Trigonometry

This can be derived easily doing an integration. Imagine slicing the cone into lots of thin, horizontal slices. The volume of each can be approximated as the area of the corresponding circle, times the thickness.

Once you learn calculus, you'll be able to solve this quickly.

If you don't know calculus yet, you may want to try slicing up a certain cone, say, into 100 thin slices of equal height. Do the calculations (an Excel spreadsheet can help), and you'll get close to the 1/3 etc.

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As the community answer above says, knowing calculus this is very easy; however, I suspect you do not know calculus. We can still find the formula using fairly hard maths (without using calculus directly), but don't let this put you off - I will try to make it as simple as possible but you will need to be able to understand algebra.

Algebra is the next level up from Arithmetic where numbers are replaced by letters and where the real language of maths is developed, creating formulae where the letters are used to stand for numbers. The formula for the volume of a cone (1/3 πr²h) is algebra.

Before I attempt the formula for a cone, I'll start with a similar problem but which is simpler which will demonstrate the techniques used for finding the formula of a cone.

Suppose you have lots of unit cubes (that is a cube where every edge is 1 unit long); you can build a pyramid where every layer is a square (made up of the cubes):

• layer 1 (the top layer) has 1 cube
• layer 2 (on which layer 1 sits) is a 2 by 2 square with 2² = 4 cubes
• layer 3 (on which layer 2 sits) is a 3 by 3 square with 3² = 9 cubes

The volume is simply the sum of the number of cubes:

• 1 layer: volume = 1² = 1
• 2 layers: volume = 1² + 2² = 1 + 4 = 5
• 3 layers: volume = 1² + 2² + 3³ = 1 + 4 + 9 = 14
• 4 layers: volume = 1² + 2² + 3² + 4² = 1 + 4 + 9 + 16 = 30
• n layers: volume = 1² + 2² + 3² + … + n² = ???

To find the volume for n layers we would need a formula; so can we find one?

Given any sequence of numbers it is possible to find a formula so that when the number of a term is put in to it the value of the term comes out. (When given a sequence to find the next number, you can create a formula using the numbers given to give any next number you want.)

Looking at the values above, they can be written as ordered pairs; (1, 1), (2, 5), (3, 14), (4, 30). These look like points on a graph, so we are looking for a formula of the form "volume = an³ + bn² + cn + d" (Isaac Newton back in the 17th century did lots of investigation into this). So let's find the formula:

We can create four simultaneous equations relating a, b, c, and d by substituting the values of n and the volumes of those pyramids, and solve them:

1. a×1³ + b×1² + c×1 + d = 1 → a + b + c + d = 1
2. a×2³ + b×2² + c×2 + d = 5 → 8a + 4b + 2c + d = 5
3. a×3³ + b×3² + c×3 + d = 14 → 27a + 9b + 3c + d = 14
4. a×4³ + b×4² + c×4 + d = 30 → 64a + 16b + 4c + d = 30
5. Subtract (1) from (2): 8a - a + 4b - b + 2c - c + d - d = 5 - 1 → 7a + 3b + c = 4
6. Subtract (1) from (3): 26a + 8b + 2c = 13
7. Subtract (1) from (4): 63a + 15b + 3c = 29
8. Subtract 2×(5) from (6): 12a + 2b = 5
9. Subtract 3×(5) from (7): 42a + 6b = 17
10. Subtract 3×(8) from (9): 6a = 2

Substitute a in (8): 12×1/3 + 2b = 5 → 2b = 1 → b = 1/3

Substitute a and b in (5): 7×1/3 + 3×1/2 + c = 4 → c = 4 - (14/6 + 9/6) = 24/6 - 23/6 = 1/6

Substitute a, b and c in (1): 1/3 + 1/2 + 1/6 + d = 1 → d = 1 - (2/6 + 3/6 + 1/6) = 1 - 1 = 0

Thus the formula is:

volume = 1/3 × n³ + 1/2 × n² + 1/6 × n

(This can be rearranged into 1/6 × n × (2n + 1) × (n + 1) which is the form most often given as the sum of the first n squares, which is what this is.)

That was your first bit of mathematical workout. Now comes the next bit - proving this formula actually works for all values of n and not just the 4 we calculated).

This we'll do using "proof by induction" which is just a fancy way of saying if it works for some value r then it also works for the value r + 1 from some starting value of r. We know it works for r = 1 since 1/3 × 1³ + 1/2 × 1² + 1/6 × 1 = 1/3 + 1/2 + 1/6 = 1 as required.

So let's calculate the volume for two pyramids: one with r layers and one with an extra layer of (r+1)² cubes (the next bigger pyramid):

• Vr = 1/3 r³ + 1/2 r² + 1/6 r
• Vr+1 = 1/3 (r+1)³ + 1/2 (r+1)² + 1/6 (r+1)

Vr+1 = 1/3 (r³ + 3r² + 3r + 1) + 1/2 (r² + 2r + 1) + 1/6 (r + 1)

Vr+1 = 1/3 r³ + r² + r + 1/3 + 1/2 r² + r + 1/2 + 1/6 r + 1/6

And rearranging the terms to bring like terms together gives:

Vr+1 = 1/3 r³ + 1/2 r² + r² + 1/6 r + 2r + 1/6 + 1/3 + 1/2

And a bit more rearranging gives:

Vr+1 = (1/3 r³ + 1/2 r² + 1/6 r) + (r² + 2r + 1)

But the first three terms are the value of Vr, so that can be put back in, and the last three terms can be factorised to give:

• Vr+1 = Vr + (r+1)²

Which says that the volume created when adding a layer of (r+1)² cubes is the same as the volume of the original plus the volume of that layer. So the formula works.

