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Double integrals for volume

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data
    find the volume of the solid that lies below the graph of z = 1/(x^2 + y^2) and that is bounded laterally by the cylinder set y =|x| and the planes y = 2 and y = 8


    2. Relevant equations



    3. The attempt at a solution
    well i kno that the z equals equation is what i will be integrating, so i was wondering the significance of the y = planes. If done on an xy plane then the y equals is the bounds and im integrating the region b/w y = 2 and 8 in regards to the y =|x|

    What i set up was the integral from 2 to 8 of the integral of -y to y of (1/x^2 + y^2)dydx

    Am i on the right track?
     
  2. jcsd
  3. Mar 24, 2009 #2
    Close, but be careful. When you're integrating with respect to y, you also have y in the bounds. You have an expression for how y varies, so use that instead.
     
  4. Mar 25, 2009 #3

    Mark44

    Staff: Mentor

    The region over which you're integrating is a trapezoid, where the parallel sides are the lines y = 2 and y = 8. The nonparallel sides come from y = |x|. Because the graph of z = 1/(x^2 + y^2) is symmetric with respect to both the x- and y-axis, you can make your integral a little simpler by integrating over half the region and multiplying your result by 2.

    In other words, one possibility for your integral is this:
    [tex]
    2\int_{y = 2}^8 \int_{x = 0}^y <your function> dx dy
    [/tex]
    You could set it up so that the order of integration is reversed, but that would make things more complicated.
     
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