Nice vid! I had fun verifying that the 3rd method does give the same answer as the 1st method for the elliptical torus [🍩] consisting of revolving an ellipse with horizontal axis a, vertical axis b, and whose left-most point is R units to the right of the origin! V=2π²abR :)
@@Jasomniac Well, a circle is a special kind of ellipse where a = b. But you're probably right that it technically may not be a torus! I meant that I liked verifying it works when you revolve any ellipse with horizontal and vertical axes around the y-axis in a similar fashion regardless of whether or not it's a circle! :)
@@MathNerd1729 sorry, I read your comment wrong, what you said is well phrased. And btw, I don't know if "general torus" is the right phrasing, but if we're gonna be very specific, I guess you could have said elliptical torus.
that would be the best method if you had a real donut (if it was waterproof), because of small irregularities on the surface of the donut that are hard to calculate
@@herman7550 Except, once waterproofed, it would probably float. You could still do it, though, if you used a very low-volume implement to force it to submerge. Fred
Very nice presentation! Thoroughly enjoyable. Some related thoughts it brings to mind: 1) For the benefit of other viewers, in your example of Pappus' Theorem - for a torus, r is called the "minor radius," and R is called the "major radius." 2) We were taught Pappus' Theorem (P.T.) in my 7th grade general math class. It was a very impressive example of mathematical beauty, as well as being useful for many otherwise very hard problems. Then, years later in calculus class, one of our exercises was to prove P.T. - the general (3D) case. 3) And just BTW, it also works for finding surface areas. You might consider also finding the *area* of the Krispy Kreme this way ;-) 4) P.T. generalizes to n dimensions, for n ≥ 2. You can find the area of a circular annulus by revolving a line segment around a point collinear with it, and outside of it. You could find the 4-capacity of a 4D solid of revolution, given the volume of the 3D solid being revolved, etc. 4a) I once used it to find the capacity of an n-ball, using an insight that a certain relation between a 2-sphere and a circular disk, might generalize to other dimensions. It involved a torus whose major and minor radii are equal (a zero-hole torus); and I was able to verify that generalization formally. I hope, one of these days, to make a YT video about that... Fred
A donut is like a cylinder which is curved into a circle of radius R....hence height of cylinder is 2πR...and if the radius of the cylinder is r then area of cross section will be πr²...hence volume of the donut will be the volume of the cylinder = (πr²)*(2πR) = 2π²r²R That's actually how I solved it😅 Edit: after watching the full video I realised that's the pappus method😅
If you morph a cylinder into a torus like that though, parts of it are going to be compressed and other parts are going to be stretched. Apparently if you choose your radius of rotation to be the centroid of the circle, the compression and stretching cancel out. It is noteworthy that you have to make sure you use the centroid though, and if you're rotating something other than a circle around a central axis, it may be difficult to figure out where the centroid is.
@@IGDZILLA That is true, but in the limit as thickness of the wedge(distance between centers of circular bases) approaches zero, the difference between outer and inner "heights" will also approach zero and the wedge will approach a disk shape
For a complete newbie in calculus, this makes so much sense, especially the Pappus Theorem! I accidentally saw it on my calculus book while cleaning my shelves, but your explanation made the theorem more exciting and sensible. Awesome content, as always
I've just had my last lesson of mathematics seminar this year on high school. It was about calculating volume of shapes with functions and integrals and a simple formula.
Its funny because I was thinking of how to solve this withouth calculus, and I thought of using the circumference of the big donut times the area of the cross section. Turns out it was the circumference of halfway through the big donut, but it was still cool that I almost figured it out on my own.
