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You're trying to compute the area of a circle?

so say you have f(x)>0 in the first quadrant

im doing the formula for surface area , made by revolving a curve about an axis .

generated i should say

Ok

I don't understand that second integral you wrote.

Oh wait, yes I do.

integral 2pi * f(x) sqrt ( 1 + f ' (x) ^2 ) dx

yeah

Polpak, when you get finished can you assist me?

so my question is , why cant we just use the easier formula
integral 2pi f(x) dx

what r u guys parternes?

Are you rotating around the x axis?

yes in this case

say y = x^2 from 0 to 2 , but trying to find surfacea area

But you still need to account for the way f(x) is changing over your dx

You would be fine if it were flat washers

But along the top edge there can be a lot of slopes that will change the area of the final surface.

thats true but

when you go to zero, dont they approach 2pi * r , the height

Yes, but they still have an angle.

You're ok computing the volume that way

but not the surface area.

actually its not volume

it would be the surface of a cylinder type figure, i guess, not sure

for example here, we have using my way integral 2pi x^2 dx from 0 to 2

the books way is integral 2pi x^2 sqrt ( 1 + (2x)^2) dx

come back

err I said that wrong

But you see what I mean.

But for the line integral it doesn't come close

yeah, why is that

Let me see if I can explain.

http://www.dabbleboard.com/draw?b=Guest666913&i=0&c=d86fb2255acc9e8490af7e397d12fe9fc183a40a

You there?

http://www.twiddla.com/529576