@@Megadumbyogyou can imagine the volume of the pyramid as being the same as a bunch of squares layered on top of each other with infinitesimally small height. Let's say each thin layer has a volume dv. dv = length * width * dh, where dh is the height of each layer. Now it's just a matter of getting the length and width as functions of the original length, width and height, then integrating both sides of the equation.
@@Megadumbyog It's basically to sum the infinite squares that form the pyramide, which is the same concept as integrating. After some logical thinking u end up with a quite ez integral that leaves u the volume formula. The cone volume can be proved through the same logic.
Integral from 0 to 1 of (x^2 dx) = 1/3. It's because doing an infinite sum over a power like x^r in calculus be ones 1/(r+1) * x^(r+1). It's just the way infinite sums behave.
Bro the way I understand this is there's a 1/4 pyramid at the corners and there are 4 corners so 4/4 but that's just a whole prism so u subtract 1/3 from 4/4 so it's 3/3 - 1/3 = 2/3 the sum of all the empty space around the pyramid. It's not actually 1/4 pyramid it's 1/6.
These are both true, but I'm not sure how much they help build an intuitive understanding of this. Explanation number 1 relies on the fact that halving the height of the pyramid halves its volume, which is true but is not immediately obvious when looking at a pyramid (as it would be for a rectangular prism, for example). Explanation #2 has the same issue with the fact that skewing a pyramid doesn't change its volume if you keep it's height constant (again this is true, but it's not something that really *feels* true without its own proof). Not to mention that there's a much easier proof using that same idea (create a square out of three pyramids whose bases sit on mutually perpendicular faces and whose points all meet in one corner, together all three fill the entire volume of the cube).
Yes your comments are fair sir. I wanted to animate the shear transformation breaking the pyramid into many many tiny cubes, then shifting them all sideways. But couldn't get manim to render that many objects. Maybe one for the future...
@@nicemathproblems I say that because I know my students struggle with that even in 2 dimensions. A lot of them seem to think that A=1/2bh only applies to right triangles. I saw someone else mention integration (as a clue to where the 1/3 comes from) and I think that's conceptually the easiest way to see why skewing doesn't affect area/volume. If you imagine breaking the shape up into a bunch of very small layers which you're adding up all that matters is how many layers there are (the height) are and how wide the layers are (the base). Shifting them side to side won't affect the area.
You can also divide the cube into 3 pyramids with their bases on the bottom, left and back side of it. It is harder to visualise tho and the first method might be the easiest to imagine
assume height is proportional to side length of cross section, and area is proportional to side length squared (both true in pyramids), then integrate with respect to height@@nicemathproblems
@@nicemathproblems x squared is the cross-sectional area of the pyramid, which is a square. integrating the cross-sectional area across the height gives the total volume
I like this way of visual reasoning, a skill we are often missing out on because schools mostly teach us how to reason algebraically, and using language. Of course he used language here to describe what was going on, but the process of seeing this if you were in the seat of someone figuring this out would involve visualizing these types of basic relationships and drawing inferences from them.
Well to explain it to my self I use integrals, saying, the base is on (0,0,0),(a,0,0),(0,b,0) if you understand coordinates its a right triangle, and for the other vertice I choos (c,d,h), so c an d are a number and h is the Height of the pyramid, if you slice the piramid you get a right triangle and if you get de functions of the respective values integrating from 0 to h respect of z the final answer its (abh)/6=(h/3)(ab/2) and (ab/2) its the area of the base, so I can say every pyramid with a base of a right triangle has this formula, but now if you can divide every convex polygon on right triangles and the formula works too, then because the cicle cant be triangulated I use anothe integral for that
Is it true that for all dimensions the hypertriangle volume would be 1/d the volume of the hyperprism? Because you can see a relationship between how in 2d it is half the area and in 3d it's a third of it.
Yes, since for the n-th dimension, you can calculate it’s volume by summing up all the squares from bottom up with the definite integral of x^(n-1)dx from 0 to the cube’s side length a, which always gives you a^n/n = V/n
I always imagined moving the peak over to a corner and fitting two more with their peaks at that same corner, and the extra pyramids added would be the same because they can be reflected over the plane from the shared edge to the opposite edge (which the shared corner is on)
The calculus proof (integrating the cross-sectional area across the height) also can be used to prove that the volume of any n-sided pyramid (or cone) is 1/3 the volume of its respective prism or cylinder since the area of any polygon is proportional to its length squared.
