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Finding the Volume of the Intersection of Two Cylinders

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the intersection of the two solid cylinders x2 + y2 ≤ 1 and y2 + z2 ≤ 1.


    3. The attempt at a solution
    Apparently this is done most easily by cartesian coordinates. I have the integral:

    [tex]\int_{-1} ^1 \int_{-sqrt(1-x^2)} ^{sqrt(1-x^2)} \int_{-sqrt(1-y^2)} ^{sqrt(1-y^2)} 1 dzdydx[/tex]But, this is disgusting to integrate (as far as I can tell). I think I either a) have the wrong bounds, or b) have missed a clue to make this problem easier.
     
  2. jcsd
  3. Dec 1, 2011 #2

    Dick

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    Changing the order of integration would help a lot. Try integrating dy last instead of dx. Order of integration can make a BIG difference.
     
    Last edited: Dec 1, 2011
  4. Dec 1, 2011 #3
    So let's see if I can do this:

    [tex]\int_{-1} ^1 \int_{-sqrt(1-y^2)} ^{sqrt(1-y^2)} \int_{-x} ^x 1 dzdxdy [/tex]
    [tex]\int_{-1} ^1 \int_{-sqrt(1-y^2)} ^{sqrt(1-y^2)} 2x dxdy[/tex]This gives an integrand of zero for the last integral. Hmm, did I use incorrect bounds?
     
  5. Dec 1, 2011 #4

    Dick

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    Umm, yes, you used the wrong bounds. Shouldn't the z bounds be determined by y^2+z^2<=1?? Why would you think they should be -x to x?
     
    Last edited: Dec 1, 2011
  6. Dec 1, 2011 #5
    I have two possible problems with those bounds: first, and without doing any actual calculations, those bounds don't make the integral look very much nicer than what I had in #1; second, I would be concerned that we "lose" information about x by not including it somewhere in the bounds (that is, I see we have z's relationship to y and x's relationship to y, but not x's relationship to z). I used -x≤z≤x because that's what I got by setting,

    x^2 +y^2 = y^2 + z^2, which suggests,

    z = +/- x.
     
  7. Dec 1, 2011 #6

    Dick

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    I have some problems with that. How does x^2+y^2<=1 and y^2+z^2<=1 imply that x^2+y^2=y^2+z^2?? I suggested integrating dy last exactly because then the x bounds depend only on y and the z bounds depend only on y. Why do think there is some other dependency?
     
    Last edited: Dec 1, 2011
  8. Dec 1, 2011 #7
    Okay. I think I understand this now. I did the calculation out and got 16/3, which I have marked as the right answer. I guess doing the integration in that order was easier.
     
  9. Dec 1, 2011 #8

    Dick

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    A LOT easier. And yes, it is 16/3.
     
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