Find the volume of the region bounded by y=e^-x, y=1, and x=3 about the line y=2

Question

I mainly think I am setting up the integral wrong because of the line it is rotating about.

Answer 1

Wyzant is very glitchy, so the video is not as smooth as I would have liked, but hopefully you get the idea.

Answer 2

1. Always draw a diagram

2 Work out what the outer and inner functions are and what the radii of rotation are from the given axis. In this case y=e^-x is farthest from axis and y=1 is the closest. (The radii are 2-e^-x and 1)

3 Figure out what kind of shape you want to sum up for the integral. In this case, we want to use vertically-oriented washers (because of the "hole" in the rotated volume) with width dx, that we integrate in x from 0 to 3

4 Write the general form of the integral: Integral from x_L to x_H of π(r_out²-r_in²)dx

5 Write out integral in terms of the y functions and their distance to the axis:

Integral from 0 to 3 of π((2-e^-x)²-(2-1)²)dx (The r's are distances, so I made the differences positive)

integral from 0 to 3 of π(3 - 4e^-x + e^-2x) dx which is reasonably integrable.

Please consider a tutor. Take care.