The volume of a cone with height $h$ and radius $r$ is $\frac{1}{3} \pi r^2 h$, which is exactly one third the volume of the smallest cylinder that it fits inside.

This can be proved easily by considering a cone as a solid of revolution, but I would like to know if it can be proved or at least visual demonstrated without using calculus.

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    +1. I always believed that a rigorous proof required calculus, but I'd love to be shown otherwise. – Larry Wang Jul 24 '10 at 03:41
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    The Egyptians new how to calculated pyramids. Turned out the shape didn't matter just the base area. Democritus put it together http://en.wikipedia.org/wiki/Democritus#Mathematics – Jonathan Fischoff Jul 24 '10 at 04:04
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    another twist on this: The other well-known calculus way to find the volume of the cone is to integrate over the disks that make up the cross sectional areas, as you go from the base at 0 to the top at h. The same argument (though it uses calculus) can show that, if you take an arbitrary region in the plane, and first form a "cylinder" of height h from it by extruding it a distance h, then form a "cone" from it by extruding and then tapering linearly, the volume of the resulting "cone" is 1/3 the volume of the resulting "cylinder" (loosely, making solids pointy nicely gives 1/3 the volume). – Jamie Banks Jul 24 '10 at 04:13
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    @Katie: Well it only takes calculus to formally prove it. Imagining a cylinder as an "infinite-a-gon pyramid" is really good for intuitively understanding thas – Casebash Jul 24 '10 at 06:48
  • Integrating the disk area by Simpson's rule is exact because Simpson's rule is exact for polynomials up to cubic. And the disk area for a cone or pyramid is only a quadratic. In fact, Simpson's rule works for many shapes, including spheres, cylinders laying on their side, and others. Simpson's rule is H*(B + 4*M + T)/6, with H height, B area of bottom, M area of middle, T area of top. You can forget several formulas if you can remember this one. – richard1941 Sep 08 '16 at 17:27
  • A related question is why the formula for the lateral area of a right cone is the same as the formula for the area of an ellipse. This question was asked by yours truly right here on MSE, here: http://math.stackexchange.com/questions/65052/how-related-are-ellipses-and-right-cones – Mike Jones Nov 01 '16 at 20:33
  • Like most answers here indicate a possible approach is through Cavalieri's principle. But this principle in essence is analytical in nature and requires calculus for a proof. This is so unlike in 2 dimensions where we can show without calculus that area of a triangle is half of that of a corresponding parallelogram. – Paramanand Singh Jun 25 '17 at 03:31
  • A proof was given by Euclid in the Elements ([Book XII Proposition 10](https://mathcs.clarku.edu/~djoyce/elements/bookXII/propXII10.html)). – Intelligenti pauca Jul 25 '21 at 16:25

11 Answers11


alt text
A visual demonstration for the case of a pyramid with a square base. As Grigory states, Cavalieri's principle can be used to get the formula for the volume of a cone. We just need the base of the square pyramid to have side length $ r\sqrt\pi$. Such a pyramid has volume $\frac13 \cdot h \cdot \pi \cdot r^2. $
alt text
Then the area of the base is clearly the same. The cross-sectional area at distance a from the peak is a simple matter of similar triangles: The radius of the cone's cross section will be $a/h \times r$. The side length of the square pyramid's cross section will be $\frac ah \cdot r\sqrt\pi.$
Once again, we see that the areas must be equal. So by Cavalieri's principle, the cone and square pyramid must have the same volume:$ \frac13\cdot h \cdot \pi \cdot r^2$

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Larry Wang
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    Great animation. I wonder if anyone makes any good (toy) physical models of such... – Jesse Madnick Jan 16 '13 at 06:04
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    But why are there two? – TonyK Jul 28 '13 at 17:41
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    There are two so you can see the animation in 3D. Place a piece of paper between the two cubes so each of your eyes can only see one of the cubes. If you sort of cross your eyes you can bring the two cubes together and make a stereo image. It is actually pretty cool! – Hucker Mar 07 '17 at 19:17
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    Holy molly, i never thought about doing a stereo image of 3D shape in 2D screen. This work for me without a piece in middle the 3D shape just float above my phone 0_0 life mind blow – Mr Rubix Nov 04 '21 at 11:34

One can cut a cube into 3 pyramids with square bases -- so for such pyramids the volume is indeed 1/3 hS. And then one uses Cavalieri's principle to prove that the volume of any cone is 1/3 hS.

