Here ... a caramel flan disappears before it can ever be measured. I think there's a physics concept built around that. (method 1 is my vote, but it's always good to work a problem from multiple paths/methods)
I liked the cone method, but it also relies on a previously found formula for the volume of a cone, so I felt that it was less of a "complete" method in a way
Wikipedia says «The term "frustum" comes from Latin frustum meaning "piece" or "crumb". The English word is often misspelled as frustrum, a different Latin word cognate to the English word "frustrate". The confusion between these two words is very old: a warning about them can be found in the Appendix Probi, and the works of Plautus include a pun on them.» en.wikipedia.org/wiki/Frustum#endnote_1 The "Appendix Probi" is from the 3rd or 4th century AD. Plautus live about 254 - 184 BC. So you're in very good company 👍
When testing the slump of concrete, we use the frustrum of a cone and fill it in 3 layers of equal volume. R = 4in, r = 2in, and H = 12". If big R is on the bottom, what are the heights of each layer?
it is also possible to first calculate the area of the quadrilateral (with sides R, r, h), and then do an integral where it is extruded radially around the axis? would it make a difference that it turns faster outside than near the axis?
So it's actually the sum of three full cones of common height h and base radii R, r & sqrt(R*r) - the geometric mean of R and r. Perhaps someone could reveRse engineer an even more elegant solution using this fact.
I kinda did it the same way (as the first method), but different way of thinking. It comes out to be exactly the same (phew..) Triple integral (cylindrical coordinates) version: integrate from 0 to 2pi integrate from 0 to h integrate from 0 to R - (R-r) * (z/h) r * dr dz dphi (yes, using r for the small radius and the variable for integration is incorrect, but youtube doesn't have subscripts) This feels more elementary to me, but it requires the knowledge of the volume element in cylindrical coordinates. If I had to derive that from scratch, I'd be slower for sure :) Volume of a rotated line segment is also kinda pretty, but again: derivation of the volume of a rotation-body takes a while.
Insert comment about how you did it faster by doing pi h/R-r integral from r to R x^2 dx, because you can just place the cone horizontally.
Any binomial expansion with R and r is never gonna be the same again
Rrrrrrrrrrrr! :)
I like the fact you began to do videos about simpler problems I can understand!
Here ... a caramel flan disappears before it can ever be measured. I think there's a physics concept built around that.
(method 1 is my vote, but it's always good to work a problem from multiple paths/methods)
I liked the cone method, but it also relies on a previously found formula for the volume of a cone, so I felt that it was less of a "complete" method in a way
To get a pyramid trunk volume, just apply cavalieri's principle! (and also I prefer the integration)
Yes I think this is equivalent to cavalieri
Thank you Dr. Peyam, it helped me at the right time.
So is it a frustrum or a frustum, since I have seen both of those elsewhere too
According to google it’s frustum apparently
Wikipedia says «The term "frustum" comes from Latin frustum meaning "piece" or "crumb". The English word is often misspelled as frustrum, a different Latin word cognate to the English word "frustrate". The confusion between these two words is very old: a warning about them can be found in the Appendix Probi, and the works of Plautus include a pun on them.»
en.wikipedia.org/wiki/Frustum#endnote_1
The "Appendix Probi" is from the 3rd or 4th century AD.
Plautus live about 254 - 184 BC.
So you're in very good company 👍
If you forgot the formula for the volume of a cone, use calculus to find it back.
Have you done surface area of frustum, to get surface area element for surfaces of revolution?
Great video ,loved the bits of humour informative and fun
When testing the slump of concrete, we use the frustrum of a cone and fill it in 3 layers of equal volume. R = 4in, r = 2in, and H = 12". If big R is on the bottom, what are the heights of each layer?
it is also possible to first calculate the area of the quadrilateral (with sides R, r, h), and then do an integral where it is extruded radially around the axis? would it make a difference that it turns faster outside than near the axis?
Dr Peyam now in 1080p60!
I love the integration method better :)
Sir please make extended videos on vectors and tensors
Frustum in 5D:
2 balls in 4 dimensions are connected to form a frustum in 5 dimensions.
What is the volume of this frustum?
Mmm flan au caramel. Cheers from baguetteland aka France !
So it's actually the sum of three full cones of common height h and base radii R, r & sqrt(R*r) - the geometric mean of R and r.
Perhaps someone could reveRse engineer an even more elegant solution using this fact.
You are awesome!
very good!!!
You could probably save a lot of writing time in the integral by doing your u-substitution for f(y) instead of expanding that in the integral.
¿Podrías activar los subtitulos es español?
You are a legend
3.141592654Rates of the carRibean
3.14159265358979323846264338327950288419716939937510rates you mean
@@utkarshsharma9563 Yes :D but I'm happy I managed to remember even 9 decimal places :D
I kinda did it the same way (as the first method), but different way of thinking. It comes out to be exactly the same (phew..)
Triple integral (cylindrical coordinates) version:
integrate from 0 to 2pi
integrate from 0 to h
integrate from 0 to R - (R-r) * (z/h)
r * dr dz dphi
(yes, using r for the small radius and the variable for integration is incorrect, but youtube doesn't have subscripts)
This feels more elementary to me, but it requires the knowledge of the volume element in cylindrical coordinates. If I had to derive that from scratch, I'd be slower for sure :)
Volume of a rotated line segment is also kinda pretty, but again: derivation of the volume of a rotation-body takes a while.
Cone method seemed like the obvious way to go lol. Why do more work that you have to?
I learned the second method when I was 13 and missed the first because I was forced to take Honor’s Calculus. I prefer the first method.
Board says "Frustrum" not "Frustum".
Both are apparently acceptable
Two cones or Theorem of Pappas before seeing what you did
Now do it using category theory
Sure, just use Yoneda’s lemma on the category of cone-o-morphisms
@@drpeyam sounds legit!
@@drpeyam HAHAHAHA!
First from Brazil !
Cringe punning gold: you do you, babe... another glorious vid.... cool.
Mann kann ja einfacher zur Lösung kommen!
Volume=[((2r(pi))*(R*H*1/2)+(r*(pi)*H)]