Thank you for the great videos! I saw your channel for the 1st time 7 mo. ago on the derivative on x^x, and I didn't even know what calculus was yet lol. After watching many of your videos, I now know lots of calculus 2 thanks to you :D
😄 I got it! But by using none of these ways, sort of (I actually did do the Pappus theorem, but solved for it first and the best part is I didn’t know it was a thing when I solved it). I basically did the washer method, except I did it the easy way. Which is by turning everything into rectangles, which ends up turning everything into cylinders. I solved for it generally, and then plugged in the numbers. The result was 2m(pi^2)(r^2) Where m is the distance from the center of the torus to the center of the solid circular part and r is the radius of solid circular part. And as expected, the approximate result was 144.343 cm^3.
I actually used this approach years ago to calculate the volume of a vase in my house. Afterward I filled the vase with water and I was only off by about 1% in accuracy. Fun stuff.
According to my mentor and professors, I am very good at calculus. I feel that way too but not as much. The thing is, bprp makes everything seem so SIMPLE!
I didn't hear about pappus theorm , I solved it by imagining a horizontal cylinder with length dL , I found that dL is length of arc with radius from center of donut to center of cross section area of the cylinder times dtheta where theta is rotation about y-axis , so small volume dv=cross section area of horizontal cylinder times arc length times dtheta , then integrate from zero to 2π and got the answer
Good video! It would be interesting to see a solution with Fubini and Tonelli theorems And maybe with the change of variables to cylindrical coordinates
I actually begin questioning into the region of integrals and limits when I questioned the volume of a cylinder in class IX. I wondered how one could find the exact volume when the circle area had infinitely small thickness and was multiplied by the height. Now in Class XI, I see how integrals can be used. Epic!
The shell method resonates best with me. The Pappus Theorem feels like a trap. Not that I think Pappus was lying; it's just that it feels like the sort of thing a person could misapply if they're not careful.
I didn't learn volume of solids by rotation until calc 2. Also, I never learned about Pappus Theorem although it showed up in my engineering statics book. I took calc 1 two years ago and got an A.
The teacher is very smart enough. I am very happy to learn from him to solve this problem. I hope the teacher can get NASA recommendation to help astronauts to solve Mars, Moon or the universe engineering calculations problems.
I was thinking Pappus theorem, I also had the top marks in all 3 OAC (grade 13) maths: Calculus, Algebra, and Finite. Ended up taking a few more Finite and Probability as electives in my undergrad. Grad school did offer a course in Numbers but it was Pentateuchal studies.
explaining to your audience why you have to use the centroid would be useful for this Pappas method. reason being that the inner radius is shorter than the outer radius makes it seem as though modelling it as a straight cylinder could be inaccurate. using the centroid sort of "averages" the over estimation of the inner radius and the under estimation of the outer radius. something like the area of a trapezoid where you average to two unequal sides. same idea in an abstract sense. it took me a little thinking to agree with this pappas method. so a better understanding, say for example if the cross section was not a circle, but say, two half circles with different radii forming a funky donut (torus). then you would have to integrate the cross sectional shape to find the centroid, then use the coords of the centroid to apply the Pappas method. nice video, I liked how you used three methods.
In one Homework in mathematics for Physicists, we had to also do that. I remember having brought this up to the table with Pappus theorem, but we couldnt use it, because it was an higher analysis course and we should have done in the 1st presented way. It was challenging but also fun, when you get your results correct :D By the way, i didnt know, that this theorem was called "Pappus Theorem"...
That use either shell or washer method. We actually didn’t learn that in calc 1 rather it was in calc 2 cause calc 1 was rushed over a summer interim session, but nonetheless it seems odd that such a simple concept requires calculus. Interesting video, the only thing I can comment on at the start is that you will have to do two times the integral of the positive half of the circle offset by some volume.
A while back I came across your world record video and always wanted to learn calculus. I think I will succeed this summer with the help of your videos! Thank you! p.s I had to laugh really hard during the wr video @4:55:55 ''Integration by parts!''
Probably some differences in naming. When I studied we called the third method "the 2nd Guldin's theorem". The 1st Guldin's theorem finds the square of the rotated curve.
