*Sorry thread title should say volume not area.
Consider the solid obtained by rotating the region bounded by the given curves about the line y = -1.
Find the volume V of this solid.
To solve this problem, I first drew the four lines and then reflected the enclosed solid over y=-1. Since the radius of the solid is a vertical line, I integrated with respect to x. I know the area of the "washer" cross section is = to PI*R^2
I set R= 1/x-1
∫ [from 1 to 5] PI(1/x-1)^2 dx
PI ∫[from 1 to 5 (1/x^2 -2/x +1)dx
I then took that antiD...
PI[-1/x-2ln(x)+x) and evaluated it from 1 to 5
PI[-1/4 - 3.218876+4] which = .53PI
My online WebAssign hw says my answer is wrong. Any ideas where I went wrong? I have tried working it several times. This is my most recent attempt!