Formula for the Volume of a ncone
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Formula for the Volume of a ncone
My calc. teacher recently gave us a paper with several geometric formulas on it and I noticed a pattern. A cone (a 3 dimensional figure) that has any figure for its base has a volume of Bh/3 and a triangle (a 2 dimensional figure)has a area of bh/2. The formula for both is the measurement of the base multiplied by the height divided by number of dimensions the figure exists within. I was wondering if this pattern held for coneequivalents in higher dimensions. If it is true, why is it true?
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Re: Formula for the Volume of a ncone
Consider how (n)area is derived  from stacking (n1)sheets on top of each other.
Consider the formula for an (n1)sheet's (n1)area.
Consider how to integrate a polynomial.
Jose
Consider the formula for an (n1)sheet's (n1)area.
Consider how to integrate a polynomial.
Jose
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Re: Formula for the Volume of a ncone
Continuing Jose's rather cryptic reply, also note that B can be expressed as B=a*h^2 for some (proportionality) constant a.

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Re: Formula for the Volume of a ncone
I have thought about it a lot and I still cannot figure out if the pattern I noticed works for higher dimensional figures.
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Re: Formula for the Volume of a ncone
jewish_scientist wrote:I have thought about it a lot and I still cannot figure out if the pattern I noticed works for higher dimensional figures.
It does, but it probably won't be intuitive why until you take calculus.
If you want to try to bend your mind:
An isosceles right triangle is half of a square. That is to say: divide a square in half along its diagonal, and you'll get two isosceles right triangles. The area of the square is bh, so the area of each triangle is bh/2.
Now, we can describe that division another way. Take one corner of the square, and look at the two edges opposite it (that is, the two edges it is not incident to). Number them Edge 1 and Edge 2. One of the triangles consists of the set of all points that are closer to Edge 1 than Edge 2, and the other triangle consists of all points that are closer to Edge 2 than to Edge 1. These two triangles have the same area by symmetry, and they divide up the square, so each has area half the square, or (1/2)bh.
This might seem like a weird way to describe the triangles, but it'll aid us when we move up to three dimensions. Take a cube, and take one corner of that cube. There are three opposite faces; we'll call them Face 1, Face 2, and Face 3. Now, divide up the cube into the points closest to Face 1 among those three faces, the points closest to Face 2, and the points closest to Face 3. Those three regions will be pyramids (that is, cones with square bases), and each one has area 1/3 of the cube, i.e. (1/3)Bh.
A similar argument holds for each higher dimension. The ncube can be divided up into n npyramids by looking at which points are closest to each of the n hyperfaces across from a given corner. Each one will have nvolume 1/n times that of the ncube, so the formula for the nvolume of an npyramid (and hence an ncone) is (1/n)Bh.
The rationale for why the formula for pyramid and cone ought to always be the same is beyond the scope of this argument. I think it really does require calculus to appreciate.
(∫p^{2})(∫q^{2}) ≥ (∫pq)^{2}
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Thanks, skeptical scientist, for knowing symbols and giving them to me.

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Re: Formula for the Volume of a ncone
Fortunately I have already taken calculus. Unfortunately, I think that argument only works for cones that have a quadrilateral as a base.
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Re: Formula for the Volume of a ncone
Yes, and then calculus shows why you can generalize.jewish_scientist wrote:Fortunately I have already taken calculus. Unfortunately, I think that argument only works for cones that have a quadrilateral as a base.
Re: Formula for the Volume of a ncone
There is no need for calculus. Just approximate the base of the cone with a bunch of little quadrilaterals (or hypercubes, in higher dimensions). In the two dimensional case, this is the same as breaking up a triangle's base into a bunch of smaller bases, cutting the whole triangle into a bunch of skinny triangles with varying degrees of tilt. In the three dimensional case, you can imagine taking the circular base of a circular cone, superimposing a square mesh on it, and taking all the squares of the mesh which are contained in the circle (for a lower bound) or which contain at least one point of the circle (for an upper bound) and joining them to the vertex of the cone. It might take some practice at visualization at first to see that all these little cones fit together to give an approximation of the circular cone.
The second thing to be able to visualize is why the amount of "tilt" of a cone with a square base doesn't matter, only the height of the vertex off the plane of the base. You can think about this as a shear transform of a more standard cone, and keep in mind Cavalieri's principle.