@@nicemathproblemsA square pyramid with height 1 in a cube with sides 1. The cross section is a square with sides that has equal length as the distance between the cross section and the top side. Using this we get that the square pyramid has a volume of int 0to1 x^2 which gives us 1/3.
ive learned alot of stuff on youtube, history, science, dinosaurs, and even electronics. i still have no idea whenever i will ever use this kind of knolage
Recently I was having a thought about a pyramid which is slanted so the apex is directly over one of the base points. I realized after some time that you could fit an entire other pyramid upside down in the same rectangular prism volume, with extra space. I was not certain but I guessed that the remaining space was of the same volume as the two pyramids. Imagine instead the principal line was the diagonal which the two pyramids share, then the half segments on either side are of equal volume to half of a pair of pyramids which point in opposite directions…
And while those two are like super obvious, my teacher tried to cut a prism into tree piramids... And the blackboard was a mess of lines and nobody understood zit 😢
Yea... Sorry... 😅 In fact, was the best teacher I had and the one who made maths interesting to me as a whole... But that day I was so dizzy... It was confusing as the three pyramids have to be deformed to fit into a prism... The ones presented here may be easier to understand even without drawing them.
If inside of a cube can have 6 pyramids with half of the height then it's simple to prove base's area × height = cube's area Small pyramid's area = base area × height ÷6 Pyramid's area = 2 small pyramids area
So we take one day from each phases in order, skipping the discount on the seed phase if it's gets to only one day, until there are no more days left to discount ?
If the apex of a pyramid is directly above its square base, and the height of that pyramid equals a side of the base, then three of those pyramids fit together to make a cube.
A square is not a rectangle. Both squares and rectangles are quadrilaterals. A rectangle is a quadrilateral with four parallel sides and two sets of lines of two different lengths. A square is a quadrilateral where each side is of equal length and 2 sets of parallel lines.
If it were 1/2 then we would not get a pyramid, but a triangular prism. For it to form a pyramud, and to get the geometric object to be pointed, it needs to be 1/3.
you have no idea how much i needed this, I literally just got a question on this today.
That's crazy! The CZcams algorithm really did its job then!
Did you pass your test??!?!?!
These are great visual proofs! The easiest way to understand it is with Calculus... bat not everyone knows Calculus.
Just curious, but what parts of calculus apply here? Would you be willing to give a short explanation at least 😂
@@Megadumbyogyou can imagine the volume of the pyramid as being the same as a bunch of squares layered on top of each other with infinitesimally small height. Let's say each thin layer has a volume dv. dv = length * width * dh, where dh is the height of each layer. Now it's just a matter of getting the length and width as functions of the original length, width and height, then integrating both sides of the equation.
@@Megadumbyog It's basically to sum the infinite squares that form the pyramide, which is the same concept as integrating. After some logical thinking u end up with a quite ez integral that leaves u the volume formula. The cone volume can be proved through the same logic.
Integral from 0 to 1 of (x^2 dx) = 1/3. It's because doing an infinite sum over a power like x^r in calculus be ones 1/(r+1) * x^(r+1). It's just the way infinite sums behave.
Bro the way I understand this is there's a 1/4 pyramid at the corners and there are 4 corners so 4/4 but that's just a whole prism so u subtract 1/3 from 4/4 so it's 3/3 - 1/3 = 2/3 the sum of all the empty space around the pyramid. It's not actually 1/4 pyramid it's 1/6.
These are both true, but I'm not sure how much they help build an intuitive understanding of this. Explanation number 1 relies on the fact that halving the height of the pyramid halves its volume, which is true but is not immediately obvious when looking at a pyramid (as it would be for a rectangular prism, for example). Explanation #2 has the same issue with the fact that skewing a pyramid doesn't change its volume if you keep it's height constant (again this is true, but it's not something that really *feels* true without its own proof). Not to mention that there's a much easier proof using that same idea (create a square out of three pyramids whose bases sit on mutually perpendicular faces and whose points all meet in one corner, together all three fill the entire volume of the cube).
Yes your comments are fair sir. I wanted to animate the shear transformation breaking the pyramid into many many tiny cubes, then shifting them all sideways. But couldn't get manim to render that many objects. Maybe one for the future...
@@nicemathproblems I say that because I know my students struggle with that even in 2 dimensions. A lot of them seem to think that A=1/2bh only applies to right triangles.
I saw someone else mention integration (as a clue to where the 1/3 comes from) and I think that's conceptually the easiest way to see why skewing doesn't affect area/volume. If you imagine breaking the shape up into a bunch of very small layers which you're adding up all that matters is how many layers there are (the height) are and how wide the layers are (the base). Shifting them side to side won't affect the area.
Moving the pyramid point to a corner and the show that 3 copies of this new pyramid makes a full cube.
