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Triple Integral-problem.

  1. Aug 21, 2005 #1
    How do I solve this triple integral,
    [tex]\int\int\int_{\Omega}^{}zdxdydz, \hspace{8} \Omega = \{(x,y,z): x^2 + y^2 + z^2 \leq 1, \hspace{6} 0 \leq z \leq \sqrt{x^2+y^2}\}[/tex]
     
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  3. Aug 21, 2005 #2

    saltydog

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    Convert to spherical coordinates. Integrate phi from 0 to pi/2, all around from 0 to 2pi, and let rho go from 0 to 1. Oh yea, then multiply by 2. Can you convert this into a real integral?
     
  4. Aug 21, 2005 #3

    Hurkyl

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    salty: you notice the region of integration isn't a sphere? :smile: (And if it was, there's an easier way to do it...)

    Where do you have problems, iNCREDiBLE?
     
  5. Aug 21, 2005 #4

    saltydog

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    No I didnt' Hurkyl. Looked like a sphere to me. Still does. Suppose I should just work it through. Sorry Incredible if I got it wrong (Hurkly is no doubt right in some kind of way I need to figure out).
    :blushing:
     
  6. Aug 21, 2005 #5
    Are you sure about this? :surprised
     
  7. Aug 21, 2005 #6

    Hurkyl

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    Note the extra condition 0 < z < &radic;(x² + y²). It's the part of the upper half-ball that's outside of a cone.
     
  8. Aug 21, 2005 #7
    I can't see what the region of integration is. But my gut says i should integrate over z first. Then convert to polar coordinates where [tex]\frac{1}{\sqrt{2}} \leq r \leq 1?[/tex]
     
    Last edited: Aug 21, 2005
  9. Aug 21, 2005 #8

    saltydog

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    Alright, it's a cone. Thanks Hurkyl. Sorry Incredible. I'll work it through for my own education but will stay out of the thread so as not to make it any worst for Incredible.
    :blushing:
     
  10. Aug 21, 2005 #9

    Hurkyl

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    You have a good gut! I too would integrate over z first, and change it into cylindrical coordinates. Now, you just have to work out the details. :smile:


    As usual, it really helps if you can draw a picture of the region of integration, or at least visualize it really well. But, we can always grind it out algebraically.


    Your region is the system of equations:

    r² + z² < 1
    0 < z < r

    So, the first question becomes "for which values of r can we solve this equation for the rest of the variables?"

    You have suggested 1/√2 < r < 1. Why?
     
  11. Aug 21, 2005 #10
    I understand that the region is
    r² + z² < 1
    0 < z < r

    But after that I'm not following you.. :cry:
     
  12. Aug 21, 2005 #11

    Hurkyl

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    You need to find the bounds: you need to know all possible values of r, then for each value of r, you need to know all possible values of z... right?
     
  13. Aug 21, 2005 #12
    1/√2 < r < 1
    0 < z < 1/√2 ?


    o:)
     
  14. Aug 21, 2005 #13

    Hurkyl

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    Okay, how did you get that?
     
  15. Aug 21, 2005 #14
    Well, Hurkyl, I think I was wrong... :wink:
    I've been thinking a lot and here's how I wanna do it now.

    Spherical coordinates:

    [tex]x = rsin\theta cos\phi[/tex]
    [tex]y = rsin\theta sin\phi[/tex]
    [tex]z = rcos\theta[/tex]

    [tex] 0 \leq r \leq 1[/tex]
    [tex] \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}[/tex]
    [tex] 0 \leq \phi \leq 2 \pi[/tex]


    [tex]\int\int\int_{\Omega}^{}zdxdydz = \int\int\int rcos\theta r^2 sin\theta drd\theta d\phi = \frac{1}{2} \int\int\int r^3 sin(2\theta) drd\theta d\phi = \pi \int\int r^3 sin(2\theta) drd\theta = \frac{\pi}{4} \int sin(2\theta) d\theta = \frac{\pi}{8}[/tex]
     
    Last edited: Aug 21, 2005
  16. Aug 21, 2005 #15

    Hurkyl

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    You were! Notice that you could still solve for z for any 0 < r < 1... it's just that 1/&radic;2 is the switching point between the two different expressions for z. (So you'd have to break the integral up into two pieces)


    So how did you get your bounds this time?
     
  17. Aug 21, 2005 #16
    We had
    [tex]x^2 + y^2 + z^2 \leq 1[/tex]
    [tex] 0 \leq z \leq \sqrt{x^2+y^2}[/tex]

    After switching to Spherical Coordinates we have

    [tex]r^2 \leq 1[/tex]
    [tex]0 \leq rcos\theta \leq \sqrt{(rsin\theta cos\phi)^2+(rsin\theta sin\phi)^2}=|rsin\theta|=rsin\theta[/tex]
    <=>
    [tex]0 \leq r \leq 1[/tex]
    [tex]0 \leq cos\theta \leq sin\theta[/tex]

    The last inequality is true for [tex]\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}[/tex]
     
  18. Aug 21, 2005 #17

    Hurkyl

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    Looks good to me. And more importantly, it looks good to you. :smile:
     
  19. Aug 21, 2005 #18
    Yeah! I'm preparing for an exam. This is a problem from a previous exam and I really hate the fact that our textbook don't discuss this sort of integrals, just those that are way simpler and not even one of our excercises has been of this sort. :devil:
     
  20. Aug 21, 2005 #19

    Hurkyl

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    IMHO you should work it out in cylindrical coordinates too. It's good exercise. :smile:
     
  21. Aug 21, 2005 #20
    The problem is that I have no clue about how the region of integrations looks like.. So when you say that I would have to break the integral into two pieces I wouldn't know why :redface:
     
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