Question

Find the volume of the solid generated when R is revolved about the line x = -2.
The region R is bounded by the x-axis, x = 1, x = 3, and y = 1/x.
Essentially, it's an integral of y = 1/x from 1 to 3, and the area under that curve comes out to 1.098 or so. I don't know what to do from there. Can anyone help?

Answer 1

It happens that the area under the curve is not really necessary here.

Have you heard of washer and disk methods for solids of revolution? One of those will help us here.

Answer 2

Heard of it, but never used it. If it helps at all, the average value of the region should be 0.549

Answer 3

So first I want to sketch our situation.


In fact it is looking like shell method would be easiest to apply here. If you have not seen the motivation of that method, we are essentially integrating:
2 pi (radius) (height)
for all radii extending from x=-2 and leading to points from x=1 to x=3.
\( \displaystyle \int_a^b 2 \pi r h \ dx \)
Radius would be distance from axis of rotation to the function. Height would be the value of the function.

I have to go for now to complete some homework and then sleep (very late here), but I hope you might be able to figure something out from here!
A nice resource for these methods is http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx
Also there may be others who are willing to help!