What is the Volume of a Donut? 2 Methods: GEOMETRY and CALCULUS
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- Опубліковано 12 чер 2024
- Through two completely different methods, you can find the volume of a donut. In both Geometry and Calculus, we must utilize infinity which makes for some interesting solutions! They each inform us about the nature of mathematics as a collaborative endeavor, that it requires multiple perspectives and approaches.
0:00 Introduction
0:26 Archimedes' Circle
1:12 Geometry Proof
3:03 Calculus Proof
6:45 Reflection
Very nice video. But actually Archimedes proved the area of circle differently, by using method of exhaustion, inscribed and circumscribed polygons. The proof you show is like an approximation to find the area empirically. And then, using fomal method he proved it. It is a masterpiece of a proof and everyone who love maths should read it, to understand the level that the mathematicians reached, 1800 years before the invention of calculus.
Great video
I was fascinated the entire time. I'm surprised I could even follow what was happening.
Thankyou! Happy to hear you could follow and found it interesting!
That was a really good video. The animations were so smooth and the explanations coherent. Thank you!
Thankyou so much!
Very solid demonstration. (pun partially intended :3)
Though, I think part of the process in the Calculus could have been simplified by recognizing that the integral from -r to r over the sqrt(r^2 - x^2) is just the area under a semicircle, which is known, and thus a lot of annoying trig could have been avoided. On the other hand, the integral for the volume of a solid of revolution is simply stated without any justification; in my opinion that might have been a better place to put in some extra insight.
Haha thanks! I really appreciate the feedback and will try to integrate (pun intended) more thorough and considered explanations in the future.
i definitely needed that reminder about trig sub tho
Actually, those Archimede style proofs are perfectly rigorous without calculus, in light of measure theory.
First of all, notice that, at each step of the sequence of "refinings", the "rearranging" map φ from the disc to the union of "rearranged" slices (defined however you like on the measure zero intersection of the slices) is measurable and measure preserving (because it's in fact an isometry when restricted to each slice). Actually, φ should be a φ_n because it depends on the step, but we'll still call it φ.
1) About the proof for the area of the disc:
You might want to consider the sequence of rectangles given, at each step, by the biggest rectangle with "vertical" sides contained in the rearranged disc (By rearranged disc I mean the figure on the right at 0:48 in your video). The inverse images of these under the φ's form an increasing sequence of subsets of the disc. The limit of the measure of these inverse images, by continuity from below, is equal to the measure of the disc.
On the other hand, the limit of the areas of the rectangles on the right is clearly equal to 2πr*r.
2) About the torus volume proof:
This is a bit more complicated. At each step, the set we consider might be the smallest "horizontal" cylinder containing the rearranged slices. Clearly, the measure of these nested cylinders tends to the volume of their intersection, that is πr^2*2πR.
Now you extend the isometry of each rearranged slice (given by the inverse of φ on each rearranged slice) to each solid angle between the pairs of planes determined by the "edge circles" of each rearranged slice. This will give you non-intersecting pieces. Then, at each step, the inverse of φ, thus extended, acts by isometries on these pieces.
So, via these isometries, you get a decreasing sequence of nested sets each of which contains the torus.
By continuity from above, their measure tends to the volume of the torus.
QED
Your point is a good one and I may have been unclear, the goal of using two methods was not intended to discredit either of them. Each individual proof is not reliant on the other. In tandem, they just display two different ways of approaching the same problem. A satisfying investigation!
My man, that was a gorgeus video. I've seen you have very few subscribers (and now I'm one of them) but I hope your channel will grow in the future, because you deserve it. Wish you the best❤
Thankyou for the kind words! I'm glad you enjoyed the video
6:52 Crimony. It's been five DECADES since I've taken a math class, and I was still able to do this.
Now I need to find my nearest Krispy Kreme to celebrate! :)
Nicely done! There’s also an interesting connection here to path integrals. Another way to think of producing the volume of a torus is to take the contour integral of the area of the little circle around the perimeter of the torus.
a great video, i was surprised to see you only had 84 subs. keep up the good work!!
Thanks to you I finally found out that the formula "for the volume of a donut" I wrote down when I was 13... Was actually right. (I didn't provide a proof though)
The integral at 4:40 can be recognized as representing the area under a semicircle, which is pir^2/2.
