I study civil engineering and every time I get stuck with integrals or whatever, I find the answers I need in your videos,for many months now. Thank You for making such a great channel
By far the simplest way of deriving area of a circle. Utilizes circumference = 2 pi r (one of the fundamental equations of a circle) as a dependent variable (and r as independent variable). The integration is Definite from 0 to R.
A very small change in the radius r which almost tends to zero, r+dr becoming zero not making any difference so the width of the elementary element within the radius is considered dr
It's the antiderivative, not the derivative. The antiderivative of r is r^2/2. Because if you take the derivative of r^2/2, you get r again. Hence, it's the antiderivative.
Regarding the ring area. Suppose the r is 3 so the area of the small circle is 9pi squared, now suppose dr is 1 then the medium circle's radius is 4 and the area is 16pi squared. 16pi - 9pi = 7pi squared which should be the area of the ring. But when I do the way u did, then 2pi·r=6·1(dr)= 6pi squared which doesn't add up - can u explain pls?
@Team IGHappiness Actually dr is supposed to mean a length which is infinitely times small( like even smaller than 0.0000000000000001) so you cant take it a big number such as 1
this is wrong you got the correct equation as the result, but your logical steps to arrive there are flawed - for instance, everything you said is _also true_ of a regular hexagon : the 'circumference' of a hexagon is related to its radius by the 'proportionality factor' of _six_ so, for a hexagon, you'd have C = 6r, then follow every step, exactly as present here, and conclude [with the exact same level of validity] that the area of a regular hexagon is 3r^2 ... which is not true you can integrate the correct formula for the area of a hexagon - or any regular polygon - if you find the 'proportionality factor' of the perimeter to the _apothem_, but not if you use the radius; the greater the number of sides of a regular polygon, the closer the apothem approaches to the radius, and you happened to pick a polygon with infinity sides, so both lengths are the same again, this is wrong because your logic is wrong and is easily demonstrated to be false for every single possible polygon other than the one you picked - it's more importantly wrong because it presents the idea that you can use these steps of logic to set up solutions, using calculus, to other questions you might encounter
no - you can take it as axiomatic that C=6*r for the 'circumference of a hexagon' and then, again, follow every one of your steps to reach the wrong conclusion, because your logic is incorrect - you can not integrate perimeters as a function of radius to get areas in the case of a regular polygon, you can integrate perimeter as a function of apothem but the more important thing is that you can NOT use the concept that the area is the sum of all the changes in perimeter in other situations - like in situations where you don't already know the answer - you got the right answer for the wrong reason, and you stopped because you already knew what you wanted the conclusion to be also, you don't need to assume C=Pi*d for this - especially for this! and especially with _diameter_! all you can assume is that the perimeter is proportional to the apothem which in this polygon is equal to the radius: C ∝r => C(r) = k * r so, the area would be equal to the, circumference as a function of the radius, times infinitely small changes to the radius, and all of those added up: area = int(k * r * dr) => area = (1/2) * k * r^2 what's the k? since C=k*r, we know there are k radians on a unit circle and a right angle has k/4 radians [which we'll use because we need a valid angle for both sin and cos] - with theta in radians: e^(i*k/4)) = cos(k/4) + i * sin(k/4) = 0 + i*1 = i i*k/4 = ln(i) k/4 = (ln(i)/i) * (i/i) = -i * ln(i) k = -4 * i * ln(i) => area = (1/2) * -4 * i * ln(i) * r^2 = -2 * i * ln(i) * r^2 = -i*ln(i^2)*r^2 area = -i*ln(-1) * r^2 [my calculator approximates -i*ln(-1) as 3.14159... something] and look at that: we used calculus, instead of geometry, in our calculus class
I'm assuming you meant to put 3:48. There's a formula for the anti-derivative, X^(n+1)/(n+1), hence the simplicity of this one. After doing them for a while, you'll start to see them easier.
@@saulremi i watched the video very closely several times and I spotted the part where you coughed and said 'hey, look over there' when you skipped a step.
@@stevematson4808 this derivation is an easy one. Look at the derivation for the area of a sphere, that one is one that takes more effort. I really do think you'll see that in time.
@@saulremi yes, I see it now. The formula for the anti-dirivitive makes more sense. This video makes more sense than some of the others ive seen on the subject.
I study civil engineering and every time I get stuck with integrals or whatever, I find the answers I need in your videos,for many months now. Thank You for making such a great channel
9 years ago 😱
Think needed to show why dr*circumference =dA by showing it same as rectangle once cut and unwound
much simpler than assuming the formulae function of a circle and the integral by dx or dy.
By far the simplest way of deriving area of a circle. Utilizes circumference = 2 pi r (one of the fundamental equations of a circle) as a dependent variable (and r as independent variable). The integration is Definite from 0 to R.
This was one of the most easiest explanation .SIR U EARNED A LIKE AND A SUBSCRIBER
Excellent.
