Without integration, why is the volume of a paraboloid half of its inscribing cylinder? (DIw/oI #8)

Rather than using integration, can we find the volume of a paraboloid? Yes, if we accept a precursor to calculus - Cavalieri's principle. Usually, integration is needed to find the volume of a paraboloid, for example using shell method, but using Cavalieri's principle, and a sneaky little trick, we can find the volume very easily - half of the volume of the circumscribing cylinder!

The idea for this video isn't actually mine, but thanks to Yehuda Simcha Waldman for suggesting the idea of this video! He emailed me about the proof, and I modified it a little bit and adapted it into this video that you are watching here.

Other than commenting on the video, you are very welcome to fill in a Google form linked below, which helps me make better videos by catering for your math levels:

If you want to know more interesting Mathematics, stay tuned for the next video!

SUBSCRIBE and see you in the next video!

If you are wondering how I made all these videos, even though it is stylistically similar to 3Blue1Brown, I don't use his animation engine Manim, but I will probably reveal how I did it in a potential subscriber milestone, so do subscribe!

Social media:

For my contact email, check my About page on a PC.

See you next time!

[mathemaniac]

EDIT: At 3:57, the solution of x should be sqrt(a/b-z/b) instead.

Thanks again to Yehuda for suggesting the idea of this video! Here, the paraboloid only means those paraboloids that can be constructed by revolving a parabola around an axis. For other paraboloids, the cross sections would generally be an ellipse instead. I don't have a solid plan for the next video yet, so do feel free to leave a comment for a possible idea.


This is because it makes the planning of future videos better in the sense that I could make the pacing better, and know what knowledge I can assume you know.

[sophieward]

That is fascinating! Really elegant, super clear and concise. I love proofs that use Cavalieri's principle for that exact reason, but it never occurred to me that the volume of a paraboloid could be derived so cleanly using it

[sirnate]

At 2:30, since pi * (1 - z) is linear, you can find the average cross-sectional area by taking the average of pi * (1 - z) evaluated at z = 0 and z = 1 (the height of the paraboloid), which is pi/2. Then multiply by the height of the paraboloid to find the total volume, in this case, pi/2.

[parthprashar]

Your fan from Quora here. Your videos are as beautiful as your answers.
They are making me love math.
Keep up the good work man.

[benYaakov]

Whoooo, you saved my day .
Very intresting and thanks to you for this ( also thanks to Yehuda😅)

[smrtfasizmu]

This viewer who sent the email has a very similar way of math thinking to me. This resonated pretty easily with me

[maxwellsequation]

Oh my godddd
This is simple and elegant and beautiful. I love it!!!

[avinashkashyapjha]

Thanks a lot for this. I needed it for my fluid mechanics course.

[forthrightgambitia]

3:54 - small nitpick but shouldn't that be sqrt(a/b - z/b)? That way you would have the reverse parabola as z = bx^2, thus x = sqrt(z/b) and the final cross section would equalise out as a/b*pi, thus final volume of the bounding cylinder as (a^2/b)*pi?

[mahxylim]

I really love this video!! It is so beautiful!

[dashnarayana]

Great site for entertainment and challenging for lovers of maths

[haniamritdas]

This is great, thank you. Galileo used the same principle to show the equivalent areas and volumes of a cone and the cylinder evacuated of a unit hemisphere. But I did not know the name of the principle or its more general application. A paraboloid, and a torus too... Very nice!

I am convinced that the geometry of the cosmos is the dual of a single point coordinatised in this framework. Oh, like a spinor I guess....along those lines. Curves. ❤️

[lyrimetacurl]

Spheroid top: 4/6 pi r^2 h
Paraboloid top: 3/6 pi r^2 h
Cone top: 2/6 pi r^2 h

[GetYourMath]

I haven't got it 2:40, I don't understand the connection created with the height of the parabolid evaluated from the top of it when it is upside down and the area of 1 of its sections. I don't get the logical step

[simjianxian]

holy fuck galaxy brain, thank you for sharing

[Djaketooth]

3:05 Tbh, I didn't see it comin' but that's so cool!!!

[EebstertheGreat]

Cavalieri's principle, or something like it, is also required to find the volume of a square pyramid (or any general cone). Three appropriately-shaped congruent square pyramids can fit together to form a cube, so their volumes must each be one third of the volume of the cube. One might hope a clever decomposition like that might exist for all pyramids, or even all polyhedra, but it does not. Dehn's negative answer to Hilbert's third problem proves that in most cases, no such decomposition exists. Therefore, to get a formula for a general square pyramid, we need to use something like Cavalieri's principle to translate the volume of the specially-shaped pyramid to the volume of any other pyramid (or indeed any cone) with the same base and height.

The ancient Greeks did not appreciate this fact, though they used something like Cavalieri's principle implicitly. Two plane figures were "equal" if they had the same area in the modern sense, and the sums of equals were equal as an axiom. But this alone is not enough to justify Cavalieri's principle, because that requires adding infinitely many areas to obtain a volume, which is nonsensical, and properties of finite sums don't necessarily hold for infinite sums anyway. It turns out that it is _impossible_ to compute or even define the volume of most solids without some notion of a continuum--something like the set of real numbers--and either infinitesimals (in non-standard analysis) or limits of sequences of real numbers. In other words, there was nothing approaching a rigorous treatment of volume until Cauchy, even for something as simple as a right square pyramid.

[Xayuap]

so, a paraboloid inside a cylinder is the same volume as the space it left

[ProofCare]

Hey I'm your new subscriber 💓 from India

[henriquecorbilopes]

Archimedes did something similar, but he employed equilibrium and center of gravity properties. But Archimedes didn't consider as a true mathematical proof, but as method to discover, to prove these properties he used the so called exhaustation method, which is attributed firstly to eudoxus of cnidus, and it is a rigorous method of mathematical proof. By means of mechanics he not only computed parabolid's volume, but as well, he could figure out area of parabolic sections, volume of spheres and ellipsoids. And it would be interesting if you brought up this archimedes' method of balance, and it's related with this video, since line sections from plane figures or area sections from solids are involved