Surface Area of Revolved Solids

[Kyle]

I'm confused about something we learned in Calc today and thought someone could explain it a little. We're doing arc lengths and now the surface area of a solid formed by revolving a curve.

 

Given any function [math]f(x)[/math] revolved about the x axis,

We treated the surface area as a lot of circumferences stacked next to each other, with the radius equal to y.

My teacher showed us the formula to be [math]\int_a^b f(x)2\pi\,ds\[/math]

I understand the concept of ds I think, but I don't understand is why you have to use ds and can't use [math]\int_a^b f(x)2\pi\,dx\[/math].

I pictured this problem as finding the circumference then multiplying it by thickness dx to make a hollow cylinder. Since dx is pretty much nothing, this "cylinder" would have a height of zero and would be just a 2-D circle. We could add up all the circumferences to get a surface area.

 

Why does arc length matter at all when your slices are infinitely thin?

[Dave]

You can use dx instead of ds if you want to; but you can't simply replace ds by dx, since [imath]ds = \sqrt{1+f'(x)^2} dx[/imath]. Take a look here for a more detailed description.