Find the region between y=e^x and y=e^-x rotated around the y axis. The bound is x=1. Would I use shell method for this or washer?
Shells would work. I don't see how "washers" would be used here.
You could also do this is two separate parts. When x= 1, y= $\displaystyle y= e^{-x}= e^{-1}$ is a bound and at x=-1, $\displaystyle y= e^x= e^{-1}$ is a bound. That is, for y= 0 to $\displaystyle y= e^{-1}$ the right and left bounds of the area are the straight lines x= 1 and x= -1. The figure is a cylinder, having area $\displaystyle \pi r^2 h= \pi(1)^2 e^{-1}= \pi e^{-1}$. You can do the volume from $\displaystyle y= e^{-1}$ up to y= 1 (where the two graphs intersect) by discs.