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Math Help - Shell method

  1. #1
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    Shell method

    Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the x-axis.

    x + y^2 = 16

    I'm not sure what to set h(y) equal to. Is it sqrt(16-x)? It can't be that because I need to have a y instead of an x, right? I understand how to find the volume and integrate from here.

    Thank you.
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  2. #2
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    In this case we're integrating along the y-axis, not the x-axis as usual. For the volume of the solid obtained by revolving the x \ge 0 portion of the curve about the x-axis, we can use the shell method, our function of height being

    x = 16 - y^2

    and the volume of each hollow cylinder being

    2\pi \cdot (16 - y^2) \cdot dy. Edit: This is incorrect.

    Since x = 0 at y = \pm 4, the total volume equals

    \int_0^4 2\pi (16 - y^2) \, dy.
    Last edited by Scott H; February 7th 2009 at 02:48 PM.
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  3. #3
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    So, could the answer really be (-7808pie)/(3)?
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  4. #4
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    nvm. THank you very much. I made an incorrect calculation.
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  5. #5
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    Sorry, I made a mistake.

    In calculating the volume of a shell, we use not only the radius of the shell but also its height. My previous formula is therefore wrong.

    The volume of one shell is

    2\pi xh(y)\,dy,

    not

    2\pi h(y)\,dy.

    Integrating, we have

    \int_0^4 2\pi(16y - y^3)\,dy.

    I apologize for the confusion.
    Last edited by Scott H; February 7th 2009 at 02:49 PM.
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  6. #6
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    So, you just multiply in an extra y?
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  7. #7
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    Correct.

    The y is for the radius of the shell, and x = 16 - y^2 is for the height.

    This was my first time doing a shell problem in a long time, and I momentarily forgot.
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