This actually helped me wrap my head around calculus quite a bit. I understood the general concept but seeing it actually worked out this way makes way more intuitive sense than someone throwing a bunch of equations I've never heard of at me.
"When you hit those infinities, that's when your approximation becomes exact." 1:36 The only reason to learn calculus is so you can say things like this.
@@somebodyiusedtoknow2012 No I mean why does it come from real analysis more so than calculus, limits are literally one of the cornerstones of calculus
@@silverspringer3577 The limits used in calculus are from Weierstrass's rigorous definition of a limit. Weierstrass is considered to be the father of modern analysis.
@@fkncompton7124 Since the volume of solid shapes don't care about orientation, so is using either method. Of course, you'll need to get your values and equations right in the first place but that's a given.
Find the value of any shape with a rectangle: 1. Make an exact replica of the figure 2. Fulfill it with water 3. Make a rectangle replica that fits exactly all the water 4. Calculate the area of the rectangle
StarGazer45 Great idea! Even easier: just get any old rectangular box that can hold the amount of water, then just calculate the area of that base * height that the water fills!
It makes sense, but I think the beauty of math is that you can find out the volume of ANY shape. What if you want to calculate the volume of a swimming pool with a weird shape? The volume of a cloud? Or of an entire planet? All you need is math and imagination :) I'm not underestimating your comment, of course. Just sharing other way to see it
@@brogant6793 IMO u r not missing anything. I think it might be innocent mistake on his part while he was explaining the relationship between r and h. This thing can happen to anyone.
I've been watching your videos since I was 10. I've now revisited your channel again and it makes me so happy you're still making great content. You've made an impact in my life and I want to say thank you.
@@samuraijosh1595 I think so too; I’m not sure if we understand the statement though. The reason I think it’s supposed to be what you said is because r/R is a ratio. So if we are working with the 0.25 and 0.75 example again, you would need to do smth like 0.25 = 1 - 0.75. Since r/R and h/H can both theoretically never exceed 1, that works. Or at least that’s what I thought. The only way to rationalize this as an incorrect solution from what I can tell is that this isn’t what he’s solving for. There’s a chance we misunderstood what he’s doing on a fundamental level.
The exercise is very well explained. I want to make a contribution: The expression of r as a function of h comes from the fact that the triangle formed by the segments h and r is similar to the triangle formed by the segments H and R. The triangles are in the position of Thales's first theorem, so the expression comes from the similarity of triangles, and then it is automatically true that h/H = r/R. I explain this because it may be a bit difficult for someone to understand the proportional relationship between the triangles. Greetings to all.
you can also use H and R to form a linear function which have the x and y intercepts of the values R and H, then take the volume of revolution by integrating the linear function squared then multiplying it by PI for the radius. Though, the way you taught it is way more intuitive and beginner friendly. It helped me learn more of what an integral really is, thank you!
I really appreciate this video! As someone who struggled with calculus in school, this breakdown of how to find the volume of any shape using calculus is incredibly helpful. The way the author explains the concepts is so clear and easy to follow. I particularly enjoyed the explanation about how hitting infinities is when the approximation becomes exact. It's amazing how math works! I also love the comment about how the only reason to learn calculus is so you can say things like that. Overall, this video has helped me understand calculus a lot better, and I can't wait to apply these concepts in real-life situations. Thank you, Domain of Science, for another excellent video!
Great video and one of my favourite pieces of math. It's also very related to my last years high school research project, where i found an elementary(using just high school math) way to derive formulas for volunes of all regular polytopes and 1 formula, that directly gives volumes of all platonic and archimedean solids except the snub cube and snub dodecahedron, that were a little trickier. Moments of inertia are even more fun.
Instead of writing r=Rh/H i think it is simpler to write r=h×C, where C is just a constant. Integration looks the same because C is a constant, and u just plug in C=r/h in the end. I also rotated the "triangle" (the side view) 90° so it actually looks like inspecting a function over h, but that probably just personal choice. Great video, i liked it a lot, and u are right, this sort of question is very good to inspire thought and understand calc.
