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Math Help - method of cylindrical shells

  1. #1
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    method of cylindrical shells

    Use the method of cylindrical shells to find the volume of the solid rotated about the line x=-1 given the conditions: y=x^3-x^2; y=0;x =0
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    Look at the picture.
    Let us keep things positive.
    -(x^3-x^2)=x^2-x^3
    Now you are rotating about x=-1
    Let us shift the coordinate axes to the right 1 unit.
    That means replace x for x+1.
    Thus, your new curve is, (x+1)^2-(x+1)^3
    And you are rotating this time about the y-axis becase we shifted the curve.
    Thus by cylindrical shells,
    2\pi \int_1^2 x[(x+1)^2-(x+1)^3]dx
    Note the new limits of integration because we shifted the curve.
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  3. #3
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    once again i thank you for your paitence. To finish the problem I have came up with the following, however I feel something is wrong.

    <br />
2\pi \int_1^2 x[(x+1)^2-(x+1)^3]dx
    =2\pi \left[-\frac{1}{3}x(x(x+3)+3)\right}_{1}^{2}\approx-19.8968<br />
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  4. #4
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    I made a mistake. When you shift the coordinate axes to the right instead of x+1 it is x-1.
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  5. #5
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    ok, based on that I have revised my anwser to this. Thoughts?

    <br />
2\pi \int_1^2 x[(x-1)^2-(x-1)^3]dx<br />
    <br />
=2\pi \left[\frac{1}{3}x((x-3)+3)\right}_{1}^{2}\approx 2.0944<br />
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  6. #6
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    Don't get disks mixed up with shells.

    \int_{0}^{1}\overbrace{(-1-x)}^{\text{radius}}\overbrace{(x^{3}-x^{2})}^{\text{height}}\overbrace{dx}^{\text{thick  ness}}=\frac{4{\pi}}{15}
    Last edited by galactus; August 30th 2006 at 08:25 AM.
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