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Calculate the volume with a rational funtion

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate the volume (V) generated by the given function:
    [itex]y^{2}=\frac{x^3}{2a-x}[/itex] about the line [itex]x=2a[/itex]

    I suppose you have to use cylindrical shells [because it's difficult to fin [itex]y[/itex]], so:

    [itex]V=2\pi(2a-x)\sqrt{\frac{x^3}{2a-x}}Δx[/itex]

    now:

    [itex]V=\int^{2a}_{0}2\pi(2a-x)\sqrt{\frac{x^3}{2a-x}}dx[/itex]

    but I tried a million things but nothing works:

    First, simplifying the integrand to this:

    [itex]V=2\pi\int^{2a}_{0}\sqrt{2a-x}\sqrt{x^3}dx[/itex]

    Next, I've been searching for any substitution function, but nothing works. If you have some idea that could help me I'd be grateful.

    Thanks
     
  2. jcsd
  3. Jan 24, 2012 #2

    SammyS

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    Hello yaakob7. Welcome to PF !

    Mostly it's just a matter of rewriting the integrand.

    The substitution, u = 2a/x will also help. Do that first.
     
  4. Jan 25, 2012 #3
    Did you mean u=2a-x??

    I tried but the expression remains almost the same:

    [itex]V=2\pi\int^{2a}_{0}\sqrt{u}\sqrt{(2a-u)^3}dx=2\pi\int^{2a}_{0}(2a-u)\sqrt{u}\sqrt{(2a-u)}dx[/itex]

    Im stuck there. I can`t find a simplify function that really works...
     
  5. Jan 25, 2012 #4
    Good topic I read and i understanding
     
  6. Jan 25, 2012 #5

    SammyS

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    Hello nazfagge. Welcome to PF !

    I'm glad that you understand this.
     
  7. Jan 25, 2012 #6

    SammyS

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    Well, u = 2a/x was a typo.

    I meant: Let u = x/(2a). Then x = (2a)u , and dx =(2a)du.

    You can then factor 2a out of a whole lot of stuff.
     
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