My intuition of what the answer would be was something like, "take a cylindrical bar of dough of that radius, bend it to circular shape where both ends meet without distorting the radius of the cylinder. The volume of that cylinder will equal the volume of the doughnut." My non-rigorous reasoning was that the outermost "dx" would stretch by the same amount that the innermost "dx" would compress. I dunno if it works exactly that way, but the cylinder idea at the end of the video felt similar to my thoughts.
I was calculating this with the first method, albeit I changed the x and y-axis. I also tried to calculate its surface, but I was not able to find the proper solution.
Can you answer me this, in what country and in what school or college do you teach math? Because it seams like your students are having a lot of fun with you
I get a giant coffee cup (although I'd have to settle for a glass coffee pot since giant coffee cups are hard to find), I fill it half way with coffee that I've measured. I then take the donut and put it into the coffee and push it down and I measure how much the coffee has risen in the glass coffee pot. I then compute the volume by: measured height difference times PI times square of the Giant Coffee Cup Radius. That gives me the volume. Another way to do it is to measure the thickness of the donut ring, and the inner and outer radius and use a little calculus. But I like the Giant Coffee Cup method better, because that way I can drink the coffee and eat the soggy donut. So why this method better; besides having breakfast? Because from a Topological point of view: there isn't any difference between a coffee cup and a donut. Works better on an old dried out Donut. It tastes better too and having food and drink and not an empty stomach helps when doing math.
Now N A S A will have easy time calculating the orbit of satellite circling around the planet. Thanks to Pappus theorem. Thanks a lot for your video. Super informative 👍👍👍👍👍
Can you also Post Videos where you Solve the Problems on a Blackboard using a Chalk from time to time, like you used to do a Few Years Ago? I miss the Blackboard-Chalk System. :(
my logic before watching: since you are rotating the circle around a larger circle that connects the center of the cross section to the center of the donut, the volume should just be the area of the cross section times the circumference, which would be (πr^2) x (2πR)=2 π^2 r^2 R. just a little bit of logic instead of actually proving it lol
For the Pappus theorem, I'd take a shortcut by replacing 2πR with πD, where D is simply 9.5 - 3 :). But I'm still missing some insight into why exactly this method is correct (although it makes intuitive sense).
That's a very good question. And you don't even need integrals to prove the formula. Just replace the thorus with a poligon made of tube, let n be the number of sides of this poligon and D the diameter of the inscribed circle in the poligon. Then increase the number of sides n. Then calculate lim f(n) for n → ∞ and you will get exactly the same formula.
Hello, thank you very much for this amazing video! I was also trying to calculate the volume of an elliptic torus shape revolved around the y axis by using the Washer method. Would this formula be applicable to ellipses as well?
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Nice vid! I had fun verifying that the 3rd method does give the same answer as the 1st method for the elliptical torus [🍩] consisting of revolving an ellipse with horizontal axis a, vertical axis b, and whose left-most point is R units to the right of the origin! V=2π²abR :)
@@MathNerd1729 I believe a circle is revolving, not an ellipse
@@Jasomniac Well, a circle is a special kind of ellipse where a = b. But you're probably right that it technically may not be a torus! I meant that I liked verifying it works when you revolve any ellipse with horizontal and vertical axes around the y-axis in a similar fashion regardless of whether or not it's a circle! :)
@@MathNerd1729 sorry, I read your comment wrong, what you said is well phrased. And btw, I don't know if "general torus" is the right phrasing, but if we're gonna be very specific, I guess you could have said elliptical torus.
@Math Nerd 1729 but when you say it works with elliptical torus, do you switch the "r^2" in the last method for a•b ?
using tripple integral to find the volume reduces to these equations depending on order of integration
How a chemist would solve it: Put in beaker and see how much the water level rises.
that would be the best method if you had a real donut (if it was waterproof), because of small irregularities on the surface of the donut that are hard to calculate
Archimedes: "Eureka!"
