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Triple Integral Problem

  1. Dec 26, 2005 #1
    Hello. This is my first of many posts at this forum. For fun recently I came across a triple integral that i would really like to know how to do.
    basically i have two equations
    if you plot these you will see that they intercept.
    i need to find the area under the graph where they intercept.
    I would like to write this as a SINGLE integral using cartesian, cylindrical and sphereical coordinates.
    anyone know how to write it any or all of these three ways. thank you very much
  2. jcsd
  3. Dec 26, 2005 #2
    Is integral necessary? (or am I making a mistake?)
    subtract these equations side by side, and you'll get
    That's where these spheres intercept.
    So put z=0 into the first equation (or the second one), you'll have:
    [tex]x^2 + y^2 = 1 (z = 0)[/tex]
    and this is the graph of interception and its area is evidently [tex]\pi.[/tex]

    I hope it helps :smile:

    If you want to calculate the "volume" of the common volume region instead of area, it's quite different. (but I guess this is not what you want.)

    The volume is calculated as

    [tex] \int_0^1 \pi (\sqrt{(1-z^2)})^2 dz + \int _{-\sqrt{2}}^0 \pi (\sqrt{(2-z^2)})^2 dz = \pi \frac {2+4 \sqrt{2}}{3}[/tex]

    It's not single integral or cylindrical or spherical, though...

    I think simple is best. :smile:
    Last edited: Dec 26, 2005
  4. Dec 26, 2005 #3
    Triple Integral

    Hello. You are absolutely right sir. I meant to ask for the volume not the area.

    How do i write that sum of those two integrals as one integral.
    Also How would I go about converting that into cylidrican and spherical coordinates. I mean I know the volume elements for spherical and cylindrical
    dV==r^2 sinphi dphi dtheta dr.
    dV==r dr dtheta dz
    thank you for all your help

  5. Dec 26, 2005 #4
    In Cartesian coordinates, the latter integral

    [tex]\int _{-\sqrt{2}}^0 \pi (\sqrt{(2-z^2)})^2 dz = \int _ 0 ^ {\sqrt{2}} \pi (2-z^2)dz[/tex]

    can be put using

    [tex]\sqrt{2} z' = z [/tex]


    [tex] \int _0^1 \pi (2- (\sqrt{2} z')^2) \frac {dz}{dz'} dz'
    = \int _0 ^ 1 \pi 2 \sqrt{2}(1-z^2) dz[/tex]

    So the total volume can be written as a single integral

    [tex]\int _0^1 \pi ((1-z^2) + 2 \sqrt{2} (1-z^2)) dz
    = \int _0^1 \pi (1+ 2 \sqrt{2}) (1-z^2) dz[/tex]

    Does it make sense? It looks like just a trick of calculation to me.

    As for polar and cylindrical coordinates, I wrote spheres as:

    [tex]r^2-2r\cos(\theta)=1[/tex] and

    [tex]r^2+z^2-2z=1[/tex] and
    [tex]r^2+z^2 = 1[/tex].

    But I haven't make integral fomulae yet.
    Last edited: Dec 26, 2005
  6. Dec 26, 2005 #5
    Yes I do understand. Thanks for your help
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