So all you really need to know is the volume of a square based cone. I really like Cauchy's dissection of a cube into cones as a visual way to understand that. Previously, the best dissection approach I had seen was a way of cutting up the ndimensional cube into n! ndimensional simplices (for each point of the cube, you decide which simplex it belongs to by checking which order its coordinates come in), which gives the formula for the volume of a cone whose base is an (n1)dimensional simplex, but this seems needlessly convoluted in comparison (most people are more familiar with visualizing cubes, or cones with cubical bases, than simplices). An alternative approach would be to divide an ndimensional cube into 2n cones, each with a vertex at the center of the cube and with base equal to one of the 2n faces  this might be even easier to visualize, since the cones involved look more regular, but it involves an extra factor of 2 since the vertex of each cone is only halfway up from any face.
The second thing to be able to visualize is why the amount of "tilt" of a cone with a square base doesn't matter, only the height of the vertex off the plane of the base. You can think about this as a shear transform of a more standard cone, and keep in mind Cavalieri's principle.
So all you really need to know is the volume of a square based cone. I really like Cauchy's dissection of a cube into cones as a visual way to understand that. Previously, the best dissection approach I had seen was a way of cutting up the ndimensional cube into n! ndimensional simplices (for each point of the cube, you decide which simplex it belongs to by checking which order its coordinates come in), which gives the formula for the volume of a cone whose base is an (n1)dimensional simplex, but this seems needlessly convoluted in comparison (most people are more familiar with visualizing cubes, or cones with cubical bases, than simplices). An alternative approach would be to divide an ndimensional cube into 2n cones, each with a vertex at the center of the cube and with base equal to one of the 2n faces  this might be even easier to visualize, since the cones involved look more regular, but it involves an extra factor of 2 since the vertex of each cone is only halfway up from any face.
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Re: Formula for the Volume of a ncone
You might not strictly need calculus, but it is certainly easier with it. Your description of "approximating the base of the cone with a bunch of little quadrilaterals" is by far the easiest if you use limits. Using exhaustion is unnecessarily complicated to say the least. But better yet, you come up with a general theorem relating the (n1)volume of the base with the radius and a constant coefficient and integrate.
Re: Formula for the Volume of a ncone
notzeb wrote:There is no need for calculus. Just approximate the base of the cone with a bunch of little quadrilaterals (or hypercubes, in higher dimensions). In the two dimensional case, this is the same as breaking up a triangle's base into a bunch of smaller bases, cutting the whole triangle into a bunch of skinny triangles with varying degrees of tilt. In the three dimensional case, you can imagine taking the circular base of a circular cone, superimposing a square mesh on it, and taking all the squares of the mesh which are contained in the circle (for a lower bound) or which contain at least one point of the circle (for an upper bound) and joining them to the vertex of the cone. It might take some practice at visualization at first to see that all these little cones fit together to give an approximation of the circular cone.
The second thing to be able to visualize is why the amount of "tilt" of a cone with a square base doesn't matter, only the height of the vertex off the plane of the base. You can think about this as a shear transform of a more standard cone, and keep in mind Cavalieri's principle.
Both of the things you described here are things I would call "calculus".
And if you've taken calculus, then the answer why the formula works is pretty straightforward. In the same way that you find the area of a region by integrating its height as you vary across the length of the region, you can find the volume of a threedimensional region by integrating the crosssectional area across the height of the object, and in general the nvolume of an nregion by integrating the (n1)volume of its hypercrosssectional slices across the remaining dimension of the object. In the case of an ncone, cross sections parallel to the base are similar figures, with the ratio of similarity in proportion to the distance from the vertex. (For example, if you slice a circular threedimensional cone parallel to its base, you get a smaller circle, and if you slice a squarebased threepyramid parallel to its base, you get a smaller square. If B is the (n1)volume of the base, then since a figure enlarged by a factor of e has its (n1)volume scaled by e^(n1), we get that the cross section a distance x from the vertex has (n1)volume B*(x/h)^(n1). So the nvolume of the ncone is the integral from 0 to h of B*(x/h)^(n1) dx, which evaluates to Bh/n.
(∫p^{2})(∫q^{2}) ≥ (∫pq)^{2}
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Re: Formula for the Volume of a ncone
Really? To me, "calculus" has always meant "a symbolic method of calculation" (some variation of this definition is called for if you want to know why "propositional calculus", "lambda calculus", or "kirby calculus" are named as they are). So solving the problem with calculus would involve setting up an integral, and then computing that integral using a list of rules for symbolically manipulating integrals (in this case, the rule for the integral of a polynomial).Cauchy wrote:Both of the things you described here are things I would call "calculus".
On the other hand, the method of exhaustion predates the integral calculus by several millennia, as does the geometric concept of areapreserving symmetries. Both are methods I would refer to as geometric (or visual) reasoning.