Good point!
Great work!
Could have also added topological methods , they're my absolute favorites. I've had quite some experience with topology and toroids together , since I've written some articles on the hypothetical formation of toroidal planets (I'm an LQG tehorist , so it kinda falls under what i do , technically).
This video is amazing!
but as a note for future videos, please use parentheses in integrals, it will make them a bit more readable, and it will be more aesthetic if the integration limits were above and below the integral, like the command $\intop$ in latex.
Thankyou for the comment and suggestions! I will enact them in future videos. My current software is quite limited so I’m learning to create videos with the Manim library where it's all possible.
Great video.
I think you could have used the geometric one to prove it.
As your proof just lacks the fact that you prove that it’s converging.
And to do it you could just have a lower and an upper bound, both converging to the solution.
Good job
amazing video!
Dude. Killer stuff!
Awesome stuff
When I was grade 8 I thought of it as the area of circle inside and times it by the middle circumference. For surface area I thought of it as the circumference of the circle and times by the middle circumference. I thought of how the shape was formed like the logic behind it before finding its volume and surface area. a donut was formed by a circle moving around in a circle.
Next -> Area of a Torus
Awesome video bro! Defo sub!
Thankyou!
What do you mean the simple one isn’t backed by mathematical proof? I would like to understand if anyone can explain it to me clearly. 😊
Sorry if it wasn’t clear! The general idea is that we are applying a method from a 2 dimensional space to a 3 dimensional problem. In this case it works, but often it doesn’t! For example, if we attempted a similar proof for the surface area of a sphere, slices would be curved and warped. There is a great explanation of this on 3Blue1Brown ua-cam.com/video/VYQVlVoWoPY/v-deo.htmlsi=Cn36nAXdSM18ZVxi
Hopefully this makes it clear why it is sometimes useful to check our answer with multiple methods in order to be confident in our findings. I'm sure there are more rigorous ways to prove this geometrically however. Perhaps it would have been better if I had said, “isn’t backed by strong mathematical proof”.
The "picture" proof in this case is perfectly rigorous even for modern standards (though of course one would have to add a lot of pedantic details, and use standard mathematical notation and language).
The reason it works is that it involves rearranging pieces by isometries (an _isometry_ being a transformation that preserves distances) and limits of sequences of nested figures of bounded area (or volume, in the case of the torus).
The example about the area of a curved surface doesn't fall under this mechanism: the presence of curvature disrupts the "isometry" aspect of it (And, also, there's no nested figures of the correct dimension).
It's nice to see these relations being applied
But I have one small correction
5:47 apparently you forgot the module, as √(x²) = |x|.
I shall use archemides principal
A truly exceptional and click inducing thumbnail with a good intuitive and easy to grasp explanation. By the way, do you have anywhere I could contact you?
Thankyou! An email is linked on the channel page if you click on more info.
@@remathematics Great!
4:20 - That is not how you expand the square of a difference.
Sorry, my notation was misleading in writing y^2. We recognised the difference of u(x) integrated and b(x) integrated as the desired area in 2 dimensions. When we apply the formula for a solid of revolution, we are using it for each function individually and then finding their difference. This creates a torus.
Is it easier to just use 2R(πr)²?
Yes that would be a totally fine formula! The arrangement of terms I show in the video aims to display the formula’s origin, from the area of the revolving circle and the circumference of the larger circle.
@@remathematics Ahh I see now, mine is a little bit harder to link to the original shape
I still don’t trust the geometry proof because I feel like it doesn’t rigorously account for the different between inner and outer circumference, ie: how one side of our cylinder gets squished and one side gets stretched (unless I missed the reasoning)
I agree! It’s hard to believe isn’t it.
@@remathematics it’s a coincidence though right? The fact that the stretching exactly cancels the squishing, that’s gotta be a product of some other property, not something that will consistently happen?
it would have been much easier to split the circle (y-R)^2+x^2=r^2 in half vertically and integrated than the method used in the video lmao
If calculus gives the same solution as the Geometrical one, it's proved that Geometry gives the correct amswer but in a shortet number of steps
Hence, we can always rest assured that Geometry will always give the right answr and the Calculus one is unnecessary.
just imagine it as a curved cylinder lol not hard