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Evaluate the double integral

  1. Apr 26, 2009 #1
    1. The problem statement, all variables and given/known data

    \int \int_R (x+y) dA R is the region y = x^2 and y = x^(1/2)

    2. Relevant equations
    I've sketched the graph and the functions are equal at (0,0) and (1,1)



    3. The attempt at a solution

    Would the limits of the integral be:

    \int_{0}^{1) \int_{0}^{1} (x+y) dydx


    regards
     
  2. jcsd
  3. Apr 26, 2009 #2
    Sorry would it be


    Would the limits of the integral be:

    \int_{0}^{1) \int_{x^1/2}^{x^2} (x+y) dydx


    regards
     
  4. Apr 26, 2009 #3

    djeitnstine

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    Gold Member

    boneill3, double integration is not like single variable integration. You need to always express your first limits in terms of your second integrating variable. Perhaps some math guru can come in here and give you a more rigorous definition.

    But here your limits are \int_{0}^{1) \int_{x^2}^{x^(1/2)} dydx

    perhaps here may help:

    http://ltcconline.net/greenl/courses/202/multipleIntegration/iteratedIntegrals.htm

    edit: I was a little too slow on the reply
     
  5. Apr 26, 2009 #4
    Not always, sometimes the two are independent, it's just in this case they are not.
     
  6. Apr 26, 2009 #5

    djeitnstine

    User Avatar
    Gold Member

    Thanks, I just know how to do it, not the reasoning behind it =]
     
  7. Apr 27, 2009 #6
    Thank you for that web link It explains it a lot better than my lecture notes.

    I see that I got the functions upper and lower boundsupside down as well.

    \int_{x^2}^{x^1/2}

    instead of

    \int_{0}^{1) \int_{x^1/2}^{x^2} (x+y) dydx
     
  8. Apr 27, 2009 #7
    Hopefully I've got this one right

    \int \int_R (xcosy) dA R in the region y = x and y = 0 and x = \pi

    This is a triangular region bounded on the right by x = \pi, below by y = 0 and above by y = x.

    So the limits are:
    0<= x <= \pi and 0 <=y <= x

    \int_{0}^{\pi) \int_{0}^{x} (xcosy) dydx

    regards
    Brendan
     
  9. Apr 27, 2009 #8
    It looks right
     
  10. Apr 27, 2009 #9
    That's right. The limits of integration are presented through the analytical expression of the domain:

    So the domain of this question would be:

    D = {(x,y)| 0<=y<=x, 0<=x<=pi }
     
  11. Apr 27, 2009 #10
    Thanks guys
     
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