# Thread: Surface Area of Revolution - Integrating over circumferences

1. ## Surface Area of Revolution - Integrating over circumferences

I understand that to find the surface area of revolution you integrate over the bounds using 2 pi f(x) arc length dx, however I do not understand why I can't simply use 2 pi f(x) dx as the integrand since this is just calculating infinite circumferences and multiplying by the height, dx, of the infinitely small cylinder created. This works for calculating volumes using areas and then multiplying by the infinitely small height, and I don't understand why I can't do the same by finding the single dimensional length of the circumference and then multiplying by the height, dx.

2. Originally Posted by dude_500
I understand that to find the surface area of revolution you integrate over the bounds using 2 pi f(x) arc length dx, however I do not understand why I can't simply use 2 pi f(x) dx as the integrand since this is just calculating infinite circumferences and multiplying by the height, dx, of the infinitely small cylinder created. This works for calculating volumes using areas and then multiplying by the infinitely small height, and I don't understand why I can't do the same by finding the single dimensional length of the circumference and then multiplying by the height, dx.
That would not take into account the fact that the graph is not just vertical but tilts.

Here's a simpler version of the same thing: find the length of the line from (0,0) to (1,1). I could argue that if I divide the interval of the x-axis from 0 to 1, each segment would have length dx and so I would integrate $\int_0^1 dx= 1$, the same as the length along the x-axis. But, of course, that is not the length of the interval from (0,0) to (1,1).