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 = -1.

Question

ANSWER:

v = ∫[1,e] π(4-(lnx+1)^2) dx

Answer 1

looks good to me. Or, using shells,

v = ∫[0,1] 2π(y+1)(e^y-1) dy

Answer 2

thank you!