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Math Help - I keep finding a negative volume for this solid of revolution. Please help?

  1. #1
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    I keep finding a negative volume for this solid of revolution. Please help?

    Find the volume of the solid of revolution formed by revolving the planar region, f(x)=x^3 and f(x)=x, about the y-axis. You may choose either the shell or the disk method.

    Here is what I did:

    v = pi*[integral from 0 to 1 of (y^2 - (cube root(y))^2)dy]
    = pi*[integral from 0 to 1 of (y^2 - y^(2/3))dy]
    = pi*(1/3y^3 - 3/5y^5/3)
    = pi*(1/3 - 3/5)
    = pi*(5/15 - 9/15)
    = -4pi/15

    I'm pretty sure I can't have a negative volume, can I? What did I do wrong?

    Also, how do you use the math tags to write a definite integral?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    If you rotate it about the x axis, the area formed by the curve f(x) = x^3 is greater than the area under the curve f(x) = x for y between 0 and 1. Hence you are getting a negative answer.

    Change your integral to :

    v = pi*[integral from 0 to 1 of ((cube root(y))^2 - y^2)dy]
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  3. #3
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    Solution

    Use shell method and \int_0^1\! 2\pi x(x-x^3)\, dx=\frac{4\pi}{15}.
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  4. #4
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    Quote Originally Posted by Unknown008 View Post
    If you rotate it about the x axis, the area formed by the curve f(x) = x^3 is greater than the area under the curve f(x) = x for y between 0 and 1. Hence you are getting a negative answer.

    Change your integral to :

    v = pi*[integral from 0 to 1 of ((cube root(y))^2 - y^2)dy]
    Ah, thank you so much. This stuff can be so tricky..

    Now, I have to revolve it around the x-axis, and I'm trying to use the shell method, but I'm getting a negative answer.

    v = 2pi*[integral from 0 to 1 of (y(y - cube root(y))dy)]
    = 2pi*[integral from 0 to 1 of (y^2 - y^4/3)dy]
    = 2pi*(1/3y^3 - 3/7y^7/3)
    = 2pi*(1/3 - 3/7)
    = -0.598

    Where did I go wrong?
    Last edited by WahooMan; June 1st 2010 at 11:28 AM.
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