I only recently discovered your account on TH-cam and it has really helped me re open my mind to the world of maths and physics as this is something I want to study, you explain things very well and I thank you for doing what you do 🎉😊
Same thing except check this out: For any related rates problem go to the top right corner of your notebook paper and write this down: Height 1, Height 2, Height 3, Change in Height 1, Change in Height 2, Change in Height 3. Or even better: H1, H2, H3, dH1, dH2, dH3. You are given a height. You are asked to find a change in height. You usually need to solve for a height using pythagorean theorm if for example, it's the ladder problem (2D object means no H3 and so no dH3). I started doing my calc 1 that way over a decade ago and I received a complement from my professor. She had never seen that before. The faculty went around to try and find who taught me that and all they could figure out is one of the professors had seen it once before, when she was in college in the late 1970s, One of the grad student teaching assistants did that. It's the superior way to do Calc 1 and I'm the only person who I know personally who does it that way. Go write down what I just told you to write down and you'll see why immeditately. It should look like how you would write sin,cos,tan, and then to the right of those three; csc,sec,cot but H1,H2,H3, dH1,dH2,dH3
@Nova Flares yeah but if the height is zero, the answer should be R and not RH (which isn't even a length anymore), it has to be r(h)=R*(1-h/H) which fits more with his explanation. When h is 25% of H, then r is going to be 1-25%=75% of R and vice versa.
Had the same thought. Yes I guess so, since when h=H, r/R should equal 0, and that wouldn't be the case when H isn't 0 if you're using the formula from the video
I remember doing this back in high school. It was magical when I did it the first time. I did it for torus too then verified the answer with Wikipedia.
This was incredibly thorough and insightful, this is basically a representation of how to actually *DO* calculus. Not just read a textbook, memorize formulas, and regurgitate them on an exam.
A student asked me a question about a tetrahedron and figuring out a proof for the varying height is exactly what I needed. Thank you! I've been working on this for a few days on and off.
I tried to get a solution using the first equation, but I didn’t get the right results. Here is what I did: First you would divide both terms by R to make r = R(H-(h/H)), then you would multiply by H to get (R/H)((H^2)-h)). Now you would create constants to make it more manageable: a = R/H and b = H^2 results in the final equation, r = a(b-h). Since we are going to square this, we will instead have a^2 = (R^2)/(H^2). For the second term, squaring gives (b^2)-2bh+(h^2). The constant (a^2) would be moved to the outside in integration, whereas we will have to integrate the other term. Using the power rule, we get (b^2)H - b(H^2) + ((H^3)/3). Expanding and simplifying constant a and b results in pi*(R^2)*((H^4/H) - (H^2) + (H/3)). This results in pi*(R^2)H*(1/3) * (3(H^2) - 3H + 1), which is not the intended result. WolframAlpha gets the same result, so I don’t think I integrated incorrectly. Any ideas?
to be 100% clear to viewers, you should include the missing index in your finite sum, as well as a definition of the sequence {h_i} for i = 1 to N, since until you get to the definition of the integral, it's not obvious that delta(h) stays constant (for any given sum) while h itself is changing (getting smaller as r gets smaller, as we move up the to the peak of the shape). In other words, work out the finite sum solution using a numerical example for some chosen delta(h) and choice of N.
I love all of these videos. My problem has always been the cognitive load of language. Listening to "as delta h gets close to zero" I realise that the image behind this language is math focussed. These terms have low cognitive load for the speaker. I just think of printing a shape in slices over time. When there is no change in the slices you get a cylinder. When there are changes in the slices you get sth else. But the changes can be categorised. If you print a shape from 0 to completion and you alter the area of the slices at a consistent positive rate, you get a disappearing cylinder like a pyramid. But you could do other things like consistently twisting about a centre. When I see it as a series of frames it's far easier to understand, and once I understand the concept the label for it can be arbitrary but we use agreed labels so we can speak and understand each other. I have always hated the tendency towards Latin and Greek words in our naming of things because they sound innately complex in English (but not so for example in Spanish). The word "integral" has one meaning as a noun and another different one as an adjective (crucial), but the word "whole" does not.
wow this helped a lot, I have already bought some of ur posters ( the physics, chemistry, math, donut of knowledge, chemistry, and math notation, lol I love ur vids )
My approach for the sphere was the formula for a circle in two dimensions, so r^2 + h^2 = H^2. The rest was just like the pyramide, but times two, because there are two half spheres in a sphere.