Grigory M
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I just did a demonstration with my class that took about 2 minutes. Granted it was just inductive reasoning but it satisfied the students for now. I had 2 pairs of students come up to the front of the class. One pair had a cone and a cylinder. One pair had a pyramid and a prism. Each pair had solids with a congruent base and height. The person with the cone had to see how many times they could fill the cone with water and fit it into the cylinder. Similarly the person with the pyramid had to see how many times they could fill the pyramid with water and fit it into the prism. Other than ensuring that the cone and the pyrmaid were not overfilled (taking into consideration that the water has a curved skin at the top) the experiment was simple and the demonstration made it easier for the students to remember the relationship. Hope this helps.


Here is a derivation of the volume of a cone which does not use calculus, Cavalieri's principle, the method of exhaustion, or any other infinitesimal arguments.

[Edit There is a flaw in this argument, see below]

[Edit 2 The flaw has been fixed, by considering the ratio of the volume of a cone to its circumscribing cylinder under different scalings]

We can split a cone horizontally into two pieces, so that the upper part forms another cone with a smaller base, and the lower part is no longer a cone but an object called a 'frustum'.

                            cone split into frustum and upper cone

For a cone with base radius $r$ and height $h$, we can use a parameter $b$ with $0 \lt b \lt 1$ to define the height of the frustum as $b h$. Because the whole cone and the upper cone form similar triangles in the vertical cross section, the upper cone with height $ (1-b) h $ will have a base radius of $ (1 - b) r $.

                                        cross-section of cone

The volume of the frustum will be equal to the volume of the original cone, less the volume of the upper cone. We don't yet know what form the function representing the volume of a cone will take, so for now we will just write $V_{cone} = V_{cone}(r,h)$ to remind us that it will be some function of the height and base radius. So the volume of the frustum is $$V_{frustum} = V_{cone}(r,h) - V_{cone}((1 - b)r,(1 - b)h)$$

At this point we make the observation that the ratio of the volume of a cone to the volume of it's circumscribing cylinder must be invariant under a scaling on the coordinates (the ratio is homogeneous of degree 0).

$$\frac{V_{cone}(r,h)}{\pi r^2 h} = \frac{V_{cone}(sr,sh)}{\pi (sr)^2 sh}$$

for all $s>0$. If we write $V_{cone} = \hat{Q}\,F(r,h)\, r^2 h$ where $F(r,h)$ is some as yet unknown function and $\hat{Q}$ is a constant, then

$$F(r,h) = F(sr,sh)$$

so $F(r,h)$ is also homogeneous of degree 0.


\begin{array}{l@{}l} V_{frustum} &{}= V_{cone}(r,h) - V_{cone}((1 - b)r,(1 - b)h) \\ &{}= \hat{Q} \, F(r,h) \,r^2 h - \hat{Q} \, F((1-b)r, (1-b)h) \, (1-b)^2r^2 (1-b)h \\ &{}= \hat{Q} \, F(r,h) \, r^2 h ( 1 - (1-b)^3) \\ &{}= Q \, r^2 h (3 b - 3 b ^2 +b^3) \end{array}

where $Q = \hat{Q} \, F(r,h)$

Now consider the following figure

                                        cone with inscribed and circumscribed cylinders of height bh

It is clear that the volume of the frustum of height $b h$ must be bigger than the inner cylinder of radius $ (1-b) r$ and height $b h$ and it must also be less than the volume of the outer cylinder with radius $ r $ and height $b h$.

$$\pi (1-b)^2r^2 b h \lt V_{frustum} \lt \pi r^2 b h$$

Substituting the expression for $V_{frustum}$ from above and dividing everything through by $ b \pi r^2 h $

$$ (1-b)^2 \lt \frac{Q (3 - 3b + b^2)}{\pi} \lt 1$$

This must hold for all $0 \lt b \lt 1$.

