Surface area of a revolution, why is this wrong?

  • #1
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Main Question or Discussion Point

Why is this way of thinking wrong?. can't I assume that when Δx tends to zero is a sufficient approximation of what I want to get? It confuses me with the basic idea of integrating a function to get the area beneath a curve of a function (which isn't also as perfect) .

PD: I put Δx tends to infinity in the image but I think the right way I was thinking is to zero.

Thank you. ç

(English is not my first language).
 

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  • #2
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This may help your understanding a bit


Notice they use the Pythagorean theorem to get the length of the dxdy triangles hypotenuse with ##dx^2 +dy^2##which gets transformed into ##1+f’^2## When you factor out the dx and then of course you take its square root and integrate it from x0 to x2.
 
  • #3
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Why is this way of thinking wrong?. can't I assume that when Δx tends to zero is a sufficient approximation of what I want to get?
Yes, your way of thinking is wrong. You have to take into account the shape of the curve. Your way would work if you were dealing with a cylinder, but it doesn't work for a surface of revolution. ##\Delta x## is not a good approximation of the distance between the points ##(x, f(x))## and ##(x + \Delta x, f(x + \Delta x)##. Take a look at the video that @jedishrfu showed.
 

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