(Throwing a wrench into the discussion, there is also Mamikon's "visual calculus", which for some confusing reason has calculus in its name despite being a visual approach to computing areas which would seem hopeless with the more traditional symbolic approach.)
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Re: Formula for the Volume of a ncone
Well "calculus" has lots of meanings, including "tartar." But obviously what we're talking about here is the mathematics of change, including more specifically integral calculus. It's what Newton called "the Calculus." It seems to me like you're pretending you don't know that.
And as I said, while you could solve this problem without calculus, it is far easier to solve it with calculus, as the OP said he is a calc student.
And as I said, while you could solve this problem without calculus, it is far easier to solve it with calculus, as the OP said he is a calc student.
Re: Formula for the Volume of a ncone
notzeb wrote:Really? To me, "calculus" has always meant "a symbolic method of calculation" (some variation of this definition is called for if you want to know why "propositional calculus", "lambda calculus", or "kirby calculus" are named as they are). So solving the problem with calculus would involve setting up an integral, and then computing that integral using a list of rules for symbolically manipulating integrals (in this case, the rule for the integral of a polynomial).Cauchy wrote:Both of the things you described here are things I would call "calculus".
On the other hand, the method of exhaustion predates the integral calculus by several millennia, as does the geometric concept of areapreserving symmetries. Both are methods I would refer to as geometric (or visual) reasoning.
(Throwing a wrench into the discussion, there is also Mamikon's "visual calculus", which for some confusing reason has calculus in its name despite being a visual approach to computing areas which would seem hopeless with the more traditional symbolic approach.)
Do limits of Riemann sums fall under integral calculus? Because that's effectively what you're doing with your square mesh. Cavalieri's Principle, that the volume of a solid can be completely calculated by knowing its crosssectional areas, is just a stone's throw away (no pun intended) from actually doing that calculation, and doing that calculation is integral calculus even if you don't explicitly set up an integral to do it.
And I thought it was obvious that when I said calculus I meant the stuff that would be taught in a calculus class in high school or college.
(∫p^{2})(∫q^{2}) ≥ (∫pq)^{2}
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Re: Formula for the Volume of a ncone
Although it might be possible to phrase what I described in terms of Riemann sums, I didn't, and I don't think it is a very productive way of approaching most geometric problems  a circle is not even the graph of a function, and it's entirely possible for the base of a cone to be stranger (think something kidney bean shaped, or with a hole cut out of the middle), so you have to go through some awkward contortions in order to force everything into the Riemann sum framework. I agree that mentioning Cavalieri's principle was borderline  I wasn't willing to assume that everyone found the fact that shear transforms preserve volume intuitive, since geometry has been watered down so heavily in the standard curriculum.Cauchy wrote:Do limits of Riemann sums fall under integral calculus? Because that's effectively what you're doing with your square mesh. Cavalieri's Principle, that the volume of a solid can be completely calculated by knowing its crosssectional areas, is just a stone's throw away (no pun intended) from actually doing that calculation, and doing that calculation is integral calculus even if you don't explicitly set up an integral to do it.
Different backgrounds, I suppose. I don't know what your calculus class was like. I am starting to get the feeling that any argument involving an approximation gets shoved under the umbrella of calculus, as does anything involving area or volume. Which is odd to me, since most inequalities are best proved algebraically, and for instance trying to compute the area of a square which has been tilted by 30 degrees from the coordinate grid using the integral calculus would be a poor approach.Cauchy wrote:And I thought it was obvious that when I said calculus I meant the stuff that would be taught in a calculus class in high school or college.
Anyway, we've gotten way off topic  I doubt the OP actually cares about the distinction between geometric reasoning and calculus, especially as it seems to be a distinction that only exists within my own head.
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Re: Formula for the Volume of a ncone
notzeb wrote:I doubt the OP actually cares about the distinction between geometric reasoning and calculus...
I am actually very interested in the difference! The reasoning behind mathematics was never taught to me outside of my geometry class and a brief section of by percalculus class (trigonometic identities). The worse result of this was when I was 'taught' all the properties of logarithms without being without being taught what a logarithm is. Even now I feel a little uncomfortable working with them because of this. The before mentioned calculus class I took was probable the first time a teacher proved an equation was true on the white board during class. Even better, he proved it for specific cases each time we learned a new tool we could use and only proved the general case during the second half of the semester.
To summarize, I am really interested in things like this and really look forward to our discussion on it!
...especially as it seems to be a distinction that only exists within my own head.
