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Double integrals + Change of variables

  1. Oct 17, 2004 #1
    Ok, i have a problem with this double integral. I am having a hard time finding the limits. The question is

    \iint \frac{dx\,dy}{\sqrt{1+x+2y}}\

    D = [0,1] x [0,1], by setting T(u,v) = (u, v/2) and evaluating the integral over D*, where T(D*)=D

    Can some one help me find the limits, and explain the process of getting those limits.

    Thanks in advance
  2. jcsd
  3. Oct 17, 2004 #2


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    Are you REQUIRED to do that change of variables?
    You could integrate it directly..
  4. Oct 17, 2004 #3
    Yes, because there are some problems which say evaluate the integral with change of variable but then check it by using an iterated integral. So the answer is yes, i have go use change of variable, and even though i dont need to use it was to get practice at it.

  5. Oct 17, 2004 #4
  6. Oct 18, 2004 #5
    Somebody is always here but not your wishing somebody.
    Can you find the integration (to x variable) of 1/root(a+x) ?
  7. Oct 18, 2004 #6


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    I THINK what you are saying is that you want to use the substitution u= x, v= 2y.
    Of course, du= dx and dv= 2dy or dy= (1/2)dv.

    In terms of u and v, the integral becomes
    [tex]\frac{1}{2}\int \frac{du\,dv}{\sqrt{1+u+v}}[/tex]

    The only problem now is finding D*. The boundaries of D are x= 0, x= 1, y= 0, y= 1.
    Okay, when x= 0 what is u? When x= 1, what is u? When y= 0, what is v? When y= 1, what is v? That gives you D* and the limits of integration.
  8. Oct 18, 2004 #7
    Oh well, sorry for my bad English, I didn't read the whole OP and thought he didn't know how to solve it, but he should say the same as you did anyway, I know that for certain :biggrin:
    Last edited: Oct 18, 2004
  9. Oct 19, 2004 #8

    I thought about the problem a little harder and i did the same exact thing you did, so thanks for your help.
  10. Jun 27, 2008 #9
    What is the non-graphing method to find the new limits ? Someone plz,,,,,...........!!!
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