You can also divide the cube into 3 pyramids with their bases on the bottom, left and back side of it. It is harder to visualise tho and the first method might be the easiest to imagine
Yes that's a good method too
can be seen through calculis too, integral of x squared is 1/3 of x cubed
Good point, but I don't see how does that relate directly to the volume of the pyramid?
assume height is proportional to side length of cross section, and area is proportional to side length squared (both true in pyramids), then integrate with respect to height@@nicemathproblems
@@nicemathproblems x squared is the cross-sectional area of the pyramid, which is a square. integrating the cross-sectional area across the height gives the total volume
Is it not easier to move the summit to one of the corner (same volume) and then see we can divide the cube in three identical part?
Yes you are right, that it is a good method
Great video
Thanks Aadi
Didn't think i would understand the formula! thanks
Glad to hear it, thanks Ctrl Z
Nice but I want the same for the volume of a cone so bad.
Sub A for πr². Badabing, badaboom.
Yea I get that but visualization would be better you know@@gachanimestudios8348
@@gachanimestudios8348 yeah I get that but the visualization is necessary you know
@@Player_is_I Turn the cube into a cylinder. Done and dusted.
Consider the volumes of decagonal pyramids and chiliagonal pyramids. They are increasingly close approximations to cones.
This made me confused on a topic ive literally never doubted or thought about, which are some of the rarest topics know to me.
I like this way of visual reasoning, a skill we are often missing out on because schools mostly teach us how to reason algebraically, and using language. Of course he used language here to describe what was going on, but the process of seeing this if you were in the seat of someone figuring this out would involve visualizing these types of basic relationships and drawing inferences from them.
Excellent video wonderful 😊😊😊😊😊😊😊❤❤❤❤❤
Aww thanks!! 🥰
Much better explanation. I was going to explain this with integrals, but this is so much easier to understand.
Well to explain it to my self I use integrals, saying, the base is on (0,0,0),(a,0,0),(0,b,0) if you understand coordinates its a right triangle, and for the other vertice I choos (c,d,h), so c an d are a number and h is the Height of the pyramid, if you slice the piramid you get a right triangle and if you get de functions of the respective values integrating from 0 to h respect of z the final answer its
(abh)/6=(h/3)(ab/2) and (ab/2) its the area of the base, so I can say every pyramid with a base of a right triangle has this formula, but now if you can divide every convex polygon on right triangles and the formula works too, then because the cicle cant be triangulated I use anothe integral for that
The integral is
[(a-(a/h)z)(b-(b/h)z)]/2 that respect to z
From 0 to h
Is it true that for all dimensions the hypertriangle volume would be 1/d the volume of the hyperprism? Because you can see a relationship between how in 2d it is half the area and in 3d it's a third of it.
Excellent question. Anyone?
Yes, since for the n-th dimension, you can calculate it’s volume by summing up all the squares from bottom up with the definite integral of x^(n-1)dx from 0 to the cube’s side length a, which always gives you a^n/n = V/n
@@janmackovcak thanks
yes, you can repeat the same construction with the 6 pyramids, but it would be 8 hyperpyramids for 4d, and so on
yep
I always imagined moving the peak over to a corner and fitting two more with their peaks at that same corner, and the extra pyramids added would be the same because they can be reflected over the plane from the shared edge to the opposite edge (which the shared corner is on)
Thanks Mango Chicken I think I can visualise what you mean 😅
Same thing applies to cylinders and cones.
The calculus proof (integrating the cross-sectional area across the height) also can be used to prove that the volume of any n-sided pyramid (or cone) is 1/3 the volume of its respective prism or cylinder since the area of any polygon is proportional to its length squared.
True. At the end of the day, Calculus is King
Cut the pyramid in half and stick those halves on two sides, then do that again for the other two sides, three pyramids in the cube in total
What's crazy is I actually did think about this earlier today
There's a word for that. Serendipity I think
I've always thought of it as taking an integral of x^2 by chopping the pyramid into horizontal slices, but this is much more intuitive!
Thanks, you know I can't visualise how the integral of x^2 would give a pyramid!
@@nicemathproblemsA square pyramid with height 1 in a cube with sides 1. The cross section is a square with sides that has equal length as the distance between the cross section and the top side. Using this we get that the square pyramid has a volume of int 0to1 x^2 which gives us 1/3.
Bro the music makes me feel like im watching a shitpost
I just did calculus. Proved the volume/surface area of a sphere and the volume of a cylindrical cone and a pyramid
ive learned alot of stuff on youtube, history, science, dinosaurs, and even electronics. i still have no idea whenever i will ever use this kind of knolage
and then the cone attacked
True, neither of those methods can be applied to a cone 😅
Wow ! I always asked this question to myself
I literally had a test last week and one of the questions was to find the volume of a square-based pyramid, WHERE WAS THIS VIDEO THEN????!!!
