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Math Help - Area of solid reflected over a line

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by Linnylou09 View Post
    *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!
    your set-up is incorrect.

    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
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