At this point, we could use the familiar argument about limits - in particular, as $b$ gets closer to zero, the lower bound approaches the upper bound of $1$, so $\frac{Q 3}{\pi} = 1$ or $Q = \frac{\pi}{3}$.

However, it is possible to find the value of $Q$ in a different way, that does not involve some limit process.

First, observe that the value of $Q$ has bounds placed on it by the geometry of the problem $0 \lt Q \lt \pi$ since the cone must have some volume, and that volume must be less than the volume of a cylinder with radius $r$ and height $h$. What we are going to show is that for all values of $Q$ in this range, with just one exception, there is a choice of $b$ with $0 \lt b \lt 1$ that causes the above inequality not to hold. In the spirit of Sherlock Holmes, '..when you have eliminated the impossible, whatever remains, however improbable [or in our case, expected], must be the truth'.

We split the problem up into two parts. The upper bound of the inequality does not hold when

$$\frac{Q (3 - 3b + b^2)}{\pi} = 1$$

Solving for $b$

$$b = \frac{3}{2}-\sqrt{\frac{\pi}{Q}-\frac{3}{4}}$$

Now introduce a parameter $\alpha$ and write $Q= \pi / (1+\alpha+\alpha^2) $. Then for $0 \lt \alpha \lt 1$ we have $\pi/3 \lt Q\lt\pi$ and the above equation reduces to $b=1-\alpha$, so $0 \lt b \lt 1$.

The lower bound of the inequality does not hold when

$$(1-b)^2 = \frac{Q (3 - 3b + b^2)}{\pi}$$

Solving for $b$

$$b = 1 - \frac{ (\frac{1}{2} + \sqrt{\frac{\pi}{Q}-\frac{3}{4}})}{\frac{\pi}{Q}-1}$$

Introduce a parameter $\alpha$ as before, but this time write $Q=\pi \alpha^2 / (1+\alpha+\alpha^2) $. Then for $0 \lt \alpha \lt 1$ we have $0 \lt Q \lt \pi/3$ and the above equation again reduces to $b=1-\alpha$, so $0 \lt b \lt 1$.

Therefore we have $0 \lt Q \lt \pi$ by the geometry of the problem, but whenever $0 \lt Q \lt \frac{\pi}{3}$ or $\frac{\pi}{3} \lt Q \lt \pi$ there exists at least one value for $b$ with $0 \lt b \lt 1$ for which the inequality does not hold. The only remaining possibility on the interval $0 \lt Q \lt \pi$ is $Q=\frac{\pi}{3}$ (for all $r,h > 0$), and so $$ V_{cone} = \frac{\pi}{3} r^2 h$$

