Let R be the region enclosed by the graphs of y = 2lnx and y = x/2 and the lines x = 2 and x = 8. Set up an integral expressing the volume of the solid generated when R is revolved about the line x = -1.

A correct answer is 2pi{(x+1)(2lnx - x/2)dx (limits from x=0 to x=8)

But what about in terms of dy? I tried to set up an integral but my calculator says it's different from the above expression.

pi{(2y+1)-(e^(y/2) + 1)dx (limits from y=1 to y=2ln8)

What's up?