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Math Help - Integral volume

  1. #1
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    Integral volume

    I have the following problem. I need to find the volume of a solid that is obtained when the region under the curve y = 1-(x^2/16) is revolved around the x axis between x=-4 and x=4

    So far i have: pi (integral 4,-4) (1-(x^2/16))^2 dx

    Any help would be appreciated
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  2. #2
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    Quote Originally Posted by menco View Post
    I have the following problem. I need to find the volume of a solid that is obtained when the region under the curve y = 1-(x^2/16) is revolved around the x axis between x=-4 and x=4

    So far i have: pi (integral 4,-4) (1-(x^2/16))^2 dx

    Any help would be appreciated
    Dear menco,

    V=\displaystyle\pi\int_{-4}^{4}\left(1-\frac{x^2}{16}\right)^2dx

    All you have to do is expand the brackets and integrate.

    V=\displaystyle\pi\int_{-4}^{4}\left(1-\frac{x^2}{8}+\frac{x^4}{256}\right)dx

    V=\displaystyle\pi\left(\int^4_{-4}dx-\int^4_{-4}\frac{x^2}{8}dx+\int^4_{-4}\frac{x^4}{256}dx\right)

    Hope you can continue.
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  3. #3
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    I went through it and as far as i know I find the derivative of the integral then substitube 4 for x and minus -4 for x

    I get a final result of 15.56919154 units
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  4. #4
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    Quote Originally Posted by menco View Post
    I went through it and as far as i know I find the derivative of the integral then substitube 4 for x and minus -4 for x

    I get a final result of 15.56919154 units
    The correct answer should be, \dfrac{64\pi}{15}=13.40412865531645
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  5. #5
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    Well i figured it out however i couldnt work out how you got (64pi/15) I just got (pi x 4.26')

    Thanks for the help
    Last edited by menco; January 28th 2011 at 12:10 AM.
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  6. #6
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    Quote Originally Posted by menco View Post
    Well i figured it out however i couldnt work out how you got (64pi/15) I just got (pi x 4.26')

    Thanks for the help
    Used a computer algebra system. You could do it by hand but it would take a lot of time because you have to deal with huge fractions.
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