J. W. Tanner
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  • Your fundamental inequality about volume of frustum of a cone assumes that $r, h$ are constant and the volume is a function of $b$. Thus your argument after that inequality does not hold. – Paramanand Singh Aug 11 '17 at 08:33
  • I understand your first sentence, it is correct. I don't see how your second sentence follows from it? – Will Aug 11 '17 at 09:31
  • @ParamanandSingh Sorry, I'm new here, didn't realise how the comments work. The $r$ and $h$ are the dimensions of the full cone, and $b$ is the proportion of the height at which the cone is cut. I don't see how your second sentence follows from that? – Will Aug 11 '17 at 09:39
  • It should clear that the volume of the conical part also depends on $b$ and as $b$ varies from $0$ to $1$ it varies from $0$ to the volume of full cone. Let this function be called $C(b) $ and you have set $C(0)$ as volume of full cone. Then volume of frustum say $F(b) $ is given by $F(b) =C(0)-C(b)$. Now your inequality is $(1-b)^{2}b – Paramanand Singh Aug 11 '17 at 17:53
  • In fact you assume that the volume of a cone is a constant multiple of product of square of radius and height and on the basis of that assumption and the inequality you try to show that the constant must be $\pi/3$. This reasoning is correct. But there is no proof that the volume of cone must be a constant multiple of $r^{2}h$. This fact does not follow from the inequality satisfied by volume of frustum of a cone. – Paramanand Singh Aug 11 '17 at 17:53
  • @ParamanandSingh $b$ cannot be 0 or 1, since then the cone has not been split into two parts. It must be on the open interval $b \in (0,1)$. My argument about the functional form is a dimensional one - the units either side of the inequality are dimensionless, therefore $V_{frustum} / (r^2 h)$ must also be dimensionless, hence $V_{frustum} \propto r^2 h$ – Will Aug 11 '17 at 18:26
  • Dimension analysis is a trick used in physics but not for getting mathematical formulas representing physical laws but rather to illustrate that units matter. In mathematics too they serve the same purpose. The dimension of $rh^{2}$ is also same as that $r^{2}h$. And getting from frustum to cone is another problem. – Paramanand Singh Aug 11 '17 at 18:46
  • I think I understand what you are saying. Essentially, $V_{frustum}$ could in principle be any function of $r,h$ as long as it is bounded above by $\pi r^2h$ (e.g. $sin(r) r^2 h$). But if we start with the assumption $V_{frustum} \propto r^2 h$, then rest of the argument follows to find the constant of proportionality. – Will Aug 11 '17 at 19:36
  • yes rest of the argument is correct. perhaps you should write this assumption in bold. Then I will reverse my downvote to an upvote. – Paramanand Singh Aug 11 '17 at 19:47
  • Pending upvote delivered. +1 – Paramanand Singh Aug 12 '17 at 06:47
  • @paramand,@will. It is easy to see that for given $r$ the volume is proportional to $h$ so $r^ah^b$ can only be $r^2h$. I find it very "picky" to disallow this as a mathematical argument, but I think that the OP only asked for a demonstration – Philip Roe Aug 15 '17 at 19:12
  • @PhilipRoe It must be some function of $r$ and $h$, but not necessarily $r^ah^b$ even though that seems most natural. Thinking about it some more, if b is fixed, then it must be $\mathcal{O}(r^2h)$ both as $r,h \rightarrow \infty$ and as $r,h \rightarrow 0$ to remain within the bounds, which are independent of $r,h$. An alternative example might be $c(1 + 1/(1+r))r^2h$ with $c$ some constant. No matter how unlikely that may be, it can't be ruled out by the argument above. – Will Aug 16 '17 at 13:33
  • @Will Would you accept the argument that if I reduce my unit of length by a factor of two, then the numerical value of the volume must increase by a factor of 8? (You can probably guess that Im an engineer) – Philip Roe Aug 16 '17 at 15:47
  • @PhilipRoe Yes I agree, that was the gist of my dimensional argument in the comment above, but I can see that it doesn't lead to a formal proof unfortunately. – Will Aug 17 '17 at 12:33
  • Great idea! However, you do not prove the constancy of the ratio $$\frac{V_{cone}}{πr^2h}.$$ You only assumed it. – Allawonder Jul 08 '20 at 23:35
  • @Allawonder The requirement is only that the ratio remains constant irrespective of the units used to measure length scales - cm, feet, light years :) the ratio will be invariant. – Will Jul 10 '20 at 12:53
  • @Will Exactly, that's my point. You only *take it as a requirement.* You do not show that this is actually the case. – Allawonder Jul 11 '20 at 13:15
  • @Allawonder So you think I also need to provide a proof of the general result that the ratio of the volumes of any two solid bodies in Euclidian space is homogeneous of degree 0? It seems self-evident to me (any solid body volume must vary as the cube of the length scale), but I will look for a proof. – Will Jul 14 '20 at 04:56
  • @Will Providing such a proof will make your answer *complete.* – Allawonder Jul 14 '20 at 08:41

I managed to find the volume of a cone without calculus using an observation that I made.

First, I put a cone on a Cartesian plane, with the tip at the origin. Thus, an equation to describe the radius(x) would be the radius over the height times x. Then, I substituted this equation into pi r squared to get cross sectional area as function of x.