@@herman7550 Except, once waterproofed, it would probably float. You could still do it, though, if you used a very low-volume implement to force it to submerge.
Fred
Chemists are water addicts
@@ffggddss It'd be trivial to correct for the volume of the implement used
Indeed. I must do calc over this summer.
Its too late lol
Gosh same
Bet you say that every summer
I watch all your videos, you are one of the best mathmatians in the world
Maybe not the best mathematician, but definitely the most entertaining 😊 (He's definitely very educated when it comes to math though)
Thank you. I am definitely not the best mathematician but I thrive to become a good math teacher/TH-camr 😃
Very nice presentation! Thoroughly enjoyable. Some related thoughts it brings to mind:
1) For the benefit of other viewers, in your example of Pappus' Theorem - for a torus, r is called the "minor radius," and R is called the "major radius."
2) We were taught Pappus' Theorem (P.T.) in my 7th grade general math class. It was a very impressive example of mathematical beauty, as well as being useful for many otherwise very hard problems. Then, years later in calculus class, one of our exercises was to prove P.T. - the general (3D) case.
3) And just BTW, it also works for finding surface areas. You might consider also finding the *area* of the Krispy Kreme this way ;-)
4) P.T. generalizes to n dimensions, for n ≥ 2. You can find the area of a circular annulus by revolving a line segment around a point collinear with it, and outside of it. You could find the 4-capacity of a 4D solid of revolution, given the volume of the 3D solid being revolved, etc.
4a) I once used it to find the capacity of an n-ball, using an insight that a certain relation between a 2-sphere and a circular disk, might generalize to other dimensions.
It involved a torus whose major and minor radii are equal (a zero-hole torus); and I was able to verify that generalization formally.
I hope, one of these days, to make a YT video about that...
Fred
Thank you, Fred!
If you're doing it by disc or shell, you can simplify the integrals by noting the symmetry of the problem.
A donut is like a cylinder which is curved into a circle of radius R....hence height of cylinder is 2πR...and if the radius of the cylinder is r then area of cross section will be πr²...hence volume of the donut will be the volume of the cylinder = (πr²)*(2πR) = 2π²r²R
That's actually how I solved it😅
Edit: after watching the full video I realised that's the pappus method😅
Yup! 😃
If you morph a cylinder into a torus like that though, parts of it are going to be compressed and other parts are going to be stretched. Apparently if you choose your radius of rotation to be the centroid of the circle, the compression and stretching cancel out. It is noteworthy that you have to make sure you use the centroid though, and if you're rotating something other than a circle around a central axis, it may be difficult to figure out where the centroid is.
@@IGDZILLA That is true, but in the limit as thickness of the wedge(distance between centers of circular bases) approaches zero, the difference between outer and inner "heights" will also approach zero and the wedge will approach a disk shape
@@JordHaj That reasoning is tricky though. That would apply to any cross-section.
@@chitlitlah I noticed it and tried to find the average of the compressed and the stretched side and realized that it IS the centroid
I absolutely prefer the Pappus Theorem, because it is the most intuitive to me (and it avoids integrals... I haven't done integrals in 20 years).
Why doesn’t this have 4 million views? It’s literally the perfect video for the algorithm
For a complete newbie in calculus, this makes so much sense, especially the Pappus Theorem! I accidentally saw it on my calculus book while cleaning my shelves, but your explanation made the theorem more exciting and sensible. Awesome content, as always
Hi man, I just want to let you know that you helped me with gaining more motivation for math. I am from the Netherlands btw. Thank you sir
I am happy to hear that my content has helped you 😃
I've just had my last lesson of mathematics seminar this year on high school. It was about calculating volume of shapes with functions and integrals and a simple formula.
Its funny because I was thinking of how to solve this withouth calculus, and I thought of using the circumference of the big donut times the area of the cross section. Turns out it was the circumference of halfway through the big donut, but it was still cool that I almost figured it out on my own.
