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Volume between given planes

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data

    (1 pt) Find the volume of the region bounded by the planes
    568127b1edc489fa46d5f8432599101.png

    2. Relevant equations
    V = ∫∫7/4-6/4y-2/4x


    3. The attempt at a solution
    Since y=x I found their values when z = 0.
    6x+2x=7, x=7/8
    y= 7/8 is the maximum value y will have in this integration as it decreases as x approaches its maximum value: 2x = 7, x=7/2.

    When I made a 2-D image of the region I would be integrating on I came up with a triangle which had two areas that needed to be integrated separately as in the first region: 0≤x≤7/8 and 0≤y≤x, and in the second region: 7/8≤x≤7/2 and 0≤y≤(7/6 - x/3).

    After integrating my result was 0.893229167.

    What I am wondering is if I am using the correct limits for my integration or if I made a mistake in my math.
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Mar 5, 2009 #2
    I would avoid double integral thus: a horizontal cross-section of the region is bounded by the x and y axes and the line

    [tex]2x+6y = 7-4z[/tex]

    The area of that right angle triangle is [tex]\frac{(7-4z)^2}{24}[/tex] for z in[tex][0,\frac{7}{4}][/tex].

    The definite integral should be [tex]\frac{7^3}{12\cdot 24}=1.19[/tex]
     
  4. Mar 5, 2009 #3
    The system still thinks that is an incorrect answer
     
  5. Mar 6, 2009 #4
    Could the system be expecting a different format than what you entered, such as different number of decimal places?
     
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