So I understand that the volume formula of a cone is: $\frac{1}{3}\pi r^2h$, and when I read about how to derive this formula, it makes sense to me. Funny thing is, I'm stuck on why it ISN'T $\pi r^2h$ when I think about deriving the volume formula in a different way.

Here's what I mean. Now we're told that to figure out the volume formula for something like a cylinder or cube, we simply multiply the area of the base by the height, and this makes intuitive sense. In the case of a cone, what if we took a triangle with base r and height h, and rotated it around some axis, producing a cone. When I think of it this way, it seems reasonable (in my mind) to calculate the volume of the produced cone by taking the area of the triangle, and multiplying it by the circumference of the circle that is the cone's base. So if the area of our triangle is: $\frac12rh$ and the circumference of the generated cone's circle-base is: $2\pi r$, then the volume should be: $\pi r^2h$. This derivation seems intuitively correct to me.

Now I know it isn't correct, and I almost feel silly asking, but I just can't see WHY it's not correct. What's the flaw in the reasoning that the volume of a cone can be derived by multiplying the area of the triangle by the circumference of its circe-base?

  • 1
    $\begingroup$ You are not alone. This is a classic fallacy. Related: to produce an uniform random point in a unit circle, you can pick a (uniform) random radius $r\in[0,1]$ and a (uniform) random argument $\theta\in[0,2\pi)$ and that is the coordinate of the point. Right or wrong? $\endgroup$
    – Gina
    Jul 9, 2014 at 2:29
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    $\begingroup$ Well, a circle is just a line segment of length $r$ rotated $2\pi$ around a point ... so the area of a circle should be $2\pi r$, right? It's not like the infinitessimal circular sectors we're integrating are all lopsided to the extent that rectangles are a poor approximation for them... $\endgroup$
    – anon
    Jul 9, 2014 at 3:46
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    $\begingroup$ Only one vertex of the triangle travels a distance of $2\pi r$. The other two travel $0$ distance. Surely it's more reasonable to take the average, $(2\pi r+0+0)/3$, which gives the volume to be... $\endgroup$
    – user856
    Jul 9, 2014 at 4:45

2 Answers 2


What you need to look at is the second theorem of Pappus. When you rotate things in a circle, you have to account for the difference in the distance traveled by the point furthest from the axis (the bottom corner of our triangle) which goes the full $2\pi$ around. However, the vertical side doesn't actually move, so it doesn't get the full $2\pi$.

Pappus' theorem says you can use the radius of the centroid as your revolution radius, and it just so happens the the centroid of this triangle is a distance $r/3$ from the axis.

Another good read is this.

  • $\begingroup$ Can you please help me to understand how second Pappus's centroid theorem can be applied here if it requires the plane to be rotated about an external axis? In case of cone the axis is not external. $\endgroup$ Nov 14, 2021 at 17:47
  • $\begingroup$ @DamirTenishev For the purpose of the theorem, this axis is external. What you need to avoid is an axis that passes through the interior of the figure, because then you will have parts of the figure moving "opposite" from the centroid and sweeping the same region of space as other parts of the figure. $\endgroup$
    – David K
    Nov 3, 2023 at 15:52
  • $\begingroup$ @DavidK, but if I got this correctly, this is exactly what we get with the cone: "an axis that passes through the interior of the figure". Isn't it? $\endgroup$ Nov 3, 2023 at 20:24
  • $\begingroup$ @DamirTenishev The triangle has base $r$ and height $h$, so I figure it is a right triangle with one leg on the axis and the other leg along a radius of the base circle. So the axis never goes into the interior of the triangle. If instead you take a cross-section of the cone through the axis, you would get an isosceles triangle with base $2r$ and height $h$, in which case the axis would indeed pass through the interior of the triangle. $\endgroup$
    – David K
    Nov 3, 2023 at 21:57

The volume of a prism is indeed base area times height. So what you are calculating with your $\frac{1}{2}hr \times 2\pi r$ is the volume of a prism with triangular base and height $2\pi r$. This is a very different beast from your cone! If you tried to do the same for a cylinder, then you'd have a rectangle of area $rh$ rotated around a circumference of $2\pi r$ and you'd get a volume of $2 \pi r^2 h$, which is twice the size it ought to be.

But that's probably not convincing enough.

So let's think about how you find the volume of a curved object. One way is to slice it up into very very thin slices. Each of these slices can be approximated by a very thin prism, with some area and a very tiny height. Adding up the volumes of all these tiny slices gives you an approximation to the full volume. The more slices you cut it into, the more accurate your approximation.

The idea here is that your very tiny heights are the full height $h$ divided by $n$, the total number of slices. If the base area of slice $i$ is $A_i$ then the volume of slice $i$ is $A_i \times \frac{h}{n}$ and the total volume is the sum of all of these. If all the $A_i$'s are the same (say equal to $A$), then you're just adding up $n$ lots of $A \times \frac{h}{n}$ which is $Ah$, so the idea works perfectly when the shape really is a prism. If all the $A_i$'s are not the same, then it won't just simply be $Ah$. Also, if you can find a formula for the base area at any given height, then doing the sum with infinitely many slices leads to an integral.

Now let's try with your idea. Essentially, you have cut your cone into a lot of tiny slices around the centre by first cutting the circumference of the circle into $n$ tiny pieces. You have multiplied the area of the slice ($\frac{1}{2}rh$) by this tiny thickness ($2\pi r / n$) and added them all up to get the volume.

HOWEVER, to multiply an area by a thickness is to assume that the slice is a prism. This is fine, but if it really was a prism, then all your slices would overlap considerably at the centre. If you add up all the volumes, then you'll include any part that overlaps more than once and the volume will be too big. With my original description of slices, the slices didn't overlap, so I could just add them up to get the total volume.

The other way to deal with it would be to declare that they are not prisms after all, but little thin wedges, but then you have to know the volume of a wedge beforehand, which is where the $\frac{1}{3}$ comes from if you do it this way.


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