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How solve this 3D integral

  1. Jul 1, 2011 #1
    find the volume enclosed by x=y z=0 and [tex]y^2+z^2=x[/tex]
    i built

    [TEX]\int_{0}^{1}\int_{x}^{\sqrt{x}}\int_{0}^{\sqrt{x-y^{2}}}dzdydx[/TEX]


    i tried to traw the projection of the top interval in the z-y plane

    [TEX]z=\sqrt{x-y^{2}}[/TEX] -> [TEX]z^2+y^2=x[/TEX]
    so we have a circle
    but the radius changes from x to [TEX]\sqrt{x}[/TEX].

    which is not possible because when we scan the ring it goes from the biiger radius to the smaller one.
    then i thought to try in anyway ,and i wrote the integral in polar
    [TEX]\int_{0}^{1}\int_{0}^{2\pi}\int_{x}^{\sqrt{x}}rdrd\theta dz[/TEX]

    but i dont know how mathmatickly to change the intervals?
    how to solve it in a polar way?
     
  2. jcsd
  3. Jul 1, 2011 #2

    SammyS

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    In the x-y plane, you are bounded by the half parabola, [itex]y=-\sqrt{x}[/itex], for y < 0, and by y = x, for y ≥ 0 .

    So, the limits on the middle integral should be: [itex]\dots\int_{-\sqrt{x}}^{x}\ \dots\ dy\,\dots[/itex]

    Added in edit:

    Sorry for the above.

    It looks like you are right.

    You have: [itex]\displaystyle \int_{0}^{1}\int_{x}^{\sqrt{x}}\int_{0}^{\sqrt{x-y^{2}}}(1)\,dz\,dy\,dx[/itex]
     
    Last edited: Jul 1, 2011
  4. Jul 1, 2011 #3

    SammyS

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    If 0 < x < 1, then x < √(x) , ... if that's what's worrying you.
     
  5. Jul 2, 2011 #4
    how to solve it?
     
  6. Jul 2, 2011 #5

    vela

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    Try using an integral where you integrate with respect to x first, z second, and y last, i.e.
    [tex]\int \int \int (\cdots)\,dx\,dz\,dy[/tex]
    If you haven't already, plot or sketch the surfaces. Setting up the integral is straightforward once you see how the surfaces intersect.
     
  7. Jul 2, 2011 #6
    why?

    i already build an integral
    how to solve it?
     
  8. Jul 2, 2011 #7

    vela

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    Then just integrate it. They're elementary integrals.
     
  9. Jul 2, 2011 #8
    have you tried?
    its not possible
    how to solve it in polar
     
  10. Jul 2, 2011 #9

    vela

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    You're right. Your integral makes a mess. If you use my suggestion, you get integrals that are easy to evaluate.
     
  11. Jul 2, 2011 #10
    I didn't try to integrate this but a couple suggestions that sometimes work for when you get stuck with messy hard to solve integrals...

    1) Try to set it up in terms of using a different order of integration...

    2) Change of variables or coordinate system. Try using polar coordinates as suggested.
     
  12. Jul 2, 2011 #11
    but i need to do addjustments to the intervals
    if i change the order of integration
    and its a 3d

    could you solve it?

    as you see my original question is about polar representation..
    could you solve it this way?
     
  13. Jul 2, 2011 #12
    it worked
    how to solve
    [tex]\int (y-y^2)^{\frac{3}{2}}[/tex]
     
    Last edited: Jul 2, 2011
  14. Jul 2, 2011 #13

    vela

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    Complete the square to write the integrand in the form [itex][a^2-(y-b)^2]^{3/2}[/itex]. Then use the appropriate substitutions.
     
  15. Jul 2, 2011 #14
    [tex]\int (y-y^2)^{\frac{3}{2}}[/tex]=[tex]\int [-(y-\frac{1}{2})^2+\frac{1}{4}]^{\frac{3}{2}}[/tex]
    what now?
     
  16. Jul 2, 2011 #15
    solved it thanks
     
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