...... Maybe we should go back to the OP instead.
notzeb wrote:Although it might be possible to phrase what I described in terms of Riemann sums, I didn't, and I don't think it is a very productive way of approaching most geometric problems  a circle is not even the graph of a function, and it's entirely possible for the base of a cone to be stranger (think something kidney bean shaped, or with a hole cut out of the middle), so you have to go through some awkward contortions in order to force everything into the Riemann sum framework. emphasis added
Why would the higher dimensional equivalents of Riemann sums be difficult for a n cone who's base is concave or have a <n hole in them?
I want to double check what I have learned so far. Cauchy proved that the general formula for a n cone's higher volume, or higher dimensional equivalent, is true for any ncone who's base is a n1 cube. notzeb then proved that this fact could be used to in a process similar to Riemann sums to solve the general case. He claimed that his proof did not use calculus. Several people commented that what notzeb described is not just similar to, but actual is the higher dimensional equivalent of Riemann sums, and therefor does require calculus. This lead to a discussion regarding exactly what 'calculus' refers to. If this is all correct, can we move on to something else?
I have never liked Riemann sums because the can only provide approximations. Is there a proof that does not rely on them?
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Re: Formula for the Volume of a ncone
Finite Riemann sums are approximations, but the limit of a (wellbehaved) Riemann sum as n goes to infinity is exact. This provides a fairly rudimentary definition for a definite integral.
I'm honestly not sure what notzeb is talking about. The shape of the base is completely irrelevant. No matter what it is, in a cone, the area of a crosssection will be proportional to the area of the base, so that's just a constant. You don't have to integrate it, and moreover it's a constant that shows up in the final expression, so you don't have to evaluate it. All you have to do is integrate the radius along the axis of the cone, a very trivial process. It's that integral that demonstrates very simply and convincingly where the 1/N factor comes from for an Ndimensional cone.
In other words, the fact that the volume is proportional to the area of the base times the height should be obvious. The fact that you divide this by N comes right out of the integral for r^{N1}. That's it.
I'm honestly not sure what notzeb is talking about. The shape of the base is completely irrelevant. No matter what it is, in a cone, the area of a crosssection will be proportional to the area of the base, so that's just a constant. You don't have to integrate it, and moreover it's a constant that shows up in the final expression, so you don't have to evaluate it. All you have to do is integrate the radius along the axis of the cone, a very trivial process. It's that integral that demonstrates very simply and convincingly where the 1/N factor comes from for an Ndimensional cone.
In other words, the fact that the volume is proportional to the area of the base times the height should be obvious. The fact that you divide this by N comes right out of the integral for r^{N1}. That's it.
Re: Formula for the Volume of a ncone
notzeb wrote:Although it might be possible to phrase what I described in terms of Riemann sums, I didn't, and I don't think it is a very productive way of approaching most geometric problems  a circle is not even the graph of a function, and it's entirely possible for the base of a cone to be stranger (think something kidney bean shaped, or with a hole cut out of the middle), so you have to go through some awkward contortions in order to force everything into the Riemann sum framework.
Just because you didn't use the words "Riemann sum" when you subdivided the base into a rectangular mesh, then found a region above each mesh piece that together approximate the volume you're looking for, and then saying that the approximation becomes exact as the mesh size goes to 0, doesn't mean you're not conducting a Riemann sum. Maybe you're upset because you don't feel you're approximating the volume so much as squeezing it between two other volumes? In that case, you're calculating lower and upper sums for the Darboux integral (which is equivalent to the Riemann integral).
I agree that mentioning Cavalieri's principle was borderline  I wasn't willing to assume that everyone found the fact that shear transforms preserve volume intuitive, since geometry has been watered down so heavily in the standard curriculum.
From the wiki article on Cavileri's principle:
wiki wrote:Today Cavalieri's principle is seen as an early step towards integral calculus, and while it is used in some forms, such as its generalization in Fubini's theorem, results using Cavalieri's principle can often be shown more directly via integration.
If you want to say that Cavalieri's principle is not integral calculus because it's merely an early step towards integral calculus, then I guess you're welcome to.
I am starting to get the feeling that any argument involving an approximation gets shoved under the umbrella of calculus, as does anything involving area or volume.
An argument involving increasingly precise approximations converging to the actual answer is an argument about limits, which are the cornerstone of integral and differential calculus, and doubly so when they involve areas or volumes. Just because you don't say the word "limit" doesn't mean you're not taking a limit.
For an argument about volume that's not in the wheelhouse of calculus, the one I provided about cutting up an ncube into n pyramids of equal volume is a very noncalculus way of determining the nvolume of one of those pyramids.