@@saulremi and what about the website where u said we can ask questions directly. the website is broke rt now
like all your video sir thanks.feels like im college again
What do you mean by the thickness dr ? A difference in the radius ? Sorry for the stupid question.. I'm very rusty
A very small change in the radius r which almost tends to zero, r+dr becoming zero not making any difference so the width of the elementary element within the radius is considered dr
Excellent video, cheers!
at 3:37 why r derative is r^2. I don't understand step 4 can please elaborate it.............. please sir
It's the antiderivative, not the derivative. The antiderivative of r is r^2/2. Because if you take the derivative of r^2/2, you get r again. Hence, it's the antiderivative.
Regarding the ring area. Suppose the r is 3 so the area of the small circle is 9pi squared, now suppose dr is 1 then the medium circle's radius is 4 and the area is 16pi squared. 16pi - 9pi = 7pi squared which should be the area of the ring. But when I do the way u did, then 2pi·r=6·1(dr)= 6pi squared which doesn't add up - can u explain pls?
@Team IGHappiness Actually dr is supposed to mean a length which is infinitely times small( like even smaller than 0.0000000000000001) so you cant take it a big number such as 1
Thanks so muchh🙏🏻
thanks dude
Nice
hello sir what pi is?
ruclips.net/video/uK2OQMUAUDQ/видео.html
The ratio of the circumference to the diameter in simple words.
food
@@lionel0353 wtf
Came here to just check my own proof and its perfectly the same... 🎆🎆
i get it!
internet high-five!
this is wrong
you got the correct equation as the result, but your logical steps to arrive there are flawed - for instance, everything you said is _also true_ of a regular hexagon : the 'circumference' of a hexagon is related to its radius by the 'proportionality factor' of _six_
so, for a hexagon, you'd have C = 6r, then follow every step, exactly as present here, and conclude [with the exact same level of validity] that the area of a regular hexagon is 3r^2 ... which is not true
you can integrate the correct formula for the area of a hexagon - or any regular polygon - if you find the 'proportionality factor' of the perimeter to the _apothem_, but not if you use the radius; the greater the number of sides of a regular polygon, the closer the apothem approaches to the radius, and you happened to pick a polygon with infinity sides, so both lengths are the same
again, this is wrong because your logic is wrong and is easily demonstrated to be false for every single possible polygon other than the one you picked - it's more importantly wrong because it presents the idea that you can use these steps of logic to set up solutions, using calculus, to other questions you might encounter
Love the confidence. But I believe you are mistaken. I took C=Pi*d as an axiom, from which the equation for the area followed.
no - you can take it as axiomatic that C=6*r for the 'circumference of a hexagon' and then, again, follow every one of your steps to reach the wrong conclusion, because your logic is incorrect - you can not integrate perimeters as a function of radius to get areas
in the case of a regular polygon, you can integrate perimeter as a function of apothem
but the more important thing is that you can NOT use the concept that the area is the sum of all the changes in perimeter in other situations - like in situations where you don't already know the answer - you got the right answer for the wrong reason, and you stopped because you already knew what you wanted the conclusion to be
also, you don't need to assume C=Pi*d for this - especially for this! and especially with _diameter_! all you can assume is that the perimeter is proportional to the apothem which in this polygon is equal to the radius:
C ∝r => C(r) = k * r
so, the area would be equal to the, circumference as a function of the radius, times infinitely small changes to the radius, and all of those added up:
area = int(k * r * dr) => area = (1/2) * k * r^2
what's the k? since C=k*r, we know there are k radians on a unit circle and a right angle has k/4 radians [which we'll use because we need a valid angle for both sin and cos] - with theta in radians:
e^(i*k/4)) = cos(k/4) + i * sin(k/4) = 0 + i*1 = i
i*k/4 = ln(i)
k/4 = (ln(i)/i) * (i/i) = -i * ln(i)
k = -4 * i * ln(i)
=>
area = (1/2) * -4 * i * ln(i) * r^2 = -2 * i * ln(i) * r^2 = -i*ln(i^2)*r^2
area = -i*ln(-1) * r^2
[my calculator approximates -i*ln(-1) as 3.14159... something]
and look at that: we used calculus, instead of geometry, in our calculus class
So at 4:48 he just pulled the anti-dirivitive out of nowhere?
I call bullshit.
I'm assuming you meant to put 3:48. There's a formula for the anti-derivative, X^(n+1)/(n+1), hence the simplicity of this one. After doing them for a while, you'll start to see them easier.
@@saulremi i watched the video very closely several times and I spotted the part where you coughed and said 'hey, look over there' when you skipped a step.
@@stevematson4808 this derivation is an easy one. Look at the derivation for the area of a sphere, that one is one that takes more effort. I really do think you'll see that in time.
@@saulremi yes, I see it now. The formula for the anti-dirivitive makes more sense. This video makes more sense than some of the others ive seen on the subject.
@@saulremi ok, I have a question for you. Does every math expression have a dirivitive and an integral- and do they always mean something?
No...pie are round...not square.