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

The region is a triangular patch with vertices at (1,0), (1,1), and (1,e)
So the volume is a stack of washers of thickness dx:
v = ∫[1,e] π(R^2-r^2) dx
where R=(1+3) and r=(lnx+3)
v = ∫[1,e] π(4^2-(lnx+3)^2) dx

Or, you can set it up as a stack of nested shells of thickness dy
v = ∫[0,1] 2πrh dy
where r=y+3 and h=x-1=e^y-1
v = ∫[0,1] 2π(y+3)(e^y-1) dy