(∫p^{2})(∫q^{2}) ≥ (∫pq)^{2}
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Re: Formula for the Volume of a ncone
Euclid proved the volume of the cone. Are you are suggesting that he knew calculus?
Euclid’s proof works by showing that if the cone were more than one third of the cylinder, then a polygonal pyramid could be found within the cone that is also more than one third of the cylinder. But the polygonal pyramid is one third of its prism, which is fully within the cylinder, so that prism would have to be bigger than something it is contained in.
And similarly if the cone were less than one third of the cylinder, then a polygonal pyramid could be found enclosing the cone which is also more than one third of the cylinder, so its prism would be smaller than the cylinder which it contains.
Euclid’s proof works by showing that if the cone were more than one third of the cylinder, then a polygonal pyramid could be found within the cone that is also more than one third of the cylinder. But the polygonal pyramid is one third of its prism, which is fully within the cylinder, so that prism would have to be bigger than something it is contained in.
And similarly if the cone were less than one third of the cylinder, then a polygonal pyramid could be found enclosing the cone which is also more than one third of the cylinder, so its prism would be smaller than the cylinder which it contains.
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Re: Formula for the Volume of a ncone
No, of course not.Qaanol wrote:Euclid proved the volume of the cone. Are you are suggesting that he knew calculus?
People have been suggesting that the way to generalize to any base shape is with calculus, and people have also described the specific sorts of operations they consider to be calculus.
Euclid used notthoseoperations to calculate notageneralized base shape, so I'm not sure how you think we're suggesting he knew calculus.
Re: Formula for the Volume of a ncone
Qaanol wrote:Euclid proved the volume of the cone. Are you are suggesting that he knew calculus?
I'm suggesting that he did what we would now put under the umbrella of calculus, whether or not he "knew calculus" or knew about calculus.
Wiki wrote:The method of exhaustion is seen as a precursor to the methods of calculus. The development of analytical geometry and rigorous integral calculus in the 17th19th centuries subsumed the method of exhaustion so that it is no longer explicitly used to solve problems.
The method of exhaustion (which you describe) is fundamentally an argument about limits. You've got two sequences, {U_n} and {L_n}: {L_n} is increasing, {U_n} is decreasing, the volume V has the property that L_n <= V <= U_n for every n, and it turns out that lim{L_n} = lim{U_n}. Then you conclude that V is equal to these common limits. (This is the discrete form of the squeeze theorem, a common syllabus point in a Calc I class.) The proof Euclid gave is even the standard calc proof: if V > lim{U_n}, then we can find k such that V > U_k, a contradiction, and similarly if V < lim{L_n}. Just because I phrased it in calculus terms and Euclid didn't doesn't mean that I'm doing calculus and he isn't. The squeeze theorem wiki article even says that Archimedes used the method to compute pi, so at least one other person agrees that ancient Greeks can have done calculus without "knowing calculus".
I guess I don't understand why people are so against Euclid having done calculus. I think it's amazing to be able to say "You know that math that we only cobbled together 400 years ago? The Greeks were doing it 2000 years before that."
(∫p^{2})(∫q^{2}) ≥ (∫pq)^{2}
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Thanks, skeptical scientist, for knowing symbols and giving them to me.
Re: Formula for the Volume of a ncone
Eh…I generally reserve ‘calculus’ to involve derivatives and integrals. Limits—especially those tractable by pure geometry—fall outside that umbrella.
Regardless, a more general line of reasoning shows that an ndimensional cone of arbitrary shape has volume 1/n times the product of its height and base.
The hypercubeslicing observation from earlier gives us the result for square pyramids in n dimension.
If we color part of the base of a square pyramid red, say p percent of it, and connect our red region to the apex with straight lines, then its crosssection is always p% of the square pyramid’s.
It does not take calculus to see that if one quantity is everywhere p% the size of another, then as a whole it is also p% of the other’s size.
Put another way, as we scale and slide the base toward the apex, our red region sweeps out p% of the nvolume that the base does, because it is always exactly p% as big as the base.
Regardless, a more general line of reasoning shows that an ndimensional cone of arbitrary shape has volume 1/n times the product of its height and base.
The hypercubeslicing observation from earlier gives us the result for square pyramids in n dimension.
If we color part of the base of a square pyramid red, say p percent of it, and connect our red region to the apex with straight lines, then its crosssection is always p% of the square pyramid’s.
It does not take calculus to see that if one quantity is everywhere p% the size of another, then as a whole it is also p% of the other’s size.
Put another way, as we scale and slide the base toward the apex, our red region sweeps out p% of the nvolume that the base does, because it is always exactly p% as big as the base.
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