I then observed how the volume of the cone could be approximated by using disks, the width of each being the height of the cone divided by the number of disks. So, the volume as a function of x would be the area as a function of x times the height divided by n, or the number of disks. However, instead of using integration to sum the volumes of all the disks, I observed that if I moved along the height in increments equal to the width of each cylinder, that the volumes of the cylinders increased in a sequence of squares, the second disk being 4 times the volume of the first, the third being 9 times, the fourth being 16 times, and so on.

To me, this showed that the second disk can be broken up into 4 cylinders equal to the volume of the first disk, the third into 9, the fourth into 16, and so on. So, the volume of a cone is equal to the volume of the first disk times the sum of all the cylinders, which we can get using the summation of squares formula. So, I got the volume of the first cylinder by putting the width of one cylinder into the volume as a function of x formula, which got pi r squared times the height over n cubed. I then multiplied this by the summation of square's formula to get: pi*r^2*h*(n(n+1)(2n+1))/(6n^3) Then, I let "N" go to infinity, which resulted in the volume of a cone being (pi*r^2*h)/3.

  • How did the ancient Greeks do it? They did not have approximation or limit theory, nor could they write and manipulate a formula like pi*r^2. For us it is easy. For them, it was formidable. – richard1941 Sep 14 '19 at 22:08

It is because a triangle in a box that has the same height and length is 1/2 if the square because it is in the second dimension so if you move in to the third dimension it will change to 1/3 and so forth.

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    This is not much of an argument. Could you provide some algebraic detail? – Simon Hayward Nov 28 '12 at 21:53
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    Yes, it's not clear what the "so forth" means -- what is the fourth dimensional counterpart? – bryn Dec 01 '12 at 09:30
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    Its not much of an argument, but it is true. An n dimensional pyramid is made from an n-1 dimensional figure where every point is joined to a single point. A 2D pyramid is a triangle. The proof is not really practical in a comment, but is a simple generalisation of the 3D case for the cone or pyramid, and ultimately the factor of 1/n derives from d(x^n)/dx = nx^(n-1) where x^n is the volume of the enclosing hypercube, just as the factor of 1/3 derives from this in the 3D case. – Peter Webb Oct 22 '14 at 11:36
  • Best answer thanks. This was the missing piece for my unsderstanding. This is so simple that no oroof are needed. Thx – Mr Rubix Nov 04 '21 at 11:43

Let $r$ & $h$ be respectively the radius & the normal height of a cone. Now place it with its geometrical axis coincident with the x-axis then the cone can be generated by rotating a straight line:$\color{blue}{y=\frac{r}{h}x}$, passing through the origin, about the x-axis. Hence, the volume of the cone $$\color{blue}{V_{cone}}=\int\pi y^2 dx=\int_{0}^h \pi\left(\frac{r}{h}x\right)^2 dx$$ $$=\frac{\pi r^2}{h^2}\int_{0}^h x^2 dx=\frac{\pi r^2}{h^2} \left[\frac{x^3}{3}\right]_{0}^h=\frac{\pi r^2}{h^2} \left[\frac{h^3}{3}\right]$$$$\color{blue}{=\frac{1}{3}\pi r^2h}$$

Similarly, the cylinder with a radius $r$ & normal height $h$ can be generated by rotating a straight line:$\color{blue}{y=r}$, parallel to the x-axis, about the x-axis. Hence, the volume of the cylinder $$\color{blue}{V_{cylinder}}=\int\pi y^2 dx=\int_{0}^h \pi\left(r\right)^2 dx$$ $$=\pi r^2\int_{0}^h dx=\pi r^2 \left[x\right]_{0}^h=\pi r^2 \left[h-0\right]\color{blue}{=\pi r^2h}$$ Thus. we find that $$\color{blue}{\text{Volume of cone}=\frac{1}{3}(\text{Volume of cylinder})}$$

Harish Chandra Rajpoot
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You can use Pappus's centroid theorem as in my answer here, but it does not provide much insight.

If instead of a cylinder and a cone, you consider a cube and a square-based pyramid where the "top" vertex of the pyramid (the one opposite the square base) is shifted to be directly above one vertex of the base, you can fit three such pyramids together to form the complete cube. (I've seen this as physical toy/puzzle with three pyramidal pieces and a cubic container.) This may give some insight into the 1/3 "pointy thing rule" (for pointy things with similar, linearly-related cross-sections) that Katie Banks discussed in her comment.

