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Need help finding the volume enclosed

  1. May 3, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the volume in octane 1 enclosed.
    x + y + 2z = 2
    2x + 2y +z = 4

    ∫∫∫ dV

    2. Relevant equations

    3. The attempt at a solution
    ∫∫∫ dv = [∫(0⇒2) [∫(2-2z)⇒(2-½z) [∫(2-2z-y)⇒(2-½z-y) dx] dy] dz]
    = [∫(0⇒2) [∫(2-2z)⇒(2-½z) [x]|(2-2z-y)⇒(2-½z-y) dy] dz]
    = [∫(0⇒2) [∫(2-2z)⇒(2-½z) (2-½z-y)-(2-2z-y) dy] dz]
    = [∫(0⇒2) [∫(2-2z)⇒(2-½z) (3/4)z dy] dz]
    = [∫(0⇒2) [¾yz]|(2-2z)⇒(2-½z) dz]
    = [∫(0⇒2) ¾z[(2-½z)-(2-2z)] dz]
    = [∫(0⇒2) ¾z^2 dz]
    = (¾)(⅓) z^3 |(0⇒2)
    = (3/12) [2^3 - 0^3]
    = (3/12) (8)
    = 2


    Am I wrong?

    I do not understand some time i got the answer above 2 when i changed dxdydz dydzdx and dzdydx

    Last edited: May 3, 2017
  2. jcsd
  3. May 3, 2017 #2
    Hi Another:

    I confess I have some trouble seeing the details of what the diagram seems to be trying to communicate, but it appears to be you are showing the two tetrahedral pyramids corresponding to the two plane defining equations. Since the planes apparently intersect only on the xz boundary plane, you might find it easier to calculate the desired volume as the difference between the two pyramid volumes. Since the two pyramids seem to share the same peak, and therefore have the same height above the xy plane, this should simplify the calculation.

    Hope this helps.

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