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Another volume of revolution around x axis

  1. Jul 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the area bounded by
    y^2 = 4ax
    and
    x=a

    rotated around the x-axis.

    2. Relevant equations
    integral of pi R^2 dh

    3. The attempt at a solution
    I just don't know how to handle the x=a part of the boundary. Any hints? thanks
     
  2. jcsd
  3. Jul 9, 2012 #2

    SammyS

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    Integrate from x=0 to x=a .

    The trickier part may be understanding how to handle the integrand .
     
  4. Jul 9, 2012 #3
    well I read lots of books on it and do lots of questions the only way to understand them sometimes is to try it and then see someone do it step by step
     
  5. Jul 9, 2012 #4

    SammyS

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    Graphing y2 = 4ax will help for this question.
     
  6. Jul 9, 2012 #5
    k so the integral of 4ax would be (4/2)ax^2 right?
     
  7. Jul 9, 2012 #6

    SammyS

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    When you write
    integral of pi R2 dh ,​
    how is R related to y ?
     
  8. Jul 10, 2012 #7
    R^2 and Y^2 end up being the same thing right?
     
  9. Jul 10, 2012 #8

    SammyS

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    (I misread this post.)

    Yes, that is the correct integral.

    What limits of integration should you use?
     
  10. Jul 10, 2012 #9
    x=0 to x=a

    but I forget how to integrate a constant (a)
     
  11. Jul 10, 2012 #10

    SammyS

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    [itex]\displaystyle \int\,a\,f(x)\,dx=a\,\int f(x)\,dx[/itex]
     
  12. Jul 13, 2012 #11
    I guess that I am not sure how to do that.

    for example, ∫(2)dx = 2x + C

    but how does that apply here? the answer doesn't have any + in it.
     
  13. Jul 13, 2012 #12

    SammyS

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    For a definite integral, the constant of integration has no effect, so it can be ignored.

    [itex]\displaystyle \int_{b}^{c}\,a\,f(x)\,dx=a\,\int_{b}^{c} f(x)\,dx[/itex]

    In your case b=0 and c=a.

    The fact that upper limit of integration is the same as the multiplicative constant is merely a coincidence.

    [itex]\displaystyle \int_{0}^{a}\,4a\,x\,dx=4a\,\int_{0}^{a} x\,dx[/itex]
     
  14. Jul 13, 2012 #13

    HallsofIvy

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    Typically, one of the very first things you learn about integrals is that the anti-derivative of [itex]x^n[/itex], for n not equal to -1, is [itex](1/(n+1))x^{n+1}+ C[/itex]. What is that for x= 0?
     
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