# Revolution Volume by Cylinder Shells

1. Nov 12, 2006

### teleport

Rotate about x-axis the region enclosed by y=e^x, y=1/x, x=1 and x=2. I can do the problem with the rings method but I don't how to even set up the integral to solve by the shells method. Help? Thanks

2. Nov 13, 2006

### HallsofIvy

The two curves cross well before x= 1 so you have a region in which y=ex is the upper curve and y= 1/x is the lower. That means that the length of your cylinder is exx- 1/x. The radius is x so you cylinder will have a surface area, for each x, of $2\pi x(e^x- 1/x)$. Multiply that by the "thickness", dx, to find the differential of volume.

3. Nov 13, 2006

### teleport

Thanks, but since the region enclosed by the boundaries given is rotated about the x-axis, then, doesn't it mean that the radius of the cylinder is y? I'm a little confused. ??