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