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

Question

I got the integral from 1 to 2.718 of pi(1)^2-pi(ln(x))^2

Is this correct or is there a trick I'm missing because that's often the case with calculus

Answer 1

what's this 2.718 stuff? That is e. Use it, just like π, without some sloppy decimal approximation.

since the axis is the line y = -3, the radius of each disc is y+3. So,

v = ∫[1,e] π((1+3)^2-(lnx + 3)^2) dx