What's the Volume of a Donut? Calculus

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  • Опубліковано 23 гру 2024

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  • @LearnPlaySolve
    @LearnPlaySolve  2 роки тому +26

    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.

    • @shikyokira3065
      @shikyokira3065 2 роки тому +4

      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

    • @marktolands9692
      @marktolands9692 2 роки тому +1

      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!

    • @LearnPlaySolve
      @LearnPlaySolve  2 роки тому

      lol no worries! Thank you so much for watching and commenting. 🙂

  • @ebrahimudaipurwala3753
    @ebrahimudaipurwala3753 11 місяців тому +5

    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 👏

  • @chalkao5071
    @chalkao5071 2 роки тому +4

    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.

  • @Laahustaja
    @Laahustaja 3 роки тому +6

    Thanks a lot, helped me tremendeously with a similar issue on integrating on a shape that is not touching x-axis.

    • @div_07
      @div_07 2 роки тому

      x-axis?
      how bout you touch some grass?
      /jk

  • @dan-lh3px
    @dan-lh3px 3 роки тому +2

    U could also use Area of a semi circle for the integral of sqrt(r^2-x^2) with limits -r to r

  • @patrickverschuren2879
    @patrickverschuren2879 2 роки тому +1

    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.

    • @LearnPlaySolve
      @LearnPlaySolve  2 роки тому +1

      It's easy to say that in hindsight, because of the result. This proves why that formula works.

    • @redtoxic8701
      @redtoxic8701 2 роки тому +3

      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

    • @smalin
      @smalin 2 роки тому +3

      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.

    • @zerglingsking
      @zerglingsking 2 роки тому

      @@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.

  • @Chalisque
    @Chalisque 2 роки тому +5

    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.

    • @JatPhenshllem
      @JatPhenshllem 2 роки тому

      How did you put that square sign on your r in 2.pi.r*pi.r?

  • @Ireikes
    @Ireikes Рік тому

    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?

  • @sinexitoalmiedo
    @sinexitoalmiedo Рік тому +1

    Hi, at minut 3:51 you factor out the 1/2 but what happen whit the one "1" before cos(u) du??

    • @LearnPlaySolve
      @LearnPlaySolve  Рік тому

      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.

    • @sinexitoalmiedo
      @sinexitoalmiedo Рік тому

      @@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

    • @LearnPlaySolve
      @LearnPlaySolve  Рік тому

      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.

    • @sinexitoalmiedo
      @sinexitoalmiedo Рік тому

      @@LearnPlaySolve yep, i see that, i realy was confusing, haha, thank you, you are the best

  • @Jasmine-ww7jl
    @Jasmine-ww7jl 3 роки тому +2

    Hi, thank you for the helpful video! Can you explain why dx=rcos theta d theta at 2:53?

    • @LearnPlaySolve
      @LearnPlaySolve  3 роки тому +3

      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.

    • @Jasmine-ww7jl
      @Jasmine-ww7jl 3 роки тому

      @@LearnPlaySolve got it, thank you!

  • @LearnMathwithMrJerry
    @LearnMathwithMrJerry 3 роки тому +1

    Thanks for the video. I searched for many videos and this one is the easiest to get!!!!Love it

  • @Sg190th
    @Sg190th Рік тому

    Would this also kind of like help that integrating the circumference gives you the area?

  • @jia-rungwang1215
    @jia-rungwang1215 Рік тому +1

    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

  • @jaked4398
    @jaked4398 2 роки тому +1

    Just helped me with a calculus project thanks!!

  • @reguret2976
    @reguret2976 3 роки тому

    underrated channel

  • @maholly2893
    @maholly2893 Рік тому +1

    hi, sir. I have a question. What's the purpose of doing Trigonometric Substitution in the integration process?

    • @LearnPlaySolve
      @LearnPlaySolve  Рік тому

      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.

  • @TristanEllison
    @TristanEllison 3 роки тому +2

    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?

    • @LearnPlaySolve
      @LearnPlaySolve  3 роки тому +1

      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.

    • @TristanEllison
      @TristanEllison 3 роки тому

      @@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.

    • @TristanEllison
      @TristanEllison 3 роки тому +1

      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

    • @LearnPlaySolve
      @LearnPlaySolve  3 роки тому +1

      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.

    • @TristanEllison
      @TristanEllison 3 роки тому

      ua-cam.com/video/ZQv-eF80FA0/v-deo.html&ab_channel=MichelvanBiezen
      A series of videos by Michel van Biezen.