What the heck, I never had pappus theorem taught in Calc 1 THAT WOULD HAVE BEEN NICE!
Thank you for the great videos! I saw your channel for the 1st time 7 mo. ago on the derivative on x^x, and I didn't even know what calculus was yet lol. After watching many of your videos, I now know lots of calculus 2 thanks to you :D
Happy to help! Thank you.
I only recently discovered your amazing channel. I'm a teacher and I would have loved to have a resource like this when I was a student. Subscribed!
I just dip in water just like Archimedes did to find the volume.............
Wish I had a math professor like him :(
Im not even in calc yet, im still in algebra. But my curiosity always push me beyond, just wandering here :))
Hey man love the vids!!!!! I'm Spanish (studying in Spain) and I got a 93/100!! Really appreciated your help with the integral videos😄😄😄
wow, never knew there is so much to learn from a donut
😄
I got it! But by using none of these ways, sort of (I actually did do the Pappus theorem, but solved for it first and the best part is I didn’t know it was a thing when I solved it). I basically did the washer method, except I did it the easy way. Which is by turning everything into rectangles, which ends up turning everything into cylinders.
I solved for it generally, and then plugged in the numbers.
The result was
2m(pi^2)(r^2)
Where m is the distance from the center of the torus to the center of the solid circular part and r is the radius of solid circular part.
And as expected, the approximate result was 144.343 cm^3.
You inspire me to learn calculus. Keep up your great videos!
Omg I was thinking for the whole video that the Pappus-Guldinus theorems would be so much easier, and then you did it; well done.
I actually used this approach years ago to calculate the volume of a vase in my house. Afterward I filled the vase with water and I was only off by about 1% in accuracy. Fun stuff.
According to my mentor and professors, I am very good at calculus. I feel that way too but not as much. The thing is, bprp makes everything seem so SIMPLE!
I didn't hear about pappus theorm , I solved it by imagining a horizontal cylinder with length dL , I found that dL is length of arc with radius from center of donut to center of cross section area of the cylinder times dtheta where theta is rotation about y-axis , so small volume dv=cross section area of horizontal cylinder times arc length times dtheta , then integrate from zero to 2π and got the answer
Well,I like the second way to solve the volume of a donut.It’s pretty clear and easy way to learn😀
Good video!
It would be interesting to see a solution with Fubini and Tonelli theorems
And maybe with the change of variables to cylindrical coordinates
You are a Great Teacher!
I actually begin questioning into the region of integrals and limits when I questioned the volume of a cylinder in class IX. I wondered how one could find the exact volume when the circle area had infinitely small thickness and was multiplied by the height. Now in Class XI, I see how integrals can be used. Epic!
Donut is a cylinder with height 2πr
Volume of cylinder = πR²(2πr)
2π²R²r.
Long ago i calculated it in my mind
π² term facinated me!!
if you had a function f for the circle, you could do pi times the integral of f^2 to also get the volume of the donut
Yes, i thought, he would mention this way, it’s the way I solved this problem at school.
how have i not heard of you sooner...this may honestly be better than khan academy
I think this would work too,
find circumference of outer and inner circle, find average, and multiply by a cross section of the donut.
The shell method resonates best with me.
The Pappus Theorem feels like a trap. Not that I think Pappus was lying; it's just that it feels like the sort of thing a person could misapply if they're not careful.
Now I will make my summer productive by choosing the donut with highest volume 😋
You are a great teacher my man
I didn't learn volume of solids by rotation until calc 2. Also, I never learned about Pappus Theorem although it showed up in my engineering statics book. I took calc 1 two years ago and got an A.
The teacher is very smart enough. I am very happy to learn from him to solve this problem. I hope the teacher can get NASA recommendation to help astronauts to solve Mars, Moon or the universe engineering calculations problems.
