Surface area of a revolution, why is this wrong?

[CollinsArg]

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).

 

[jedishrfu]

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.

 

[Mark44]

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.