area volume

srry if i m not generous ...

btw u actually havent yet understood what my doubt is

i want to use your same arguement "discs are infinitesimally small in thickness" and hence write area ∫2πydx

even buttons work for me !!

clearly you have just by hearted the integration and all......

thanx man[1].....

u got eyes or buttons?
in the same post i also mentioned how integration makes correction for it!

but i don't care to explain you the basics now!

i m very generous these days.....so for sake of ur weak eyes i m HIGHLIGHTING how integration makes correction:

discs are infinitesimally small in thickness

"as integration would automatically make the correction"

clearly you have just by hearted the integration and all......
can you explain how does it corrected?
and why doesnt it correct the area if u write area as 2Ï€ydx??????

of course there is an answer to it ...but i havent got a convincing one yet

i dont have common sense like you and i cant keep anything in mind that isnt convincing ...may be you got the ability of by hearting even wat you dont understand completely

yaar discs are infinitesimally small in thickness so we need not worry about end portions of the disc as integration would automatically make the correction.

bhai use common sense

at last ... someone did help !! thanx nishant bhaiyya and pritish !!!

i got the points in #13 & #14 ,

nishant sir i agree that area shud always be ∫2πydL

an easier proof goes like this

dA can be treated as an area vector and it has 2 components, 2Ï€ydx & 2Ï€ydy

hence |dA |= ( (2Ï€ydx)²+ (2Ï€ydy)² )^1/2 = 2Ï€ydx (1+(dy/dx)²)^1/2

and this is exact and without any approximation ,

but in calculating volume we make an approximation ,
we say that the volume of disc will be Ï€y²dx ,

but that isnt even a perfect disc, and we are neglecting some volume , i.e 12 x (2πydy) x (dx) = πydydx

why does approximation work in volume calc but not in area ...

( in area if we use ∫2πydx the error adds up to πR(L - H) !!! )

Well..if Nishant sir says it's genuine, my apologies :) I never did this concept before my first sem of engineering, hence the misunderstanding.
You do know where ∫2Ï€y (1 + (dy/dx)²)^1/2 dx came from right?
If not, let me explain...this may be unnecessary, but it may help you in understanding the basic concept behind this. Suppose we have a curve AB defined by y = f(x), assuming it doesn't cut the X-axis. We take two points P(x, y) and Q(x + dx, y + dy) on it. Now the difference in the methods of calculating volume and area is this :

For area, we take the length of the arc PQ as ds and that of AP as s(making AQ = s + ds). If we drop perpendiculars C from A, M from P, N from Q and D from B, then 'S' denotes the curved surface area generated by revolving the area CMPA about the X-axis. dS is the curved surface generated by revolving MNQP. We assume that the curved area generated by revolving MNQP lies between curved surfaces of right circular cylinders whose radii are PM and NQ, which are of the same thickness ds.
which => 2Ï€yds < dS < 2Ï€(y + dy)ds (obviously!!)
=> 2Ï€y < dSds < 2Ï€(y + dy)

As Q→P or as ds→0, dy→0.
Thus we get dSds = 2Ï€y
=> dS = 2Ï€yds
We integrate within curve limits to get S, which is what we need.
∫2Ï€yds = ∫2Ï€ydsdxdx which gives your relation as dsdx = √1 + (dydx)².
Now you can apply this to the specific case of a cone, which is obtained by revolving a right angle triangle about it's perpendicular.

We use a different procedure for volume calculation where we get dVdx = πy² and πy² < dV/dx < π(y + dy)².

It's much simpler to think the way Nishant sir has, though :P

pleajjj helpppppp

nishant sir pls help !!

no one?

prophet sir + nishant sir + kaymant sir helpp