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Find the volume of the following solid of revolution.

The region bounded by \(\displaystyle y=\frac{1}{x^2}\), y=0, x=2, x=8, and revolved about the y-axis.

I am trying to use the shell method to solve this, as it seems the best scenario in this situation. This is my set-up,

\(\displaystyle 2\pi\int_2^8 (radius)(height)dx\)

\(\displaystyle 2\pi\int_2^8 (x+2)(\frac{1}{x^2})dx\)

\(\displaystyle 2\pi\int_2^8 (\frac{1}{x}+\frac{2}{x^2})dx\)

Now here is where I am getting into a snag. Following this through I am coming up with a completely different solution to the manual, and the solution manual is showing the first integration step as \(\displaystyle 2\pi\int_2^8 (\frac{1}{x})dx\), but I can't figure out how they are coming up with that integral. It is showing a final solution of \(\displaystyle \pi\ln(16)\)

Any help, or a kick in the right direction, would be greatly appreciated.

Mac