find volume when the region bounded by the curves y = Ln(x), y = 2, and x = 1 is revolved around the line y = −2.

Answer 1

The corners of the roughly triangular region are at

(1,0),(0,2),(e^2,2)

So, using thin washers of thickness dx,

v = ∫[0,e^2] π(4^2-(2+lnx)^2) dx
= 95.298

Using shells of thickness dy, we have

v = ∫[0,2] 2π(2+y)(e^y-1) dy
= 95.298