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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
I don't understand that second integral you wrote.
Oh wait, yes I do.
integral 2pi * f(x) sqrt ( 1 + f ' (x) ^2 ) dx
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
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.
the books way is integral 2pi x^2 sqrt ( 1 + (2x)^2) dx
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
err I said that wrong
But you see what I mean.
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
But for the line integral it doesn't come close
yeah, why is that
Let me see if I can explain.