I was thinking Pappus theorem, I also had the top marks in all 3 OAC (grade 13) maths: Calculus, Algebra, and Finite. Ended up taking a few more Finite and Probability as electives in my undergrad. Grad school did offer a course in Numbers but it was Pentateuchal studies.
Yooo congrats bprp on the sponsor!
Thanks.
explaining to your audience why you have to use the centroid would be useful for this Pappas method. reason being that the inner radius is shorter than the outer radius makes it seem as though modelling it as a straight cylinder could be inaccurate. using the centroid sort of "averages" the over estimation of the inner radius and the under estimation of the outer radius. something like the area of a trapezoid where you average to two unequal sides. same idea in an abstract sense. it took me a little thinking to agree with this pappas method. so a better understanding, say for example if the cross section was not a circle, but say, two half circles with different radii forming a funky donut (torus). then you would have to integrate the cross sectional shape to find the centroid, then use the coords of the centroid to apply the Pappas method.
nice video, I liked how you used three methods.
In one Homework in mathematics for Physicists, we had to also do that. I remember having brought this up to the table with Pappus theorem, but we couldnt use it, because it was an higher analysis course and we should have done in the 1st presented way.
It was challenging but also fun, when you get your results correct :D
By the way, i didnt know, that this theorem was called "Pappus Theorem"...
That use either shell or washer method. We actually didn’t learn that in calc 1 rather it was in calc 2 cause calc 1 was rushed over a summer interim session, but nonetheless it seems odd that such a simple concept requires calculus. Interesting video, the only thing I can comment on at the start is that you will have to do two times the integral of the positive half of the circle offset by some volume.
I got an A in my calc I class - 45 years ago!
48 for me. (Had to pull out a calculator to figure that).
i could solve it by the pappus theorem before watching the video:)
Finally some real word applications for maths
A while back I came across your world record video and always wanted to learn calculus. I think I will succeed this summer with the help of your videos! Thank you!
p.s I had to laugh really hard during the wr video @4:55:55 ''Integration by parts!''
Exact value: 14.625pi^2
Save this man
Probably some differences in naming. When I studied we called the third method "the 2nd Guldin's theorem".
The 1st Guldin's theorem finds the square of the rotated curve.
My intuition of what the answer would be was something like, "take a cylindrical bar of dough of that radius, bend it to circular shape where both ends meet without distorting the radius of the cylinder. The volume of that cylinder will equal the volume of the doughnut." My non-rigorous reasoning was that the outermost "dx" would stretch by the same amount that the innermost "dx" would compress. I dunno if it works exactly that way, but the cylinder idea at the end of the video felt similar to my thoughts.
That's exactly what the theorem ends up doing.
I was calculating this with the first method, albeit I changed the x and y-axis. I also tried to calculate its surface, but I was not able to find the proper solution.
⭐ What a brilliant video! ⭐
Practical integrals at its best!
Finally some actual real world use for math!
Can you answer me this, in what country and in what school or college do you teach math? Because it seams like your students are having a lot of fun with you
I've always wondered how loud they were.
For the confused ones, Pappus' theorem is also known as Guldin's theorem.
would have been nice to add in the cm after the numbers, so we can really get the cm3 :)
You can also measure volume by dunking donut into a cup and you can see the difference between before and after
I get a giant coffee cup (although I'd have to settle for a glass coffee pot since giant coffee cups are hard to find), I fill it half way with coffee that I've measured. I then take the donut and put it into the coffee and push it down and I measure how much the coffee has risen in the glass coffee pot. I then compute the volume by: measured height difference times PI times square of the Giant Coffee Cup Radius. That gives me the volume. Another way to do it is to measure the thickness of the donut ring, and the inner and outer radius and use a little calculus. But I like the Giant Coffee Cup method better, because that way I can drink the coffee and eat the soggy donut. So why this method better; besides having breakfast? Because from a Topological point of view: there isn't any difference between a coffee cup and a donut. Works better on an old dried out Donut. It tastes better too and having food and drink and not an empty stomach helps when doing math.
