Some people have commented that they don’t understand the point or that this solution is far too complex, since the torus is just a cylinder bent into a full circle. It’s a good thing to recognize that similarity and then form a hypothesis that their volumes would therefore be the same. But to automatically assume that as fact without first proving it is a dangerous thing. There are many examples in mathematics of things that seem obvious but end up being very counterintuitive. One example is the paradox described in my “Belt Around the Earth” video. I don’t really expect anyone to actually use this method if they genuinely need the volume of a doughnut. This is just a fun proof that demonstrates many important concepts in calculus. Thank you all for watching and for all the supportive comments.
My first thought was that too, but after giving a second thought, I realize that the inner circle have smaller volume than the outer circle, hence you can't just think of it like a typical cylinder The correct way of viewing it is a cylinder with 1 side higher than the other, it would therefore won't be a cylinder And using this logic, u can actually find out the volume. Since the outer side of a torus has more volume than the inner side of the torus, we can just use the mean of both Rs (inner R and outer R), and use the mean R to get the mean R perimeter. This is the reason the big R in the video is in the middle So the breakdown of the volume is z=2πr V=πr²*z z is perimeter of the mean R. Exactly like the formula in the video
hey LearnPlaySolve Im sorry for posting that comment against you at first I thought this was a like one of those click bait 10,000,000 view videos but you actually did a really good job explaining the integral calculus of solving volumes when thier solutions might not be so intuitive. Keep it up!
Beautifully derived! Thanks! I have not worked on this before, but from general volume formula (Base area * Height), if we cut the donut and straighten it up in cylinder shape, the Base area becomes the area of a small circle where as the Height that passes through the center of the donut cylinder becomes the circumference through the center of the donut. That seems to finalize the donut formula in short in seconds.
This was the most complex explanation I ever saw, for a very simple problem. Just cut the donut such that you will get a Cylinder with a base area of pi*r^2 and a length/height of 2*pi*R.
Yeah, the thing is to also prove that it works that way. Because visual representation isn't always enough, sometimes it can lead you to wrong formulas
The inside (minimum) radius of a torus, Rmin, is less than R, and the outside (maximum) radius, Rmax, is greater than R, so we know that the volume of the torus has to be between (pi*r^2)*(2*pi*Rmin) and (pi*r^2)*(2*pi*Rmax), but why would we expect it to be exactly (pi*r^2)*(2*pi*R)? If you can explain that in a simple but convincing way, then I will agree that it is a very simple problem.
@@smalin yeah I had the same thought before starting the problem. I thought the different perimeters outside and inside the circle would have an effect on the volume so I integrated horizontal slices of the torus: rings of different width, which gave me the same result in the end.
It had me wondering why it is exactly this nice 2πr*πr² which would seem like a naïve first guess. Then it occurred to me to think in terms of a bunch of circles, and noting that the circumference of a circle is a linear function of its radius. Thus, for each circle of radius R+r, there is a corresponding circle of radius R-r, whose combined circumference is 2π(R+r+R-r)=4πR. Essentially, we can treat the torus as if it is all concentrated exactly on the circle of radius R at the centre of the torus, and get the right answer.
Can you explain why you can change the limits like you do at 3:10? Is it literally just because r was used as the limit for the initial integral in terms of x?
It's funny I did it the other way around : I integrated "vertically" rings of area 4*pi*R*(sqrt(r^2-y^2)) where y varies between -r and r and i get the same result! I thought integrating along the circle would maybe cause problems since the "speed" at which each side of circle move but I guess not 😅
Yea I can see how that's a little confusing now that you point it out. I probably should've put parentheses around the (1+cosu). The differential du always belongs to every part of the integral, not just the last term.
@@LearnPlaySolve mmmm, i mean, if you factor out 1/2 of "1 + (cos u)/2" the resull is 1/2(2 + cos us). But you factor out 1/2 of "1 + (cos u)/2" and you let as "1+ cos u" I m sorry, that realy confuses me. It something that i am not seeing why
You don't factor it out of the (1+cosu) because it was never distributed to those terms. The 1/2 is only with the du, not the cosine. Its just a matter of moving the 1/2 from the end of the integral to the beginning.