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It is easy to imagine that the volume of a Cone is equal to the volume of a square pyramid as stated in other answers.

Now we only have to show that the volume of square pyramid is 1/3 b h.

For that, consider this cube made of 6 square pyramids: enter image description here

Let the height of the pyramid be h.

So the side length of the cube is 2h.

The volume of the cube

= base $\times$height of the cube

= base $\times$ $2h$

From the figure,

The volume of the pyramid

= $\frac{1}{6}$ $\times$ volume of the Cube

= $\frac{1}{6}$ $\times$ base $\times$ $2h$

= $\frac{1}{3}$ $\times$ base $\times$ $h$

Abhinav P B
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Imagine a pyramid inside a cube; One of the point of the pyramid is touching the top face of the cube, the point can be anywhere as long as it is on the top face of the cube, and still not change the volume.

Imagine that the point i just mentioned went to the corner of the cube Cut the top half of that pyramid, it would look exactly like the pyramid, except that the volume would be exactly $\frac {1}{8}$ of the original.

Now let's look at the lower half, you would probably notice that you can cut a part of it to get the exact same shape as the top half. Cutting it so you have $2$ of those small pyramids. The remaining object will have a volume $\frac {1}{4}$ of the cube, the two small pyramids is $\frac {1}{8}$ of the original.since you have 2 of them. The two parts combined will be $\frac {1}{4}$ of the original pyramid, Which means the remaining bit is 3/4 of the original pyramid,which is 1/4 of the cube the 2 parts are a 1/3 of the remaining part so if we add them together $\frac{1+1/3}{4}$=$\frac 1{3}$

This might be a little confusing But it works :P

Some dude
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$\def\Vol{\text{Vol}}$Here is one more approach.

Take a cone with radius $r$ and height $h$ and chop it into four pieces of height $h/4$; call these pieces $F_1$, $F_2$, $F_3$ and $F_4$ from narrowest to widest. Let $C$ be a cylinder of height $h/4$ and radius $r/2$. We claim that $$\Vol(F_1) + \Vol(F_4) = \Vol(F_2) + \Vol(F_3) + \Vol(C). \quad (\ast)$$ To see this, slice through the five solids at height $y$ from the top. The areas of the slices through $F_1$, $F_2$, $F_3$ and $F_4$ are $\pi r^2 (y^2/h^2)$, $\pi r^2 ((y+h/4)^2/h^2)$, $\pi r^2 ((y+2h/4)^2/h^2)$ and $\pi r^2 ((y+3h/4)^2/h^2)$ respectively, and the area of the slice through $C$ is $\pi (r/2)^2 = \pi r^2 (h^2/4)/h^2$. We check that $$(y + 3 h/4)^2 + y^2 = (y + 2 h/4)^2 + (y + h/4)^2 + h^2/4.$$ Multiplying by $\pi r^2/h^2$, we deduce the claim by Cavalieri's principle.

Now, let $K$ be the original cone. The volume of a cone must be proportional to $r^2 h$, so we have $$\Vol(F_1) = (1/64) \Vol(K)$$ $$\Vol(F_2) = (8/64-1/64)\Vol(K) = (7/64) \Vol(K)$$ $$\Vol(F_3) = (27/64-8/64)\Vol(K) = (19/64) \Vol(K)$$ $$\Vol(F_4) = (64/64-27/64)\Vol(K) = (37/64) \Vol(K).$$

Plugging this into $(\ast)$, we have $$\tfrac{1+37}{64} \Vol(K) = \tfrac{7+19}{64} \Vol(K) + \Vol(C)$$ $$\tfrac{3}{16} \Vol(K) = \Vol(C)$$ and $\Vol(K) = \tfrac{16}{3} \Vol(C)$. Now, remember that $C$ had radius $r/2$ and height $h/4$. So the volume of $K$ is $1/3$ the volume of a cylinder of radius $r$ and height $h$, as desired.

David E Speyer
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