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Double Integral (underneath a surface and above a square)

  1. Jan 10, 2012 #1
    1. The problem statement, all variables and given/known data

    The volume underneath the surface z= y/ (1+xy) and above the square {(x,y)| 2≤x≤3 , 3≤
    y≤ 4} is:


    2. Relevant equations

    Please see attachment.



    3. The attempt at a solution

    Please see attachment for solution.

    My professor had provided us with 8 possible solutions (where only one is correct). However the answer I had produced did not match with any of the 8. Any insight to where/how I went wrong is greatly appreciated!

    Thanks in Advance!
     

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  3. Jan 10, 2012 #2

    SammyS

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    Your solution looked good to me.

    What were the eight choices given?

    Added in Edit:

    There is a mistake in doing each of the last two integrals.

    You should have coefficients of 1/3 & 1/2 respectively.
     
    Last edited: Jan 10, 2012
  4. Jan 11, 2012 #3
    Could you explain this mistake to me? As I am not seeing how I come to get the coefficients.

    But yes you are right, when you factor in the coefficients it does give me one of the possible answers the prof gave us!
     
  5. Jan 11, 2012 #4

    SammyS

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    [itex]\displaystyle \int\ln(1+3y)dy[/itex]

    Let u = 1+3y → du = 3 dy → [itex]\displaystyle dy=\frac{1}{3}du[/itex]

    Can you take it from there?
     
  6. Jan 11, 2012 #5
    Yes, that makes perfect sense. Thank you!
     
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