Surface Area of a donut 

regular people: it's a donut math students: it's a toroid topologist: it's a coffee mug

Bakery: May I help you? Me: would like to have a glazed donut in multivariable calculus by Dr. Peyam? Bakery: Sure

Ilove the way you just say thanks for watching at the very beginning of the video, like now I'm gonna watch it entirely even I don't know how to integrate.

Lol I attempted to solve it using single variable calculus before watching. Once I was done, I thought, “that can’t be right, it’s like it didn’t even matter that it was a donut”. Then I saw the intuitive method and my mind was blown because it was the same thing I got. Thank you for your video!

0/20: Forgot the sugar on the donut. Sugar would be dropped downward alongside the revolution axis uniformly; but you would have to compute the effective surface that is touched by sugar: \pi (R + Delta)^2 - \pi (R - Delta)^2 = \pi ( 4RDelta). This is the area that is touched by sugar. Now you need to compute how many crystals of sugar did touch the donut. Introduce the parameter D (number of sugar per unit of surface of the emitter of sugar-particles on top of the donut) of the uniform distribution, there are \pi 4RDDelta crystals of sugar, on average. Now you need to compute the added surface of those sugar drops, let's say one crystal has an exterior surface of x; then the new surface of sugar is \pi 4RDxDelta, on average. However, you need to remove the surface that is in contact with the donut, because it would be counted twice, both donut and sugar. Now you can imagine all sorts of shape for the sugar: a cube would add, on average, one of its face L*l, but times the number of sugar crystals on the donut; a sphere would need to consider a penetration radius in which the sugar goes inside the donut (-> slice of a sphere). Total exterior surface of a donut: Surface of the raw donut ( a torus) + Exterior surface of each crystal of sugar one the top of the donut - Surface of the sugar touching the donut. = 4\pi^2 R Delta + \pi 4RDxDelta - Depends on the geometry One could argue that you could put other stuff on your donut, or that you don't need sugar on your donut, which is clearly a mistake of gastronomy and/or might be a source of diabetes. Also, I rewatched the video "Volume of a ball in n dimensions " dated 29/12/2017, which is one of my favorite of your channel, and I wondered if in this problem, what you called there disk method would be: - the method of calculating the perimeter of the circle, then integrating d phi or - the method that is described here: split in cartesian coordinates Part of me wants to believe it would be the second one, because the method in Volume of a ball in n dimensions is taking a slice of the ball alongside cartesian coordinates; and it is makes sense that the description of cartesian coords at 4:10 fits with the split between x1, x2, ..., xn. However, I wonder if the method of calculating the perimeter of the circle, then integrating d phi would be valid as "disk method" because it is the same under polar coords: you slice through d\theta, then through d\phi, just that this is no in cartesian coords. It is not really important though, just a thought.

You can do it quicker if you inscribe torus in coordinate system and think of average distance between the torus cross section circle and the torus axis. In the horizontal direction the average is the torus radius and in perpendicular is 0 so the av. dist. is radius. Then count a simple line integral and done. sorry for my eng

Dr peyam: Donut, surface area... integrals!! Mechanical engineers: Let's just use Pappus' theorem instead

Totally delicious, going to watch the volume video now. I was very excited by this problem when in calc II, right after learning about double and triple integrals, surface area etc. Honestly, I never worked this out but knew I had the tools to do it finally. Its rewarding to see each step, thank you! Of course now I'm interested, for two-holed tori I could do this same process as long as I account for the middle area being counted twice right? and 3-holed tori etc. using the inclusion-exclusion principal for counting the cardinality of overlapping sets.

The intuitive method can in fact be rigorous. It is the Pappus's centroid theorem. Great video!

How many languages do you speak Dr. Peyam?

Nice video. Now I want to go out and buy a dozen at Dunkin Donuts.

Dr Peyam: I have a question at 4:40 min. Alpha and theta should be big than 0 and less or equal to 2pi , isn't? Because of the parametrization is a function

I used simple *Method of exhaustion* to find the formulae for both volume and surface area.

Mais pourquoi PAS faire simple? À mon âge, je préfère les choses qui ne sont pas très difficiles.

I'm french and agree 😂 pourquoi faire simple quand on peut faire compliqué ?

Why does the magnitude of the cross product vector need to be taken? Isn't the cross product of the two partial derivative vectors both the area of their parallelogram AND its normal vector magnitude already?

What definition of surface area are you using?

Thanks for another fun video. Still only 10% thumbs up, come on all you great subscribers, lets get to 90% thumbs up, it'll make Dr. Peyam's day! :)

Why is the multiplication of the perimeters intuitive? I don't have that intuition, by analogy to "straight" objects like a cylinder I can see how one might get there but I do not understand why it's that way for a curved object