Just to be clear, you are meaning the surface area of the revolved region, not the volume?

I'm not very familiar with finding surface areas of revolved regions, I've mainly only dealt with volumes of solids, but I think I can offer some information.

I don't think this problem is quite as simple as you are remembering. You suggested that the surface area would be the integral of 1/x times 2pi. Try to think about this graphically: as we integrate with respect to x (assuming that's how you're going to integrate this function) the radii of the cross-sections of the solid are not all the same (2pi). The radii of the cross-sections are getting smaller as we move from left to right.

Here is how I look at it. The changing circumference is the fact that 1/x is always decreasing from 1 to infinity. If I take an infinitesimal slice at any point it will be a tiny band.

The reason I remember it so vividly is that when the teacher wrote the problem on the board, I was immediately taken aback that the area was infinity because LN(infinity) is infinity.