So how does this help us to find the formula for the volume of a cone?

Just as the volume of the pyramid was the sum of its layers, if we split the cone into layers parallel to the base of the cone (all of the same height - just like the baby/toddler toy that consists of different sized discs on a spindle), the volume of the cone is the sum of the layers. In the pyramid each layers was a square, but here each layer is a disc, or (little) cylinder for which the formula for the volume is known (volume_cylinder = πr²h).

Suppose we split the cone into n discs. We can easily calculate the height (H) or each disc as it is simply the height (h) of the cone divided by n, ie H = h/n. But what about the radius (R) of each disc?

If we cut the cone in half by slicing it perpendicular to the base and through the apex of the cone we see that it create an isosceles triangle with height the same as that of the cone (h) and base the same as the diameter of the base of the cone (2r).

Next, considering only the triangle if we draw in the height (h) from the base to the top it creates two right angle triangles, each with a base the radius (r) of the cone.

Considering one of these two right angled triangles, we can see that the width is proportional to the height (from the top): at the top the radius is 0, half way down (at height h/2) the radius is half that of the base (r/2), three quarters of the way down, (at height 3h/4) the radius is three quarters of the base (3r/4), etc.

As there are n discs, we want the values at the heights from the top of h/n, 2h/n, 3h/n, …, hn/n (this last value is the same as h) which gives us R = r/n, 2r/n, 3r/n, …, r

The volume of each disc is πR²H, and their total volume is:

volume = π(r/n)²(h/n) + π(2r/n)²(h/n) + π(3r/n)²(h/n) + … + πr²(h/n)

→ volume = π(r/n)²(h/n)(1² + 2² + 3³ + … + n²)

What's the sum of 1¹ + 2² + 3² + … + n²? We found this earlier when we found the volume of the pyramid. We can use it again, and this volume becomes:

volume = π(r/h)²(h/n)(1/3 × n³ + 1/2 × n² + 1/6 × n)

→ volume = π × r²/n² × h/n × (1/3 × n³ + 1/2 × n² + 1/6 × n)

→ volume = πr²h × (1/n³) × (1/3 × n³ + 1/2 × n² + 1/6 × n)

→ volume = πr²h(1/3 + 1/(2n) + 1/(6n²))

If we use only 1 disc, we have a cylinder that has the radius of the cone and its volume is very much bigger than that of the cone. If we split it into two discs, each disc (cylinder) has a height that is half the height of the cone, but the top disc has a radius of half that of the cone, so the volume of the two discs is less than before, but still greater than the volume of the cone. The more discs we use, the smaller their total volume becomes however it is always greater than that of the cone, but it also does get closer and closer.

As the number of discs (n) gets larger and larger, 1/(2n) and 1/(6n²) get smaller and smaller, and have less and less effect on the (1/3 + 1/(2n) + 1/(6n²)) part of the formula. For large values of n (and a desired accuracy), we could ignore 1/(2n) and 1/(6n²).

What is the largest possible value of n?

This is the hardest maths in finding the formula for the volume of the cone (and is a basis of calculus): it is the concept of limits.

If you draw the graph of y = 1/x you will notice that as x gets larger and larger, the value of y gets smaller and smaller heading towards the x-axis (y=0) but never quite reaching it. What is said in this case is that as x tends towards ∞, y tends towards 0; or it can be expressed as the limit as x tends to infinity of 1/x is 0.

So as the number of discs gets larger and larger, n is tending towards infinity, the values of 1/(2n) and 1/(6n²) both tend towards 0. Thus if there were an infinite number of discs, the values of 1/(2n) and (1/6n²) would both be zero, the sum of the volumes of the discs would equal the volume of the cone, and the formula becomes:

volume = πr²h(1/3 + 1/(2n) + 1/(6n²))

→ volume = πr²h(1/3 + 0 + 0)

→ volume = πr²h(1/3)

→ volume = 1/3 πr²h

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If we were to use calculus, we would consider the graph of y = (r/h)x between x = 0 and x = h, and rotating this about the x-axis would create a cone and its volume is given by:

volume = π ∫ y² dx (between x = 0 and x = h)

→ volume = π ∫ ((r/h)x)² dx

→ volume = π ∫ (r/h)² x² dx

→ volume = π (r²/h²) ∫ x² dx

→ volume = π (r²/h²) × 1/3 x³ {between x = 0 and x = h}

→ volume = π (r²/h²) × 1/3 × (h³ - 0³)

→ volume = 1/3 π r²/h² × h³

→ volume = 1/3 π r²h

The derivation of the formula for the volume of rotation comes from the work shown above in the specific case of the cone, but for a general curve y = f(x).