My idea for SurfaceArea (Sphere). (i know , it is illogical, but works on somewhere). What is the simplest object in 3 Dimension world ? it is a Triangular, has only 4 vertexes and 4 sides. ..... SurfaceArea(Triangular) = Area(Polygon of tri) x4 sides. Now, we think, this most simple object as a kind of "neutral element" in 3D. Next, Sphere is also a simplest object in 3D. So, we can think it is a variant of "neutral element", = a variant of Triangular ( or Philosophical Triangular). So, it must have 4 sides. ..... SurfaceArea(Sphere) = Area(Polygon of circle) x4 sides. --> (pi x r^2) x 4
The Egyptians used the first couple of steps of this method to work out the internal volume of their pyramids.. I think they turned it into a right angle triangle by moving the slices to left-align.
Could have set up as - V = π h/r Σ (r² Δr) And ended up in the same place, and then noticed that - h/r = tanθ _(θ=the slant angle)_ And then ended up with - V = ⅓ π tanθ r³ Everyone likes to say that you need height but it's not strictly true. You need one more piece of information beyond the base radius and that can either be the height or the slant angle. If you think in terms of the slant angle it's easier to scale the volume up or down for questions about doubling volume, for example, because then you can just increase height by a factor of ³√2.
About finding the volume of cones and pyramids, one of my math teachers once said: "If you can't sit on it, divide it by 3" Although it does not tell the reason behind the formula, it is a great way to remember it
Excellent video, but the bit with replacing r with an expression in terms of h was unnecessarily convoluted. It only works if you hold r and h in constant proportion (r/h = k for any r and h on the cone), which is true due to similar triangles. However, once you've made that assumption, all that math becomes redundant anyway. You can just replace r with kh, and substitute (R/H) in for k after doing the integration.
Can you explain why not? If he didnt use 30 60 90, how did he even know the relationship between r and h right from beginning when discussing “ half way up h is half way r” im paraphrasing.
8:14 I think the 'h' in the second situation is different from the 'h' in the first situation. In fact, the 'h' in the second situation is 'H-h' from the first situation. I don't quite understand this part. If the blogger sees this, could you please reply when you have time?
For the cone's volume, the middle of h doesn't not necessarily correspond with the middle of r as you can have "flatter" cones or "pointy/sharper" cones. I suggest you try to explain the calculation of cone volume by spinning a right triangle rather than stacked cylinders. Anyway, it's a very good job you are doing!
It's funny how using cylindrical cross sections approximate (converge to) the volume of a cone (and other objects) but do not approximate surface area. You need frustums to do that.
Can someone explain 1) How he knew “h as half way up the big H height coincided with an r thats half the big R” 2) How could he know any of these relations without using basic trig? He never gives credit to the use of trig at all.
Ooh. So calculus is the name of THIS method in english. I know that most people don't like calculus but for me it's much easier than most of geometry. I actually had to do my final exam in math about calculus and it was one of my easiest exams.😅
Opened my eyes to help with optimization problems, but my prof said my algebra is lacking and😢 rip lol, ill have to retake calc again. I can say i felt a bit motivated to do calculus after watching this, but its the weekend rn
Serveral non calculus ways offered here by our esteemed viewers..perhaps they are missing the point ."find the shape using calculus' is the task at hand.
just a little confused,,, how do you know that at half the height of the cone it will be half the radius? doesn’t this idea only apply to this specific cone with a gradient of 1? what if a cone has a different steepness with a gradient of 2 or 1/2? how would you go about adjusting the width of each disk as the cone thins out? or is a gradient of 1/1 part of the definition of a cone? thanks!
The easiest way to imagine this (imo) is with similar triangles. We have a similar right triangle for every triangle with r smaller than R. Each of these triangles is within the initial triangle sharing 2 sides (the definition of a similar triangle). So it follows that r/R = h/H, as it does in any similar triangle where the proportion of height to the base is preserved.
