and an 𝑛 squared on the bottom that will cancel and I’ve got four on the top; if I divide that by four, I get one and then twelve on the bottom; if I divide that by four, I get three. And the second term is times negative one, but it’s negative. So if we’re taking away the negative of that expression, then we’re adding that expression.

So this is what we’ve ended up. This is our grand formula for the volume of a cone: 𝜋𝑟 squared ℎ divided by three or a third 𝜋𝑟 squared ℎ plus 𝜋𝑟 squared ℎ over twelve 𝑛 squared. Now I think what we’d agreed earlier was that the more discs that we put on this stack — that the more that we break the cone down into — the more accurate our answer would be. So really we want 𝑛 to be a thousand or a million or a billion or a trillion or even better still infinity. And when we do that, what we’ll find is this second time here will have basically a very, very, very, very large number; it’s timesed by itself. So that’s an incredibly large number on the denominator of that fraction. So it doesn’t really matter what the radius or what the height are and we’re gonna be dividing by an infinite number. So this term here, so it’s gonna become pretty much close to zero or as close to zero as we can get. So as long as we agree that we’re gonna split this down into as many discs as possible to make this as accurate as possible, this is the formula we’ve come up with for our volume of a cone: a third 𝜋𝑟 squared ℎ.

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