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Volume of a cone covered with a plane

  1. Mar 23, 2013 #1
    1. The problem statement, all variables and given/known data

    I need to find the volume of a cone covered with a plane z=h using multiple integrals. The scheme is something like this:

    attachment.php?attachmentid=57019&stc=1&d=1364060517.png

    2. Relevant equations

    Formula of the cone x^2-y^2-z^2=0

    3. The attempt at a solution

    I tried to integrate ∫∫∫(x^2-y^2-z^2)dxdydz in all different kinds of bounds, for example {x,0,L}, {y,-R,R}, {z,-R, H}, but the answer is always negative. I can't really get my head around the error, help please?
     

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    Last edited by a moderator: Mar 23, 2013
  2. jcsd
  3. Mar 23, 2013 #2
    What does this formula even mean?
     
  4. Mar 23, 2013 #3

    LCKurtz

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    And remember if you are calculating a volume with a triple integral you always use a formula$$
    Vol = \iiint_R 1\, dV$$
    where ##R## is the region in question and ##dV## is ##dxdydz## in some order and ##R## is described by putting appropriate limits on the integrals. The integrand is always ##1##.
     
  5. Mar 23, 2013 #4

    haruspex

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    Your first step is to choose a way to slice up the volume into laminae whose areas you can hope to calculate. I would suggest slices parallel to the xz plane, so the area will be a function of y.
     
  6. Mar 23, 2013 #5

    LCKurtz

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    Also notice that either the picture or the equation is incorrect. The cone that is pictured would be a portion of ##x^2-y^2+z^2=0##.
     
  7. Mar 26, 2013 #6
    The integral is very nasty if it were to be done by hand. Is it a homework question?
     
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