Hi, I know you made this 4 years ago and it's quite a stretch to even try asking, but do you think you could do this with say, a cow? Could you tell me the volume of a cow with calculus?
we could simply calculate the area of the triangle (1/2 × r × h). after that we make a revolution on the vertical axis at 360 degrees to find the exact volume. so ( 1/2 × r × h ) × ( π × r2 ) keep it simple lol
Although I can loosely grasp the concepts at my early stage of learning, I have a hard time believing I could have confidence in maintaining accuracy and logic throughout all the steps. In other words, it just looks like a lot of opportunity for a little mistake to slip in and ruin everything.
I get anxiety whenever I look at people holding pencils with their index finger all bend and crocked applying a lot of force. Why do people grab the pencils like that? Can you guys grab the pencils with a gentle finger?
This actually helped me wrap my head around calculus quite a bit. I understood the general concept but seeing it actually worked out this way makes way more intuitive sense than someone throwing a bunch of equations I've never heard of at me.
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"This is Math; we can do whatever we like." -somewhere around 5:00
4:55 *
Learning calculus during lockdown, Great timing
Nice one, great to see people making good use of lockdown.
this is what i did during last three months hahaha
Its how Newton invented it.
U remind me 3blue1brown
@Gente que le encanta estar mamando
"When you hit those infinities, that's when your approximation becomes exact." 1:36
The only reason to learn calculus is so you can say things like this.
It comes from Real Analysis.
@@metakatana What do you mean
DynastyGuy big brain math
@@somebodyiusedtoknow2012 No I mean why does it come from real analysis more so than calculus, limits are literally one of the cornerstones of calculus
@@silverspringer3577 The limits used in calculus are from Weierstrass's rigorous definition of a limit. Weierstrass is considered to be the father of modern analysis.
I so love your videos! Thank you. Geometric speaking: a pyramid always has a squared base. The shape with the triangular base is a tetrahedron.
you can also take a triangle through a y=mx+c function and then a volume of revolution of 360° about the x axis.
Wouldn't it have to rotate about the y axis
@@fkncompton7124 Since the volume of solid shapes don't care about orientation, so is using either method. Of course, you'll need to get your values and equations right in the first place but that's a given.
i 'm doing it like that. more simple
My trig precalc knowledge is weal but can you tell me what the formula would look like? Do we need to use unit circle?
Can you explain what you mean by taking a triangle “through” y=mx + b?
Find the value of any shape with a rectangle:
1. Make an exact replica of the figure
2. Fulfill it with water
3. Make a rectangle replica that fits exactly all the water
4. Calculate the area of the rectangle
Well that's easy.
StarGazer45 Great idea! Even easier: just get any old rectangular box that can hold the amount of water, then just calculate the area of that base * height that the water fills!
It makes sense, but I think the beauty of math is that you can find out the volume of ANY shape. What if you want to calculate the volume of a swimming pool with a weird shape? The volume of a cloud? Or of an entire planet? All you need is math and imagination :)
I'm not underestimating your comment, of course. Just sharing other way to see it
Erm, I think you meant 'volume'.
If you can't fulfill the water, you can also satisfy it.
IMO the expression for non inverted cone should be
r/R = 1 - (h/H)
Ranadeep i thought the same
Ranadeep his formula doesn’t give r=0 for h=H so it isn’t correct it gives r=HR-R which is wrong (unless I’m missing something here)
@@brogant6793 Exactly, must use the one from Ranadeep
@@brogant6793
IMO u r not missing anything. I think it might be innocent mistake on his part while he was explaining the relationship between r and h.
This thing can happen to anyone.
true!
I've been watching your videos since I was 10. I've now revisited your channel again and it makes me so happy you're still making great content. You've made an impact in my life and I want to say thank you.
Ah that's so nice of you to say. Thank you!
@@domainofscience At 6:51 shouldn't it be r/R = (1-h/H) I'm so confused...😭😭
@@samuraijosh1595 if R =1 and H=1 then you would be right. But he just used some value to better picture what he was doing.
@@samuraijosh1595 I think so too; I’m not sure if we understand the statement though.
The reason I think it’s supposed to be what you said is because r/R is a ratio. So if we are working with the 0.25 and 0.75 example again, you would need to do smth like 0.25 = 1 - 0.75. Since r/R and h/H can both theoretically never exceed 1, that works. Or at least that’s what I thought.
