find the volume of a solid formed when the region bounded by y = e^(x-3) and y = 1 + ln(x) is rotated about the y-axis.

To find the volume of the solid formed when the region bounded by the curves y = e^(x-3) and y = 1 + ln(x) is rotated about the y-axis, we can use the method of cylindrical shells.

First, let's find the points of intersection between the two curves. Set e^(x-3) equal to 1 + ln(x) and solve for x:
e^(x-3) = 1 + ln(x)

Taking the natural logarithm of both sides:
x - 3 = ln(1 + ln(x))

Now, let's solve for x numerically using a graphing calculator or software. The solution is approximately x ≈ 2.887.

Next, we need to determine the bounds of integration. We want to rotate the region from x = 2.887 to x = a, where a is the upper bound. To find the upper bound, set e^(x-3) equal to 1 + ln(x) and solve for x:
e^(x-3) = 1 + ln(x)
x - 3 = ln(1 + ln(x))

Solve for x numerically to find the upper bound.

Now, we can calculate the volume using the formula for volume by cylindrical shells:

V = ∫[a to 2.887] (2πx)(f(x) - g(x)) dx

where f(x) is the equation of the upper curve and g(x) is the equation of the lower curve.

In this case, f(x) = e^(x-3) and g(x) = 1 + ln(x).

Substituting these values into the formula:

V = ∫[a to 2.887] (2πx)(e^(x-3) - (1 + ln(x))) dx

Evaluate this integral to obtain the volume of the solid.

Note: The upper bound, a, needs to be determined numerically as it depends on where the two curves intersect.