Draw the region. y = 2e-x and y = 4 intersect when 2e-x = 4. Solving the equation, we get x = -ln2.
Using the disk method, the volume of a typical disk is π(4-2e-x)2Δx
So, the volume of the solid is ∫(from -ln2 to 6)(4-2e-x)2dx.
= ∫(from -ln2 to 6)[16-16e-x+4e-2x]dx
If you need help in evaluating the integral, let me know.