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Homework Help: Evaluating triple integral

  1. Jan 12, 2006 #1
    Hi,

    I don't know whether this is the best place to ask, anyway, I would like to check my results in Maple and don't know how to evaluate something like this there:

    [tex]
    \iiint_{M} z^2\ dx\ dy\ dz\mbox{ , where M = [x,y,z] \in \mathbb{R}^3, x^2+y^2+z^2 \leq R^2, x^2 + y^2 + z^2 \leq 2Rz}
    [/tex]

    There is function Tripleint in Maple but I don't know how to properly pass the bounds..

    Thank you
     
    Last edited: Jan 12, 2006
  2. jcsd
  3. Jan 12, 2006 #2

    benorin

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    What version of Maple R U using?
     
  4. Jan 12, 2006 #3
    It's version 9.5
     
  5. Jan 12, 2006 #4

    benorin

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    Me too. Have you tried using the MultiInt function call with spherical coordinates? (trying to figureit out also)
     
  6. Jan 12, 2006 #5
    I didn't know about this function, thank you.

    After converting to spherical coords, I got

    [tex]
    \iiint r^4\sin^2 \phi \cos \phi\ dr\ d\phi\ d\theta
    [/tex]

    Anyway, I'm still not sure about the bounds..
     
  7. Jan 12, 2006 #6

    benorin

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    the bounds are two spheres, the first is centered at the origin with radius R, the second is [itex]x^2+y^2+(z-R)^2\leq R^2[/itex], which is a sphere of radius R centered at (0,0,R). These spheres overlap and have a curve of intersection which is a circle contained in some plane z=k, for some constant k. But that is not the point; the region is bounded below by the shifted sphere and bounded above by the other sphere.
     
    Last edited: Jan 12, 2006
  8. Jan 12, 2006 #7

    benorin

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    I get

    [tex]\int_{\theta =0}^{2\pi}\int_{\phi =0}^{\pi} \int_{r =2R\cos\phi}^{R} r^4\sin^2 \phi \cos \phi\ dr\ d\phi\ d\theta=\frac{4\pi R^5}{15}[/tex]
     
    Last edited: Jan 12, 2006
  9. Jan 12, 2006 #8
    Thank you benorin. First, I didn't know how to write the bounds in spherical coordinates so I tried to compute in cartesian coordinates. But I got a different result. Here's how I approached:

    When I draw it, it's clear that for [itex]z \in [0,\frac{R}{2}][/itex] I have

    [tex]
    A:= {[x,y] \in \mathbb{R}^2, x^2 + y^2 \leq 2Rz - z^2}
    [/tex]

    which I can imagine as area of circle with radius [itex]\sqrt{2Rz - z^2}[/itex].

    For [itex]z \in [\frac{R}{2}, R][/itex] I have

    [tex]
    B:= {[x,y] \in \mathbb{R}^2, x^2 + y^2 \leq R^2 - z^2}
    [/tex]

    ie. area of circle with radius [itex]\sqrt{R^2 - z^2}[/itex].

    So I can write the original integral as sum of these two:

    [tex]
    I = \int_0^{\frac{R}{2}}\left(\iint_{A} 1\ dx\ dy \right)\ dz + \int_{\frac{R}{2}}^{R}\left(\iint_{B} 1\ dx\ dy \right)\ dz
    [/tex]

    [tex]
    = \int_{0}^{\frac{R}{2}} z^2\pi(2Rz - z^2)\ dz\ +\ \int_{\frac{R}{2}}^{R} z^2\pi(R^2 - z^2)\ dz = ... = \frac{59}{480} \pi R^5
    [/tex]

    I think it's ok, isn't it?
     
    Last edited: Jan 12, 2006
  10. Jan 12, 2006 #9

    benorin

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    Better to use cylindrical coordinates, I think:

    [tex]\int_{\theta =0}^{2\pi}\int_{r=0}^{\frac{\sqrt{3}}{2}R} \int_{z =R-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}} z^2r dz dr d\theta=\frac{59}{480}\pi R^5[/tex]

    Yep: same as you. :biggrin:
     
  11. Jan 12, 2006 #10
    Nice :) Anyway, could you tell me please how did you get the bounds when you were computing it with the spherical coordinates? I mean the inner integral with bounds [itex][2R\cos \phi, R][/itex]...
     
  12. Jan 12, 2006 #11

    benorin

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    I did it wrong. But substitute [itex]x^2+y^2+z^2=r^2\mbox{ and }z=r\cos \phi[/tex] into the equations of the spheres.
     
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