Question

Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = -3.

Answer 1

Using washers, the volume is given by

$\displaystyle\pi\int_{x=1}^{x=e}((1+3)^2-(\ln x+3)^2)\,\mathrm dx$
$=\displaystyle\pi\int_{x=1}^{x=e}(7-6\ln x-(\ln x)^2)\,\mathrm dx$