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Triple Integral, already solved, need checked

  1. Dec 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Let S be the region in the first octant under the plane 3x + 2y +z = 4. Find the volume of S.

    2. Relevant equations

    3. The attempt at a solution

    [tex]\int^{\frac{4}{3}}_{0}\int^{\frac{3}{2}x + 2}_{0}\int^{-3x - 2y + 4}_{0}dzdydx[/tex]

    = [tex]\int^{\frac{4}{3}}_{0}\int^{\frac{3}{2}x + 2}_{0}(-3x - 2y + 4)dydx[/tex]

    = [tex]\int^{\frac{4}{3}}_{0}(-\frac{15}{2}x^{2} - 6x + 4)dx[/tex]

    = 448/27
    Last edited: Dec 16, 2009
  2. jcsd
  3. Dec 16, 2009 #2
    Sorry, took a little while to figure out the symbols and dealies, pretty sure it's good to go right now.
  4. Dec 16, 2009 #3


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    Mumble, mumble..gotta check the bounds:

    The upper line in the z=0 plane obeys the equation 3x+2y=4

    [tex]0\leq{x}\leq\frac{4}{3}, 0\leq{y}\leq{2}-\frac{3}{2}x[/tex]
    Thus, you've got a sign wrong in the y-bound, according to my view.
  5. Dec 16, 2009 #4
    You're so right, thankyou!
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