If your integral is an algebraic expression that "resembles" a trigonometric identity, then you can perform a substitution which turns it into a trigonometric expression. This allows you to exploit certain properties of whatever identity it is, and rewrite it in a way that is far easier to integrate. I hope that helps.
Thank you for your question. Anytime you solve an integral via substitution, you must remember to also change the differential. In this case, since we substitute x with rsin(theta), we take the derivative of both sides of that substitution to get a new differential. x becomes dx, and rsin(theta) becomes rcos(theta)dtheta. Now we have something in terms of theta to replace the dx. Remember, the derivative of sine is cosine. I hope this helps.
Great video! I need to calculate a volume for a sector not a full circle. Example would be a pizza slice from point R to 1pm and 2pm (on a clock face) then rotated around the x asix just like the torus. Any idea for that formula or where I could look for help?
Thank you very much! That’s an interesting problem, and it sounds like a fun one to solve. You would have to separate it into two separate integrals at the point where the straight line of the sector (radius) intersects the curved part (arc). It’s kinda hard explain in just words. I might make a video for that kind of problem in the future.
@@LearnPlaySolve www.dropbox.com/s/i38t3g3ugupfbhd/Circle%20Sector%20Torus.PNG?dl=0 That's a quick sketch I've just made, in case my technical terms about pizzas wasn't clear :) I can easily calculate the volume with CAD but I want to be able to do it with formulas. Your video is the best explanation I've found with the step by step workings, but I'm stumped on how to apply it to my problem.
Pappus-Guldinus Theory, that's what I need. I can calculate the area, find the centroid point rotate it into its correct position then just do V = A 2pi y where y is the centroid distance from the x axis
To be honest, I’m not familiar with the Pappus-Guldinus Theorem, but I’m excited to learn something new about solids of revolution, especially if the theorem provides a faster and easier way of finding their volumes. So thank you for bringing that to my attention.
Very helpful video, thank you. May I use some of the images and drawings from this video for a school project? They are the best ones I've found so far and it would help me a lot.
The part where you describe plugging the equations for the top and bottom half of the circle was confusing; it was not clear what "it" you were referring to when you said "under it".
Sorry about that. I should have explained better. The top equation represents the top half of the circle. So using it in the equation means you are taking all of the area under the the top half of the circle, all the way down to the x-axis, and revolving that area around the x-axis to end up with a certain volume. Likewise, if you do the same thing with the bottom half of the circle, and then subtract that from the previous answer, you will be left with the inside of the circle rotated about the x-axis. I hope that makes more sense.
Hmmm, if I have a hosepipe of internal radius r and it is laid in a straight line of length 2.pi.R, then its volume would be 2.pi.R.pi.r.r - what your result is. If I arrange the hosepipe in a circle so that its end touches its start, the outer edge travels more than 2.pi.R going around the circle and in inner edge less than 2.pi.R (actually 2.pi.(R+r) and 2/pi/(R-r)). It seems that the lesser volume of the inner section (less than R from the centre of the circle) is exactly compensated by the extra volume from the section greater than R from the centre of the circle. A lucky coincidence?
Probably not a coincidence. However, what if you made that assumption before you really knew whether it was true or not. How would you prove it? Many things in math may seem obvious but turn out to be counterintuitive.
That’s a very good question, and the answer is yes, but only under very specific conditions. If you had an equilateral triangle and it was oriented just the right way, and the radius went up to the exact center of the triangle, then yes it would work. But such restrictive conditions in my opinion make it an unreliable formula for anything other than a circle.
@@LearnPlaySolve funnily enough, I’ve actually tested it using spaceclaim (3D modelling software with a volume function) and the equation works for any triangle and I think any shape. You do need additional information as you suggested. The surface area of the shape is one, but the most important information is the centre of mass of the shape. So for a triangle the centre of mass is H/3, so if your triangle is pointing out of the donut it’s R+H/3 in if the triangle is pointing in its R+2H/3. Where R is the radius of the donut in your equation. Thanks for your response.
Very interesting. Thank you for sharing it. I tried with an isosceles right triangle oriented with the hypotenuse down. It works, but if I rotate it so one of the legs is down, my radius has to change to accommodate the new location of the triangle’s center.
