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Triple integral?

  1. May 4, 2006 #1
    I like to use cartesian coords

    Find the region to the intersecting cyclinders x^2+y^2<=a^2 and x^2+z^2<=a^2

    What I have trouble finding is the domain of integration

    Currently I have
    a to -a for dx
    -srt(a^2-x^2) to srt(a^2-x^2) for dy
    -srt(x^2+y^2) to srt(x^2+y^2) for dz

    But this integration turned out to be too complicated to evaluate hence probably wrong.
  2. jcsd
  3. May 4, 2006 #2


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    "Find the region to ..." ? I assume you mean to find the volume of that region.
    It's not at all clear to me where you would get "-srt(x^2+y^2) to srt(x^2+y^2) for dz". I don't see any reason to think that z2 is ever equal to x2+ y2. The projection of the figure down on to the xy-plane is the circle x2+ y2= a2 so, yes, the outer integrals would be
    [tex]\int_{x= -a}^a \int_{y=-\sqrt{a^2-x^2}}^{\sqrt{a^2- x^2}} dydx[/tex]

    But, for every (x,y), z ranges between [itex]-\sqrt{a^2- x^2}[/itex] and [itex]\sqrt{a^2- x^2}[/itex] also. Looks to me like the inner integral should be
    [tex]\int_{z= -\sqrt{a^2-x^2}}^{\sqrt{a^2-z^2}}dz[/tex]
    Of course, that could be done simply as the double integral
    [tex]2\int_{x= -a}^a \int_{y=-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\sqrt{a^2- x^2}dydx[/tex]

    I think it would be even simpler in polar coordinates
  4. May 5, 2006 #3
    I see how you got your solution which yields the correct solution, 16a^3/3.

    I eventually used a different method but is slightly wrong with a factor of 'a' too much. I integrated along one cyclinder along one axis and then specified the limits of the other two axis as -a and a. My integral from the inside out is
    From -(x^2+z^2) to x^2+z^2 along y
    From -a to a along x
    From -a to a along z

    I got 16a^4/3. Where did I go wrong?
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