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`