🤣
Recently I was having a thought about a pyramid which is slanted so the apex is directly over one of the base points. I realized after some time that you could fit an entire other pyramid upside down in the same rectangular prism volume, with extra space. I was not certain but I guessed that the remaining space was of the same volume as the two pyramids. Imagine instead the principal line was the diagonal which the two pyramids share, then the half segments on either side are of equal volume to half of a pair of pyramids which point in opposite directions…
Yes you're absolutely right, I should have included that visualisation
I pretty much did that to figure out why a right cone is Pi *h* r^2 *1/3
OHHHH I BEEN WONDERING ABT IT FOR A LONG TIME THANKS
There’s my thought process
Find an equation relates A and h
A=k*h^2
Then do integral((k*h^2)dh)
=1/3k*h^3
=1/3(k*h^2)h
=1/3A*h
bro is the next 3blue1brown
everyone is
i think there's a subtle aspect here about why moving the apex parrallel to the base doesn't change the volume...
Because bh is constant. It's circular reasoning but here we are
Awesome
Another way is to fill the triangle with water and pour it in the rectangular prism, it will only fill 1/3 of the prism.
This whole video i was just like 👁 👄 👁
Thanks and I like your use of emojis 🤣
And while those two are like super obvious, my teacher tried to cut a prism into tree piramids... And the blackboard was a mess of lines and nobody understood zit 😢
Haha go easy on your teacher. Neither of these diagrams would be easy to visualise on a whiteboard! 😅
Yea... Sorry... 😅
In fact, was the best teacher I had and the one who made maths interesting to me as a whole...
But that day I was so dizzy... It was confusing as the three pyramids have to be deformed to fit into a prism... The ones presented here may be easier to understand even without drawing them.
If inside of a cube can have 6 pyramids with half of the height then it's simple to prove
base's area × height = cube's area
Small pyramid's area = base area × height ÷6
Pyramid's area = 2 small pyramids area
Thank you now I don't understand even more... 💀💀
My brain went:
Can you try explaining it a bit slowly ,maybe make a longer video
So we take one day from each phases in order, skipping the discount on the seed phase if it's gets to only one day, until there are no more days left to discount ?
i would only focus on proofs that can generalize to higher dimensions, like your first one
Ok that's amazing
Just move the apex to one of the cube's top corners.
If the apex of a pyramid is directly above its square base, and the height of that pyramid equals a side of the base, then three of those pyramids fit together to make a cube.
Directly above a corner of its square base....
The pyramid formula keeping you big as hell
Great video, but i feel like it was kinda all over the place. Could have been simpler but love the bisual proofs
All squares are rectangle not all rectangles are squares. That's what i was told in class
Your teacher did not lie! 😊
A square is not a rectangle. Both squares and rectangles are quadrilaterals. A rectangle is a quadrilateral with four parallel sides and two sets of lines of two different lengths. A square is a quadrilateral where each side is of equal length and 2 sets of parallel lines.
twice the volume of 1/6 of a cube
If it were 1/2 then we would not get a pyramid, but a triangular prism. For it to form a pyramud, and to get the geometric object to be pointed, it needs to be 1/3.
"Understandable, have a great day"😂
(Don't take this seriously😅)
I saw the pyramids, then I saw a mystery, after that I solved it, but I forgot he same time? Are the Egyptians coursed the pyramids or not?
I feel like explains how the integration method works with this is easier.
I learned more than 10 years school
can you do that for a cone?
Do it when the base is an n sided regular polygon, then let n go to infinity, and you've proven it for a cone
I proved this almost a year ago but these were way more satisfying than my proof, very cool lol
Cone please?
I dare you to prove the surface area of a sphere
🙀😅
It's just the cone formula applied for a square bade
Base
Calculus is best way to understand
I am not convinced. I want more math to prove that statement. What is h in the equation? Why not just V=1/3A?
In the formula on top of the screen at the end, aren't you implying 1/4 + 1/3 = 1/2? Correct me if I'm wrong but I'm confused.
Imagine watching this high? 😅
The first one wasn't that intuitive 😕
Integrating -> OP
So real
Nah...
My brain is to small for wibbly wobbly lubba dub dub bullshit. 💀💀💀
geometry dash
Or just use algebra with integrals to get this formula
but i think it can't be generalized
Heh
questionable assumptions. they are true maybe but voming from space
That doesn't make sense