The way I did it was by thinking of it as a flexible tube, cutting it in one place then finding the area of this cylinder, which has a length of 2piR and a cross-sectional area of pir^2, thus giving 2R(pi r)^2
That's a great point! Although we can substitute anything for x that helps us evaluate the integral, it would have been less confusing to use r cos theta. Thanks for the input.
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
Some people have commented that they don’t understand the point or that this solution is far too complex, since the torus is just a cylinder bent into a full circle. It’s a good thing to recognize that similarity and then form a hypothesis that their volumes would therefore be the same. But to automatically assume that as fact without first proving it is a dangerous thing. There are many examples in mathematics of things that seem obvious but end up being very counterintuitive. One example is the paradox described in my “Belt Around the Earth” video. I don’t really expect anyone to actually use this method if they genuinely need the volume of a doughnut. This is just a fun proof that demonstrates many important concepts in calculus. Thank you all for watching and for all the supportive comments.
My first thought was that too, but after giving a second thought, I realize that the inner circle have smaller volume than the outer circle, hence you can't just think of it like a typical cylinder
The correct way of viewing it is a cylinder with 1 side higher than the other, it would therefore won't be a cylinder
And using this logic, u can actually find out the volume. Since the outer side of a torus has more volume than the inner side of the torus, we can just use the mean of both Rs (inner R and outer R), and use the mean R to get the mean R perimeter. This is the reason the big R in the video is in the middle
So the breakdown of the volume is
z=2πr
V=πr²*z
z is perimeter of the mean R. Exactly like the formula in the video
hey LearnPlaySolve Im sorry for posting that comment against you at first I thought this was a like one of those click bait 10,000,000 view videos but you actually did a really good job explaining the integral calculus of solving volumes when thier solutions might not be so intuitive. Keep it up!
lol no worries! Thank you so much for watching and commenting. 🙂
I was working with tori for a math paper and I must say I have not found a derivation that is this well explained! Kudos 👏
Beautifully derived! Thanks! I have not worked on this before, but from general volume formula (Base area * Height), if we cut the donut and straighten it up in cylinder shape, the Base area becomes the area of a small circle where as the Height that passes through the center of the donut cylinder becomes the circumference through the center of the donut. That seems to finalize the donut formula in short in seconds.
Thanks a lot, helped me tremendeously with a similar issue on integrating on a shape that is not touching x-axis.
x-axis?
how bout you touch some grass?
/jk
This was the most complex explanation I ever saw, for a very simple problem. Just cut the donut such that you will get a Cylinder with a base area of pi*r^2 and a length/height of 2*pi*R.
It's easy to say that in hindsight, because of the result. This proves why that formula works.
Yeah, the thing is to also prove that it works that way. Because visual representation isn't always enough, sometimes it can lead you to wrong formulas
The inside (minimum) radius of a torus, Rmin, is less than R, and the outside (maximum) radius, Rmax, is greater than R, so we know that the volume of the torus has to be between (pi*r^2)*(2*pi*Rmin) and (pi*r^2)*(2*pi*Rmax), but why would we expect it to be exactly (pi*r^2)*(2*pi*R)? If you can explain that in a simple but convincing way, then I will agree that it is a very simple problem.
@@smalin yeah I had the same thought before starting the problem. I thought the different perimeters outside and inside the circle would have an effect on the volume so I integrated horizontal slices of the torus: rings of different width, which gave me the same result in the end.
U could also use Area of a semi circle for the integral of sqrt(r^2-x^2) with limits -r to r
It had me wondering why it is exactly this nice 2πr*πr² which would seem like a naïve first guess. Then it occurred to me to think in terms of a bunch of circles, and noting that the circumference of a circle is a linear function of its radius. Thus, for each circle of radius R+r, there is a corresponding circle of radius R-r, whose combined circumference is 2π(R+r+R-r)=4πR. Essentially, we can treat the torus as if it is all concentrated exactly on the circle of radius R at the centre of the torus, and get the right answer.
How did you put that square sign on your r in 2.pi.r*pi.r?
Just helped me with a calculus project thanks!!
I’m pleased to hear that!
