Area of a Circle (equation derived with calculus)

Equation for the area of a circle is derived with integration.

[diskospira]

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

[michaelpurtell]

Think needed to show why dr*circumference =dA by showing it same as rectangle once cut and unwound

[nickbarton]

much simpler than assuming the formulae function of a circle and the integral by dx or dy.

[vratislavgoldie]

This was one of the most easiest explanation .SIR U EARNED A LIKE AND A SUBSCRIBER

[qualquan]

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.

[Gypsy_Danger_TMC]

What do you mean by the thickness dr ? A difference in the radius ? Sorry for the stupid question.. I'm very rusty

[kryptonnite]

like all your video sir thanks.feels like im college again

[teamighappiness]

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?

[homiedeesvideos]

at 3:37 why r derative is r^2. I don't understand step 4 can please elaborate please sir

[randomdosing]

Came here to just check my own proof and its perfectly the same... 🎆🎆

[joelabraham]

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

[stevematson]

So at 4:48 he just pulled the anti-dirivitive out of nowhere?
I call bullshit.