# Surface area of a revolution, why is this wrong?

• I

## 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).

#### Attachments

• 16.1 KB Views: 308

jedishrfu
Mentor
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.
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.