The only way to rationalize this as an incorrect solution from what I can tell is that this isn’t what he’s solving for. There’s a chance we misunderstood what he’s doing on a fundamental level.
Man keep this up, I’ve been learning a lot from this channel 👍
The exercise is very well explained. I want to make a contribution: The expression of r as a function of h comes from the fact that the triangle formed by the segments h and r is similar to the triangle formed by the segments H and R. The triangles are in the position of Thales's first theorem, so the expression comes from the similarity of triangles, and then it is automatically true that h/H = r/R. I explain this because it may be a bit difficult for someone to understand the proportional relationship between the triangles. Greetings to all.
Thanks a lot sir
I dont understand - if R and H are 4, then r is 1 then shouldnt h be 3? How come they are equal?
When I was in high school, this was one of my fav things about calculus. I was like " why didnt you teach us this earlier!".
you can also use H and R to form a linear function which have the x and y intercepts of the values R and H, then take the volume of revolution by integrating the linear function squared then multiplying it by PI for the radius. Though, the way you taught it is way more intuitive and beginner friendly. It helped me learn more of what an integral really is, thank you!
I really appreciate this video! As someone who struggled with calculus in school, this breakdown of how to find the volume of any shape using calculus is incredibly helpful. The way the author explains the concepts is so clear and easy to follow. I particularly enjoyed the explanation about how hitting infinities is when the approximation becomes exact. It's amazing how math works! I also love the comment about how the only reason to learn calculus is so you can say things like that. Overall, this video has helped me understand calculus a lot better, and I can't wait to apply these concepts in real-life situations. Thank you, Domain of Science, for another excellent video!
Great video and one of my favourite pieces of math.
It's also very related to my last years high school research project,
where i found an elementary(using just high school math) way to derive formulas for volunes of all regular polytopes
and 1 formula, that directly gives volumes of all platonic and archimedean solids except the snub cube and snub dodecahedron, that were a little trickier.
Moments of inertia are even more fun.
He has made a simple idea into a complicated one
An inconvenient into a possible one
That's math
It’s a basic calculus 1 problem
Well u can actually derive it non-calculusly but that is shitty as hell and require like 40 min proof so not anywhere simpler
@@davidaugustofc2574 inconvenient isn’t impossible in the first place
Instead of writing r=Rh/H i think it is simpler to write r=h×C, where C is just a constant. Integration looks the same because C is a constant, and u just plug in C=r/h in the end. I also rotated the "triangle" (the side view) 90° so it actually looks like inspecting a function over h, but that probably just personal choice.
Great video, i liked it a lot, and u are right, this sort of question is very good to inspire thought and understand calc.
I only recently discovered your account on TH-cam and it has really helped me re open my mind to the world of maths and physics as this is something I want to study, you explain things very well and I thank you for doing what you do 🎉😊
Same thing except check this out: For any related rates problem go to the top right corner of your notebook paper and write this down: Height 1, Height 2, Height 3, Change in Height 1, Change in Height 2, Change in Height 3. Or even better: H1, H2, H3, dH1, dH2, dH3. You are given a height. You are asked to find a change in height. You usually need to solve for a height using pythagorean theorm if for example, it's the ladder problem (2D object means no H3 and so no dH3). I started doing my calc 1 that way over a decade ago and I received a complement from my professor. She had never seen that before. The faculty went around to try and find who taught me that and all they could figure out is one of the professors had seen it once before, when she was in college in the late 1970s, One of the grad student teaching assistants did that. It's the superior way to do Calc 1 and I'm the only person who I know personally who does it that way. Go write down what I just told you to write down and you'll see why immeditately. It should look like how you would write sin,cos,tan, and then to the right of those three; csc,sec,cot but H1,H2,H3, dH1,dH2,dH3
At 6:38: shouldnt the formula be r/R = 1 - h/H?
H/H-h/H = 1-h/H . @snyper BRO
Nanak its not H/H - h/H, just H - h/H
@Nova Flares yeah but if the height is zero, the answer should be R and not RH (which isn't even a length anymore), it has to be r(h)=R*(1-h/H) which fits more with his explanation. When h is 25% of H, then r is going to be 1-25%=75% of R and vice versa.