Now N A S A will have easy time calculating the orbit of satellite circling around the planet. Thanks to Pappus theorem. Thanks a lot for your video. Super informative 👍👍👍👍👍
National donut day is a Friday in June so it's appropriate.
great.I like the second method
Could you please do a video where you calculate an egg's volume?
The integral of (arctan(1/x))/x
lol. That end ist good "Just make it easy i bored you"
Love this idea!
Measuring in centimeters? Thought you were in the US where they will use anything to avoid metric!
Hahaha
he seems to be an abnormal
Wow! Just amazing!
This is a very creative idea
哈哈哈好眼熟 指考生寫積分的時候有寫到🤓
I will just say.....maths is smthg which is endless🥰
Washer: π*(r)^2
Shell: 2π*r*h
Just some notes for me
The last method was definitely the best, since you didn't have to use integration!
Can you explain when and why you pull out dy/dx? Never really understood it
Well, since a donut is a circle, you could calculate the volume of the circle and then subtract it from the hollow circle in the middle of the donut
"Volume of a circle" is meaningless, since a circle is a 2D shape, and volume is a 3D concept.
Can you make a video on the volume of a Croissant?
I thought that the third method would involve taking an integral in polar coordinates. Instead, it was just simple multiplication. Why is that?
Can you also Post Videos where you Solve the Problems on a Blackboard using a Chalk from time to time, like you used to do a Few Years Ago? I miss the Blackboard-Chalk System. :(
Maybe sometimes in the summer when I visit some schools again.
@@blackpenredpen Thank You. And a Great Video as Always. 😃
I'm deeply impressed that in the US, you've got rulers divided in cm!
my calc teacher did this for us too
What about the volume of a donut using the water method (first discovered by Archimedes)?
Definitely not water. Maybe coffee is okay. 😆
water? custard!
won't work, the donut will soak up the liquid ruin the result 😜
it will be more accurate because a donut is not a perfect Taurus and there is air in the donut that you might not want to include.
Will you still upload over the summer?
Yes
Or just use the Archimedes method with water and a paralellipedical tank
Yes, if you have an utterly nonporous, nonabsorbent doughnut.
Me: *blending the krispy kreme fine and using a measuring cup*
Engineers: Welcome to the approximation appreciation group
my logic before watching: since you are rotating the circle around a larger circle that connects the center of the cross section to the center of the donut, the volume should just be the area of the cross section times the circumference, which would be (πr^2) x (2πR)=2 π^2 r^2 R.
just a little bit of logic instead of actually proving it lol
5:35 I think this must be x = 3,25 - (1,5^2 - y^2)^1/2 not x = 3,25 - (1,5 - y^2)^1/2 (this is next to donat hole).
For the Pappus theorem, I'd take a shortcut by replacing 2πR with πD, where D is simply 9.5 - 3 :). But I'm still missing some insight into why exactly this method is correct (although it makes intuitive sense).
That's a very good question.
And you don't even need integrals to prove the formula. Just replace the thorus with a poligon made of tube, let n be the number of sides of this poligon and D the diameter of the inscribed circle in the poligon. Then increase the number of sides n. Then calculate lim f(n) for n → ∞ and you will get exactly the same formula.
So much work for a donut. It’s worth it, though
😆
dip it in a water pool, displacement
Calculus: Shells. Non-calculus: Area x Distance.
I really hoped you would dunk the donut in water to measure the volume at the end 🤣
Now, if you knew rho, the density of the cooked donut flour, you could then calculate the weight of said donut!
breathtaking ,talk you later- 🤠
😆
As an always hungry mathematician
That was so delicious 👏
It's winter here in Brazil
Hello, thank you very much for this amazing video! I was also trying to calculate the volume of an elliptic torus shape revolved around the y axis by using the Washer method. Would this formula be applicable to ellipses as well?