  • @mostafaelsokkary2158
    @mostafaelsokkary2158 3 роки тому +2

    great animation and thanks for your help

  • @dave_lawrence
    @dave_lawrence 2 роки тому +1

    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?

    • @LearnPlaySolve
      @LearnPlaySolve  2 роки тому +1

      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.

    • @dave_lawrence
      @dave_lawrence 2 роки тому +1

      @@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?

    • @LearnPlaySolve
      @LearnPlaySolve  2 роки тому

      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.

  • @DerLiesl
    @DerLiesl 2 роки тому +1

    Now I want to know how to calculus the volume of a donut made from donut (let's call it a tube torus)

  • @Ambigious
    @Ambigious 2 роки тому +1

    Why am I watching this, I already know this from way back... But btw good explaination

  • @the_eternal_student
    @the_eternal_student Рік тому

    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".

    • @LearnPlaySolve
      @LearnPlaySolve  Рік тому +1

      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.

  • @zerglingsking
    @zerglingsking 2 роки тому +2

    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 😅

  • @pitbull_cruel
    @pitbull_cruel 3 роки тому +1

    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.

    • @LearnPlaySolve
      @LearnPlaySolve  3 роки тому

      Thank you for your kind words. I would be honored if you used it. Go right ahead!

    • @pitbull_cruel
      @pitbull_cruel 3 роки тому

      @@LearnPlaySolve, thank you very much. How should I quote you?

    • @LearnPlaySolve
      @LearnPlaySolve  3 роки тому +2

      Here’s the link to an article I found on how to properly cite youtube videos: chat.library.berkeleycollege.edu/faq/166951

    • @pitbull_cruel
      @pitbull_cruel 3 роки тому +2

      @@LearnPlaySolve thank you again

  • @ridwan6695
    @ridwan6695 2 роки тому +1

    Bro I loved your explanation 🥰

  • @George6r4
    @George6r4 3 роки тому

    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?

    • @LearnPlaySolve
      @LearnPlaySolve  3 роки тому +1

      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.

    • @George6r4
      @George6r4 3 роки тому

      @@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.

    • @LearnPlaySolve
      @LearnPlaySolve  3 роки тому +1

      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.

  • @Marco_Lucca
    @Marco_Lucca 2 роки тому +1

    Thank you so much

  • @I_brahimyalcin
    @I_brahimyalcin 3 роки тому +1

    clear and nice

  • @jbrathod5206
    @jbrathod5206 Рік тому +1

    Very helpful thx for the video❤

  • @r.guerreiro140
    @r.guerreiro140 Рік тому +1

    Thank you :)

  • @veronicalin2492
    @veronicalin2492 11 місяців тому

    why x=rsin theata not rcos?

    • @LearnPlaySolve
      @LearnPlaySolve  11 місяців тому

      Either one would work. I just like to end up with a positive derivative.

  • @vinuthomas7193
    @vinuthomas7193 2 роки тому

    As diagrammed, isn't x = r cos theta? The answer would still be the same, of course.

    • @vinuthomas7193
      @vinuthomas7193 2 роки тому

      When factoring out the 1/2, shouldn't the 1 change to 1/2?

    • @LearnPlaySolve
      @LearnPlaySolve  2 роки тому

      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.

  • @Mathematics_tv
    @Mathematics_tv 2 роки тому +1

    You are genius

  • @joonatan003
    @joonatan003 3 роки тому +1

    Very helpful :)

  • @aliberro
    @aliberro 2 роки тому

    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

  • @matr1x_glitch
    @matr1x_glitch 8 місяців тому +1

    I SAY WE MAKE DONUTS SQUARESSSSSSS

  • @rbettsx
    @rbettsx 2 роки тому

    Or.. just cut through the donut, bend it straight, and take the volume of the cylinder?

    • @LearnPlaySolve
      @LearnPlaySolve  2 роки тому +2

      Yes... this demonstrates why that works. Thank you.

  • @circleoffifth9048
    @circleoffifth9048 2 роки тому

    Just cut it and make it a cylinder
    Volume =base*altitude
    =πr2* 2π*(Ro-R1)/2

  • @marktolands9692
    @marktolands9692 2 роки тому +1

    its just a cylinder idk what the big deal is lmfao

  • @mujtabaandrabi5410
    @mujtabaandrabi5410 2 роки тому

    gurvir singh lol, your 7months l8. But cheers m8