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anonymous

  • 5 years ago

revolve region bounded by y=(x+2)^2 , y=1, about y axis, using shell method

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  1. anonymous
    • 5 years ago
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    hey , i get a negative volum

  2. nowhereman
    • 5 years ago
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    laplace, you might get a negative volume, because the area you are rotation is on the left side of the y-axis. Try to mirror it, i.e. you \(y≤(x−2)^2\) and the interval \([0,2]\) Also, could you please change that picture, it is not appropriate for some people here!

  3. anonymous
    • 5 years ago
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    ok sorry

  4. anonymous
    • 5 years ago
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    one sec

  5. anonymous
    • 5 years ago
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    so what conditions do we have, when do we have to worry about this happening, negative volume.

  6. anonymous
    • 5 years ago
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    for example , we dont have to worry about this in the disc method, because of pi*r^2

  7. nowhereman
    • 5 years ago
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    There are only problems, when you apply a formula to a situation it is not made for. The formula for the shell method is made for interval on the positive part of the x-axis. So if you have the situation where the area is on the negative part, you have to mirror it. The solid of revolution is the same, as it is symmetric, but you can only then apply the formula.

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