@Nova Flares He said at 5:50, thanks, I was having the same question. Time-saving.
Had the same thought. Yes I guess so, since when h=H, r/R should equal 0, and that wouldn't be the case when H isn't 0 if you're using the formula from the video
I remember doing this back in high school. It was magical when I did it the first time.
I did it for torus too then verified the answer with Wikipedia.
Can u derive the surface area of torus?
This was incredibly thorough and insightful, this is basically a representation of how to actually *DO* calculus. Not just read a textbook, memorize formulas, and regurgitate them on an exam.
A student asked me a question about a tetrahedron and figuring out a proof for the varying height is exactly what I needed. Thank you! I've been working on this for a few days on and off.
I’ve tried to understand calculus using TH-cam. Honestly, this is the first video I’ve seen that tells us what calculus actually does.
I tried to get a solution using the first equation, but I didn’t get the right results. Here is what I did:
First you would divide both terms by R to make r = R(H-(h/H)), then you would multiply by H to get (R/H)((H^2)-h)). Now you would create constants to make it more manageable: a = R/H and b = H^2 results in the final equation, r = a(b-h). Since we are going to square this, we will instead have a^2 = (R^2)/(H^2). For the second term, squaring gives (b^2)-2bh+(h^2). The constant (a^2) would be moved to the outside in integration, whereas we will have to integrate the other term. Using the power rule, we get (b^2)H - b(H^2) + ((H^3)/3).
Expanding and simplifying constant a and b results in pi*(R^2)*((H^4/H) - (H^2) + (H/3)). This results in pi*(R^2)H*(1/3) * (3(H^2) - 3H + 1), which is not the intended result. WolframAlpha gets the same result, so I don’t think I integrated incorrectly. Any ideas?
to be 100% clear to viewers, you should include the missing index in your finite sum, as well as a definition of the sequence {h_i} for i = 1 to N, since until you get to the definition of the integral, it's not obvious that delta(h) stays constant (for any given sum) while h itself is changing (getting smaller as r gets smaller, as we move up the to the peak of the shape). In other words, work out the finite sum solution using a numerical example for some chosen delta(h) and choice of N.
Can you explain it like im five? Having trouble
Following what your caveat is.
I love all of these videos. My problem has always been the cognitive load of language.
Listening to "as delta h gets close to zero" I realise that the image behind this language is math focussed. These terms have low cognitive load for the speaker.
I just think of printing a shape in slices over time. When there is no change in the slices you get a cylinder. When there are changes in the slices you get sth else. But the changes can be categorised. If you print a shape from 0 to completion and you alter the area of the slices at a consistent positive rate, you get a disappearing cylinder like a pyramid. But you could do other things like consistently twisting about a centre.
When I see it as a series of frames it's far easier to understand, and once I understand the concept the label for it can be arbitrary but we use agreed labels so we can speak and understand each other.
I have always hated the tendency towards Latin and Greek words in our naming of things because they sound innately complex in English (but not so for example in Spanish). The word "integral" has one meaning as a noun and another different one as an adjective (crucial), but the word "whole" does not.
" Beautiful "
Love how you explain things. What kind of pencil is that? It's gorgeous.
I think it's MUJI Hexagonal Wooden Pencil
Of the top of my head, cylinder is base x height, cone and pyramids are 1/3(base x height) , circle is 4/3*pi*r^3
wow this helped a lot, I have already bought some of ur posters ( the physics, chemistry, math, donut of knowledge, chemistry, and math notation, lol I love ur vids )
My approach for the sphere was the formula for a circle in two dimensions, so r^2 + h^2 = H^2. The rest was just like the pyramide, but times two, because there are two half spheres in a sphere.
My idea for SurfaceArea (Sphere).
(i know , it is illogical, but works on somewhere).
What is the simplest object in 3 Dimension world ?
it is a Triangular, has only 4 vertexes and 4 sides.
..... SurfaceArea(Triangular) = Area(Polygon of tri) x4 sides.
Now, we think, this most simple object
as a kind of "neutral element" in 3D.
Next, Sphere is also a simplest object in 3D.
So, we can think it is a variant of "neutral element",
= a variant of Triangular ( or Philosophical Triangular).
