# Thread: Area of solid reflected over a line

1. ## Area of solid reflected over a line

*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.

y=1/x
y=0
x=1
x=5

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!

2. Originally Posted by Linnylou09
*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.

y=1/x
y=0
x=1
x=5

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!
washer method for the described region in quad I about the line $y = -1$ ...
$V = \pi \int_1^5 \left[\frac{1}{x} - (-1)\right]^2 - [0 - (-1)]^2 \, dx$