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Math Help - Integrate the function

  1. #1
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    Integrate the function

    Hi, I am kinda having trouble trying to find the limits for the triple integral.
    Please help....

    Thanks.

    Integrate the function over the solid given by the figure below, if P=(-4,-4,2), Q=(-3,-3,2), R=(3,3,2), and S=(4,4,2).
    integral =
    (You should find the actual value when you evaluate the integral!)
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  2. #2
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    Are we to assume the bottom is in the xy-plane? Use cylindrical coordinates, of course.
    \theta goes from \pi/4 to 5\pi/4 (arctan(4/4)= arctan(1) to arctan(-4/4)), r from 3\sqrt{2} to 4\sqrt{2}, and z from 0 to 2.
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  3. #3
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    Thanks a lot for the reply...

    I tried to integrate the function and getting 2960/3 which is not the right answer...

    okay.. first I converted to cylindrical coordinate!!!

    Given, -5x+5y

    so -5(r*cos(t))+5(r*sin(t))

    now integrated the following function:

    => (-5(r*cos(t))+5(r*sin(t)))*r*dz*dr*dt

    => (740/3)*sqrt(2)*sin((1/4)*Pi)+(740/3)*sqrt(2)*cos((1/4)*Pi)-(740/3)*sqrt(2)*sin((5/4)

    => 2960/3
    Last edited by mr fantastic; March 21st 2009 at 02:08 PM. Reason: Merged posts
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  4. #4
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    finally..

    the answer is -2960/3 ... so I guess its not a volume...

    Thanks again....
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  5. #5
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    what would make you think it was a volume?
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  6. #6
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    volume!!!

    Well, double and triple integral could be volume if( f(x)>=0), so I assumed that they wanted me to find the volume... but I was wrong...
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