So, it must have 4 sides.
..... SurfaceArea(Sphere) = Area(Polygon of circle) x4 sides.
--> (pi x r^2) x 4
The Egyptians used the first couple of steps of this method to work out the internal volume of their pyramids.. I think they turned it into a right angle triangle by moving the slices to left-align.
BRILLIANT... so the answer still looks like hieroglyphs... what is the volume in cm2?
Could have set up as -
V = π h/r Σ (r² Δr)
And ended up in the same place, and then noticed that -
h/r = tanθ _(θ=the slant angle)_
And then ended up with -
V = ⅓ π tanθ r³
Everyone likes to say that you need height but it's not strictly true. You need one more piece of information beyond the base radius and that can either be the height or the slant angle. If you think in terms of the slant angle it's easier to scale the volume up or down for questions about doubling volume, for example, because then you can just increase height by a factor of ³√2.
About finding the volume of cones and pyramids, one of my math teachers once said: "If you can't sit on it, divide it by 3" Although it does not tell the reason behind the formula, it is a great way to remember it
How could I not have heard of this channel?! Im in *LOVE* ! 🥰🥰
The real miracle of Calculus is how Liebniz and Newton developed it at the same time independently!
Glad TH-cam recommended me this video ❤
Excellent video, but the bit with replacing r with an expression in terms of h was unnecessarily convoluted. It only works if you hold r and h in constant proportion (r/h = k for any r and h on the cone), which is true due to similar triangles. However, once you've made that assumption, all that math becomes redundant anyway. You can just replace r with kh, and substitute (R/H) in for k after doing the integration.
aha, I got lost in his derivation (and I think there's a mistake???) but this explanation is so clear and simple.
Seeing the illustrations of shapes is so cool. I wish i understood a quarter of what he’s explaining. I extol mathematicians.
This was great, actual practical application to really drill a few things home.
Thanks for helping more people get to appreciate how calculus is smart and beautiful
Cool
And the less weird shaped 3 makes it 10 times better
Hey man nice video. How is the name of the operation you've done with the r/R = h/H ? What's the logic behind that?
Great vid! You could also use the 30, 60, 90 triangle to find the relationship between r and h.
Lol, no. Not all of this are 30 60 90 ones....
Can you explain why not?
If he didnt use 30 60 90, how did he even know the relationship between r and h right from beginning when discussing “ half way up h is half way r” im paraphrasing.
You know you're early when there are no views and no likes.
No shit, Sherlock
wow I never knew that, thanks
@@chrissmith1152 🤣🤣
one of the best video i ever seen! do more videos like this!
I really like a soft music in the back, helps me focus
8:14 I think the 'h' in the second situation is different from the 'h' in the first situation. In fact, the 'h' in the second situation is 'H-h' from the first situation. I don't quite understand this part. If the blogger sees this, could you please reply when you have time?
For the cone's volume, the middle of h doesn't not necessarily correspond with the middle of r as you can have "flatter" cones or "pointy/sharper" cones.
I suggest you try to explain the calculation of cone volume by spinning a right triangle rather than stacked cylinders.
Anyway, it's a very good job you are doing!
Nice! At first I was thinking that you were going to take the solid of revolution route but this works as well :)
My trig precalc knowledge is weal but can you tell me what the formula would look like? Do we need to use unit circle?
My man just explain high school student calculus better than their teacher
Thanks very ensightful!
BTW 08:19 has to do with Triangle Proportionality Theorem right?
This idea works only for a smaller class of all possible shapes (except when one defines shapes using boring functions only, i.e. smooth, integrable).
It's funny how using cylindrical cross sections approximate (converge to) the volume of a cone (and other objects) but do not approximate surface area. You need frustums to do that.
Spherical coordinates for the win! (when calculating the volume of a sphere)
no need, the volume of the sphere is 0
@@mirijason Sphere, ball, to-mah-to, tomato
Can someone explain
1)
How he knew “h as half way up the big H height coincided with an r thats half the big R”
2)
How could he know any of these relations without using basic trig? He never gives credit to the use of trig at all.
Please, always list the music played in the video! (With links if possible.)
Great Job!
