A general question about Surface Area

This question is about the surface area of a function defined as f(x) rotated around the x-axis

Now I understand how the ACTUAL integral works to find the surface area, but i'm wondering why a different integral wouldn't work...

[tex]\int[/tex] 2[tex]\pi[/tex]*f(x) dx

wouldn't this add up the differential circumferences, so shouldn't this equation work? why do you have to use the Pythagorean theorem integral version to find the surface area?

thanks if you can answer this question :)
 

LCKurtz

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That doesn't work for the same reason such a technique doesn't work for calculating arc length. Look at the graph of y = x from 0 to 1. We know its length is sqrt(2). But suppose you make a staircase function going up that line and use the sum of the delta-x's as your approximation to the length. Even when you take a finer and finer staircase, the sum of the delta-x's adds up to 1. The horizontal steps even though they get closer and closer to the straight line, don't approach it in length. Your surface area example is the same thing in one higher dimension.
 
That doesn't work for the same reason such a technique doesn't work for calculating arc length. Look at the graph of y = x from 0 to 1. We know its length is sqrt(2). But suppose you make a staircase function going up that line and use the sum of the delta-x's as your approximation to the length. Even when you take a finer and finer staircase, the sum of the delta-x's adds up to 1. The horizontal steps even though they get closer and closer to the straight line, don't approach it in length. Your surface area example is the same thing in one higher dimension.
very interesting... i really never thought of it like that. damn, thank you!

you really know your stuff :)
 

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