影片的化簡、算式很厲害，花非常多時間處裡畫面的流暢
用來複習很方便；但對初學者來說，有點吃力，要常按暫停思考
underrated channel
Wow thank you very much!!
Bro I loved your explanation 🥰
Thanks for the video. I searched for many videos and this one is the easiest to get!!!!Love it
great animation and thanks for your help
Why am I watching this, I already know this from way back... But btw good explaination
Now I want to know how to calculus the volume of a donut made from donut (let's call it a tube torus)
Very helpful thx for the video❤
Can you explain why you can change the limits like you do at 3:10? Is it literally just because r was used as the limit for the initial integral in terms of x?
It's funny I did it the other way around : I integrated "vertically" rings of area 4*pi*R*(sqrt(r^2-y^2)) where y varies between -r and r and i get the same result! I thought integrating along the circle would maybe cause problems since the "speed" at which each side of circle move but I guess not 😅
That's actually very cool out-of-the-box thinking! 😃
Hi, at minut 3:51 you factor out the 1/2 but what happen whit the one "1" before cos(u) du??
Yea I can see how that's a little confusing now that you point it out. I probably should've put parentheses around the (1+cosu). The differential du always belongs to every part of the integral, not just the last term.
@@LearnPlaySolve mmmm, i mean, if you factor out 1/2 of "1 + (cos u)/2" the resull is 1/2(2 + cos us). But you factor out 1/2 of "1 + (cos u)/2" and you let as "1+ cos u" I m sorry, that realy confuses me. It something that i am not seeing why
You don't factor it out of the (1+cosu) because it was never distributed to those terms. The 1/2 is only with the du, not the cosine. Its just a matter of moving the 1/2 from the end of the integral to the beginning.
@@LearnPlaySolve yep, i see that, i realy was confusing, haha, thank you, you are the best
hi, sir. I have a question. What's the purpose of doing Trigonometric Substitution in the integration process?
If your integral is an algebraic expression that "resembles" a trigonometric identity, then you can perform a substitution which turns it into a trigonometric expression. This allows you to exploit certain properties of whatever identity it is, and rewrite it in a way that is far easier to integrate. I hope that helps.
Would this also kind of like help that integrating the circumference gives you the area?
Hi, thank you for the helpful video! Can you explain why dx=rcos theta d theta at 2:53?
Thank you for your question. Anytime you solve an integral via substitution, you must remember to also change the differential. In this case, since we substitute x with rsin(theta), we take the derivative of both sides of that substitution to get a new differential. x becomes dx, and rsin(theta) becomes rcos(theta)dtheta. Now we have something in terms of theta to replace the dx. Remember, the derivative of sine is cosine. I hope this helps.
@@LearnPlaySolve got it, thank you!
Thank you so much
Great video!
I need to calculate a volume for a sector not a full circle.
Example would be a pizza slice from point R to 1pm and 2pm (on a clock face) then rotated around the x asix just like the torus.
Any idea for that formula or where I could look for help?
Thank you very much! That’s an interesting problem, and it sounds like a fun one to solve. You would have to separate it into two separate integrals at the point where the straight line of the sector (radius) intersects the curved part (arc). It’s kinda hard explain in just words. I might make a video for that kind of problem in the future.
@@LearnPlaySolve www.dropbox.com/s/i38t3g3ugupfbhd/Circle%20Sector%20Torus.PNG?dl=0
That's a quick sketch I've just made, in case my technical terms about pizzas wasn't clear :)
I can easily calculate the volume with CAD but I want to be able to do it with formulas.
Your video is the best explanation I've found with the step by step workings, but I'm stumped on how to apply it to my problem.
Pappus-Guldinus Theory, that's what I need.
I can calculate the area, find the centroid point rotate it into its correct position then just do
V = A 2pi y where y is the centroid distance from the x axis
To be honest, I’m not familiar with the Pappus-Guldinus Theorem, but I’m excited to learn something new about solids of revolution, especially if the theorem provides a faster and easier way of finding their volumes. So thank you for bringing that to my attention.
ruclips.net/video/ZQv-eF80FA0/видео.html&ab_channel=MichelvanBiezen
A series of videos by Michel van Biezen.
Thank you :)
You are genius
clear and nice
Very helpful video, thank you.
