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Math Help - Washers

  1. #1
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    Washers

    Find the volume of the solid of revolution generated when the given region of the figure is revolved about the indicated line:
    The given is :
    lines: x = 4, y = 8
    The curve is: y^2 = x^3

    Question:
    Find the volume of The region below the line y = 8 (meaning inside the parabola) revolving about the line x = 4


    I cant hit the answer key and i always end up with 1024/35 pi cubic units ...

    the answer is suupposed to be 3456/35 pi cubic units from the answer key

    help! and thanks alot!
    Last edited by ^_^Engineer_Adam^_^; April 2nd 2007 at 04:02 AM.
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  2. #2
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    Hello, ^_^Engineer_Adam^_^!

    Your sketch may be incorrect . . .


    Find the volume of the solid of revolution generated
    when the given region is revolved about the indicated line.

    Region bounded by: .y≤ = x≥, .x = 0, .y = 8
    . . revolved about x = 4.

    Answer: .3456π/35 units≥
    Code:
            |
           8+ - - - - -*
            |::::::::::|
            |:::::::::*|
            |::::::::* |
            |::::::*   |
            |:::*      |
          --*----------+--
            |          4

    I prefer to use Cylindrical Shells.

    . . V .= .
    (radius)(height) dx

    The radius is: .4 - x
    The height is: .8 - x^
    {3/2}


    We have: . V . = .
    (4 - x)∑(8 - x^{3/2}) dx

    . . . . . . . . . . .= .
    ∫ (32 - 8x - 4x^{3/2} + x^{5/2}) dx

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  3. #3
    Senior Member
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    Is there a way to use
    the volumes by slicing and rotating method?

    \int{\pi (f(x))^2}dx
    Last edited by ^_^Engineer_Adam^_^; September 20th 2007 at 05:43 PM.
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  4. #4
    MHF Contributor
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    Find the volume of The region below the line y = 8 (meaning inside the parabola) revolving about the line x = 4

    It is the thing inside the parabola.

    Then the radius about the vertical line, the axis of rotation, x=4 is not a single radius. There are two radii involved---the inner radius and the outer radius. That is why you mentioned "washers".

    inner radius = (4 -x)
    outer radius = 4 -0 = 4

    So your infinitesimal volume, or dV, is a washer whore radii are those, and whose thickness is dy.
    Hence, dV = {pi[(outer r)^2] -pi[(inner r)^2]}*dy
    dV = pi[(outer r)^2 -(inner r)^2]dy
    dV = pi[4^2 -(4-x)^2]dy
    dV = pi[16 -16 +8x -x^2]dy
    dV = pi[-x^2 +8x]dy

    But dy has nothing to do with x's in integration, so convert the x's into their equivalent y's.

    y^2 = x^3
    Raise both sides to their 1/3 powers,
    y^(2/3) = x
    And x^2 = y^(4/3)

    So,
    dV = pi[-y^(4/3) +8y^(2/3)]dy ----(i)

    Boundaries for dy are y=0 and y=8

    So, intregrating both sides of (i),
    V = (pi)INT.(0-->8)[-y^(4/3) +8y^(2/3)]dy
    V = (pi)[-(3/7)y^(7/3) +(24/5)y^(5/3]|(0-->8)
    V = pi{[-(3/7)(8^(7/3)) +(24/5)(8^(5/3))] -[0]}
    V = pi{-(3/7)(2^7) +(24/5)(2^5)}
    V = pi{-384/7 +768/5}
    V = pi(3456/35)
    V = (3456/35)pi cu.units --------------answer.
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