Volume of an ice cream cone
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- Опубліковано 21 сер 2018
- In this video, I calculate the volume of an ice cream cone that is obtained as the region between a cone and a hemisphere. Here I present a direct approach, without spherical coordinates. In a future video I’ll show you how to do this with spherical coordinates.
You always make me smile when I'm watching your videos :)
Edit: Moments like 10:35 for example :D
I work in an ice cream shop and that's not a particularly realistic ice cream cone. :P More rather it's a mostly licked-down ice cream cone where the hemisphere intersects with the cone that is itself not easy to hold properly. What you really need is a cone with a steeper slope to its sides (imagine if you will that if you sliced the cone through its center and projected it onto a 2D plane, it might resemble a function similar to y=|2x| or y=|3x|), and the sphere needs to rest on top such that it is tangent to the sides of the cone, *then* calculate the volume of the space enclosed by the two volumes. ;)
yes
This is some really amazing stuff. As usual, thanks so much, this is so interesting and it always puts me in a good mood
I like your presentation about any mathamatical expression and now in this video you had done marvelous!!!!!!!!
😀😀😀😀😀😊🤔😄🙋🙏👍👍👍👌👆👌👌🤘
Lol do an ice cream cone with ten scoops where each scoop is ten percent the radius of the previous scoop, or something like that would be fun and interesting and have a cool picture to go with it, likely easier to do with geometry than calculus at that point though 0.o
Classic multi variable question.
What if the ice cream was a minimal surface ice cream; would this affect the volume calculation?
This is way easier with geometry though but okay
Before watching and without any calculus I'd say it's the revolution body about the z-axis of the region between z=r and z^2+r^2=8, so a cone with base area 8pi and height sqrt(8), plus half a sphere of radius sqrt 8: so volume = 1/3 • 8•pi • sqrt(8) + 2/3• 8•pi•sqrt(8) = 16•sqrt(2)•pi. Amazing that the cone is half the volume of the hemisphere!
Edit: no, I was wrong, it should have taken only the top slice of the sphere, and also less of the cone! What I calculated here correspond to the much bigger volume between a translated cone z = sqrt(x^2+y^2) - sqrt(8) or z=r-sqrt(8) and the complete hemisphere, a much bigger shape with slightly more resemblance to a true ice cone shape (yay!) but still too widely open.
this is PEYAMazing :)
Glad you watched it!!! 😄
The cone of an ice-cream transitions to the sphere at a tangent circle that's less than the diameter of that sphere. Usually? If all you do is place a sphere inside a cone, there is a negative volume in the transition for which you haven't accounted. It's a ring with almost a triangular cross-section.
Good morning, Peyam, can you please make a video of calculating integral of {1/x} from 0 to 1 (like you did with tangent). I believe the answer is 1-γ, where γ is the Euler-Mascheroni constant. Very beautiful imho. And it's also quite nice because it's obvious that this integral is somewhere on the interval [0;1], so 0
I think you’ve read my mind, but there will be a video on this coming either in September or October 🤗
@@drpeyam wow thanks, you are the best math channel i've ever seen :)
what is the definition of disk?
is it a circle but filled? and then the circle is just the shape?
like sphere and ball?
In a usual plane, given a point O and a distance R
the circle of center O and radius R is the set of points M such that the distance between O and M is equal to R
the disk of center O and radius R is the set of points M such that the distance between O and M is equal to or less than R
Yea you can say that the circle only refers to the boundary of the disc
*Volume of an ice cream boi
but shouldn't a constant be added to the hemisphere equation so that it lies on the cone??
No, not really
@@drpeyam also try calculating the volume without calculus
Why could you multiply the integrand by r at 9:28?
When you change from cartesian to polar coordinates, an extra factor of r appears. So it is: dxdy -> rdrdθ. This is just how it is when you either do it using the Jacobian or using vector calculus and getting the cross product directly.
@@John75ify Got it, I forgot about the dxdy change!
Y u no make reference to silver ratio(you do realize the reciprocal of the silver ratio has just appeared
?
I’m referring to how sqrt(2)-1=the reciprocal of the silver ratio. You should hVe made a reference to that
Bcos the answer is 32/3*pi*1/(silver ratio)