Sorry for a belated question, but when you set up the equation r/R =... should it not have been r/R = (H-h)/H = 1 - h/H ?
6:36 r/R=H-h/H 3/6=6-3/6 if H=6 and R=6 7:49 r/R=h/H so 3/6=3/6
Hey, could you please post/do a video on map of scientific research? Thanks for the other map videos you guys have posted!
Ooh. So calculus is the name of THIS method in english. I know that most people don't like calculus but for me it's much easier than most of geometry. I actually had to do my final exam in math about calculus and it was one of my easiest exams.😅
Opened my eyes to help with optimization problems, but my prof said my algebra is lacking and😢 rip lol, ill have to retake calc again. I can say i felt a bit motivated to do calculus after watching this, but its the weekend rn
Sir,your video impressive.
Can you guide.
How can I use in finding volumes and areas of my daily life problems.
Or some other work using this calculus
Wow, this is pretty beautiful!!
Serveral non calculus ways offered here by our esteemed viewers..perhaps they are missing the point ."find the shape using calculus' is the task at hand.
just a little confused,,, how do you know that at half the height of the cone it will be half the radius? doesn’t this idea only apply to this specific cone with a gradient of 1? what if a cone has a different steepness with a gradient of 2 or 1/2? how would you go about adjusting the width of each disk as the cone thins out? or is a gradient of 1/1 part of the definition of a cone? thanks!
The easiest way to imagine this (imo) is with similar triangles. We have a similar right triangle for every triangle with r smaller than R. Each of these triangles is within the initial triangle sharing 2 sides (the definition of a similar triangle). So it follows that r/R = h/H, as it does in any similar triangle where the proportion of height to the base is preserved.
Very tricky setting up a ration to represent the relationship between radius and height
Why is this tricky?
Hi, I know you made this 4 years ago and it's quite a stretch to even try asking, but do you think you could do this with say, a cow? Could you tell me the volume of a cow with calculus?
we could simply calculate the area of the triangle (1/2 × r × h).
after that we make a revolution on the vertical axis at 360 degrees to find the exact volume.
so ( 1/2 × r × h ) × ( π × r2 )
keep it simple lol
What is the general procedure to find the volume of a general 3D solid shape that isn't symmetrical about any axis? (i.e. just a blob in space)
Nice explanation, Thanks!
that is fucking incredible
14:15 oh no thank you for making the vid
V=(h/6)(A1+A2+4Am)
Works in any solid shapes.
What separates r from dr and h from dh? Aren't they both smaller pieces of the larger constant: R and H? Why not do the sum of pi(dr)^2dh?
So clear and awesome!
Skip to 5:46 for the real essence of life!
Doubt- if R and H are 4, and r is 1 then h should be 3. For eg. How come r/R = h/H? Please help
Although I can loosely grasp the concepts at my early stage of learning, I have a hard time believing I could have confidence in maintaining accuracy and logic throughout all the steps. In other words, it just looks like a lot of opportunity for a little mistake to slip in and ruin everything.
Then you work it back and figure out your mistake! You’re not in an exam you’ve got all the time you need. Mistakes are normal 😅
For a sphere, what is the relationship between r and h? Is it r = sqrt(R^2 - h^2)
so the volume of a 4d square pyramid is a^3*h/4
What abt more complex and irregular shapes?
Dead useful vid, thanks mister. keep gooing 🎉🎉🎉
Which software are using for video and animation for figure
Is there any point in doing this in a GCSE exam? Or should I remember the formulas for method marks .3.
This is awesome, great work
I get anxiety whenever I look at people holding pencils with their index finger all bend and crocked applying a lot of force. Why do people grab the pencils like that? Can you guys grab the pencils with a gentle finger?
Hey bro can u make a mind map of electromagnetism pls
You can also do this with double integrals
They said for volume , use double integration . Is this a new approach ?
Thanks ..but i always find any music or TV on in the background distracting and hard to concentrate.
Please make the Map of Philosophy
prep for jee and this will be child's game then
To find the expression we can use Thales
thank you so much for the video....
Great video sir! Its rare that by the end of a math video I say, "that is cool." Inspiration for me, thank you.
then what about objects with more complex shapes like tetrahedral, trigonal biprymidal