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

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

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

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

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!

To find the volume generated by revolving the region bounded by the curves about the x-axis, we can use the washer method or the shell method. Let's determine the appropriate method and find the boundaries for this specific problem.

First, let's graph the region bounded by the curves x=1, y=e^(x/2), and y=e^x.

To determine which method to use, we should check if the region is better represented by horizontal or vertical slices.

By observing the graph, we can see that the region is more naturally portrayed with horizontal slices. So the washer method will be more suitable.

Now, we need to find the boundaries of the region along the x-axis. Let's solve the equations to find their intersection points:

1. x = 1 (a vertical line)
2. y = e^(x/2) (represents the graph)
3. y = e^x (represents the graph)

To find the boundaries, we need to find where these curves intersect.

1. x = 1: This is a vertical line at x=1, which serves as the starting and ending point for the region.

Next, we equate the two equations representing the graphs:

2. e^(x/2) = e^x
Taking the natural logarithm of both sides:
ln(e^(x/2)) = ln(e^x)
(x/2) = x
(x-x)/2 = 0
0 = 0

Since 0=0 is always true, the two equations y=e^(x/2) and y=e^x intersect at all points, and their intersection boundaries are infinite.

Therefore, the region has no upper or lower bounds along the x-axis. Thus, the region extends indefinitely.

So, to find the volume generated by revolving this region about the x-axis, we integrate using the washer method. The integral will be:
V = π * ∫[(outer radius)^2 - (inner radius)^2] dx

The outer radius is given by y=e^x, and the inner radius is given by y=e^(x/2). Since the region extends indefinitely, we need to integrate from x=1 to x=∞:

V = π * ∫[e^(2x) - e^x] dx

Please note that integrating from x=1 to x=∞ can be challenging, as it deals with improper integrals. Some additional techniques such as substitution or partial fractions may be required to evaluate this integral.

Finally, you can solve this integral to find the volume generated by revolving the region.