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.

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.

great animation and thanks for your help

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

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

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

Bro I loved your explanation 🥰

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.

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

Very helpful thx for the video❤

Just helped me with a calculus project thanks!!

clear and nice

Thank you so much

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.

Very helpful :)

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

Thank you :)

underrated channel

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 😅

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.

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

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

You are genius

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?

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

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

I SAY WE MAKE DONUTS SQUARESSSSSSS

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?

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

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".

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?

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

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

why x=rsin theata not rcos?

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

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

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

its just a cylinder idk what the big deal is lmfao