Find the volume generated by revolving the region bounded by x=1, y= e^x/2, y= e^x about the x-axis?

Question

Using the washer method or would it be the shell method. I'm having trouble finding that the boundaries would be? Thank you.

Answer 1

I assume you mean e^(x/2)
You can use either method. Using washers of thickness dx, the ring of each washer is just the distance between the curves. Both graphs start at (0,1), so
v = ∫[0,1] π(R^2-r^2) dx
where R=e^x and r=e^(x/2)
v = ∫[0,1] π((e^x)^2-(e^(x/2))^2) dx
you can simplify that to
v = ∫[0,1] π(e^(2x) - e^x) dx = π/2 (e-1)^2

Answer 2

What about the x=1 part does that just create the boundary or do I shift the graph to the y-axis[ left 1]?

Answer 3

the line x=1 sets the right limit of integration. The region is roughly triangular, with vertices at
(0,1), (1,√e), (1,e)
integrating in the x direction just stacks up all those washers horizontally.

You can use shells of thickness dy, but it gets a bit more complicated, since the right-hand boundary changes at (1,√e). From y=1 to y=√e, the height of the horizontal cylinders is the distance between the two curves
x = 2lny and x = lny
From y=√e to y=e, the height is just 1-lny
The volume of a shell is 2πrh, but h changes as we pass y=√e
v = ∫[1,√e] 2πy(2lny-lny) dy + ∫[√e,e] 2πy(1-lny) dy
you have to use integration by parts to get
∫y lny dy = y^2/4 (2lny-1)
Putting all that together, we have the final result of
v = 2π(y^2/4(2lny-1))[1,√e] + 2π(y^2/2 - y^2/4 (2lny-1))[√e,e]
Yes, I did the algebra, and it again comes out π/2 (e-1)^2

go with washers!