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Double Integral evaluation help

  1. May 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate the double integral [tex]\int \int_{R} ln(xy) dA[/tex] where [tex]R[/tex] is the rectangle bounded by [tex]x=e, x=e^2,y=1,y=e[/tex].

    2. Relevant equations
    [tex]ln (xy) = ln x + ln y[/tex]
    3. The attempt at a solution

    I was just wondering, do I need to do anything other than take the integral with respect to x, plug in the bounds, and then take the integral with respect to y, and plug in the bounds? I know to use integration by parts for ln x, and I know to break the ln xy up, but is it fine just to treat the ln y as a constant even if I do integration by parts? That would make it become x ln y. Thanks.
     
  2. jcsd
  3. May 3, 2009 #2

    Mark44

    Staff: Mentor

    If you're integrating with respect to x, then yes, ln y can be treated as a constant.

    But what is "it" that would become x ln y?
     
  4. May 3, 2009 #3
    I was referring to the constant when I integrate with respect to x. Sorry, I was a little vague.

    Thank you, though, I just was a little unsure on whether having a "u" substitution would change things up, but I should just treat it as a constant.

    I ended up with the result [tex]2e^3 - 2e^2[/tex].
     
  5. May 4, 2009 #4
    Just do the integral with respect to why then x, so it will be written dy dx,

    for dy the bounds will just be from 1 to e,

    then for dx it will go from e to e^2
     
  6. May 4, 2009 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [tex]\int_{x=a}^b \int_{y= c}^d f(x)+ g(y) dy dx= \left(\int_{x= a}^b f(x)dx\right)\left(\int_{y= c}^d dy\right)+ \left(\int_{x=a}^b dx\right)\left(\int_{y= c}^d g(y)dy\right)[/tex]
    [tex]= (d-c)\int_{x=a}^b f(x)dx+ (b-a)\int_{y= c}^d g(y)dy[/tex]
     
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