May I use some of the images and drawings from this video for a school project? They are the best ones I've found so far and it would help me a lot.
Thank you for your kind words. I would be honored if you used it. Go right ahead!
@@LearnPlaySolve, thank you very much. How should I quote you?
Here’s the link to an article I found on how to properly cite youtube videos: chat.library.berkeleycollege.edu/faq/166951
@@LearnPlaySolve thank you again
The part where you describe plugging the equations for the top and bottom half of the circle was confusing; it was not clear what "it" you were referring to when you said "under it".
Sorry about that. I should have explained better. The top equation represents the top half of the circle. So using it in the equation means you are taking all of the area under the the top half of the circle, all the way down to the x-axis, and revolving that area around the x-axis to end up with a certain volume. Likewise, if you do the same thing with the bottom half of the circle, and then subtract that from the previous answer, you will be left with the inside of the circle rotated about the x-axis. I hope that makes more sense.
Very helpful :)
Hmmm, if I have a hosepipe of internal radius r and it is laid in a straight line of length 2.pi.R, then its volume would be 2.pi.R.pi.r.r - what your result is.
If I arrange the hosepipe in a circle so that its end touches its start, the outer edge travels more than 2.pi.R going around the circle and in inner edge less than 2.pi.R (actually 2.pi.(R+r) and 2/pi/(R-r)). It seems that the lesser volume of the inner section (less than R from the centre of the circle) is exactly compensated by the extra volume from the section greater than R from the centre of the circle. A lucky coincidence?
Probably not a coincidence. However, what if you made that assumption before you really knew whether it was true or not. How would you prove it? Many things in math may seem obvious but turn out to be counterintuitive.
@@LearnPlaySolve I love maths for this reason. Next postulate: is the volume of the hosepipe always 2.pi.R.pi.r.r, independent of it's shape?
That's a good question, and it really depends on what you mean by r. But, if by pi*r*r you mean the area of the cross section, then yes.
If we rotated a triangle around a point to make triangular looking donut, would the volume than be V=(base*height)/2 * 2*pi*R?
That’s a very good question, and the answer is yes, but only under very specific conditions. If you had an equilateral triangle and it was oriented just the right way, and the radius went up to the exact center of the triangle, then yes it would work. But such restrictive conditions in my opinion make it an unreliable formula for anything other than a circle.
@@LearnPlaySolve funnily enough, I’ve actually tested it using spaceclaim (3D modelling software with a volume function) and the equation works for any triangle and I think any shape. You do need additional information as you suggested. The surface area of the shape is one, but the most important information is the centre of mass of the shape. So for a triangle the centre of mass is H/3, so if your triangle is pointing out of the donut it’s R+H/3 in if the triangle is pointing in its R+2H/3. Where R is the radius of the donut in your equation. Thanks for your response.
Very interesting. Thank you for sharing it. I tried with an isosceles right triangle oriented with the hypotenuse down. It works, but if I rotate it so one of the legs is down, my radius has to change to accommodate the new location of the triangle’s center.
The way I did it was by thinking of it as a flexible tube, cutting it in one place then finding the area of this cylinder, which has a length of 2piR and a cross-sectional area of pir^2, thus giving 2R(pi r)^2
Read the pinned comment
I SAY WE MAKE DONUTS SQUARESSSSSSS
why x=rsin theata not rcos?
Either one would work. I just like to end up with a positive derivative.
As diagrammed, isn't x = r cos theta? The answer would still be the same, of course.
When factoring out the 1/2, shouldn't the 1 change to 1/2?
That's a great point! Although we can substitute anything for x that helps us evaluate the integral, it would have been less confusing to use r cos theta. Thanks for the input.
Or.. just cut through the donut, bend it straight, and take the volume of the cylinder?
Yes... this demonstrates why that works. Thank you.
Just cut it and make it a cylinder
Volume =base*altitude
=πr2* 2π*(Ro-R1)/2
Read the pinned comment
gurvir singh lol, your 7months l8. But cheers m8
its just a cylinder idk what the big deal is lmfao
I was also thinking the same...😂😂
@@circleoffifth9048 hes out here acting like he solved E =MC^2 and its just a tube fr lmfao
Please see my pinned comment
lol