Question

Let R be the region in the xy-plane between the graphs of y = ex and y = e-x from x = 0 to x = 2.
a) Find the volume of the solid generated when R is revolved about the x-axis.

Answer 1

I'm guessing that you want the region between\[e^x\text{ and }e^{-x}\]revolved about x, right?

Answer 2

disk method seems the way to go here, but since we will have a hole in each disk because of the gap between e^(-x) and the x-axis we will have to adapt disk method to become 'washer method'
each washer will have this cross-sectionthe area of each washer as a function of x will be\[A(x)=\pi(r_{outer}^2-r_{inner}^2)=\pi[(e^x)^2-(e^{-x})^2]\]the volume will be the sum (integral) of all these washers from x=0 to x=2, so those will be your bounds of integration