im stumped calculus circumference arc. how come we cant use integral C(x)dx to find surfacea area. C(x) is the circumference so its 2pi*r , and dx is the thickness.
so why isnt surface area integral 2pi * r * dx
instead it is integral 2pi y sqrt ( 1 + dy/dx^2)dx

- anonymous

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- anonymous

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

- anonymous

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

- anonymous

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

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## More answers

- anonymous

generated i should say

- anonymous

Ok

- anonymous

I don't understand that second integral you wrote.

- anonymous

Oh wait, yes I do.

- anonymous

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

- anonymous

yeah

- anonymous

Polpak, when you get finished can you assist me?

- anonymous

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

- anonymous

what r u guys parternes?

- anonymous

Are you rotating around the x axis?

- anonymous

yes in this case

- anonymous

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

- anonymous

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

- anonymous

You would be fine if it were flat washers

- anonymous

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

- anonymous

thats true but

- anonymous

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

- anonymous

Yes, but they still have an angle.

- anonymous

You're ok computing the volume that way

- anonymous

but not the surface area.

- anonymous

actually its not volume

- anonymous

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

- anonymous

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

- anonymous

See you're way makes the assumption that along the top edge for a tiny dx, that the length of the line of the function = dx.

- anonymous

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

- anonymous

come back

- anonymous

but it doesn't. The length of the line for a small dx is proportional the the square root of the square of dy^2 + dx^2

- anonymous

err I said that wrong

- anonymous

But you see what I mean.

- anonymous

We are ok with the area under the curve assuming that as we take smaller and smaller values for dx, the area approaches f(x) * dx

- anonymous

But for the line integral it doesn't come close

- anonymous

yeah, why is that

- anonymous

Let me see if I can explain.

- anonymous

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

- anonymous

You there?

- anonymous

http://www.twiddla.com/529576

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