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Double integral

  1. Jan 4, 2012 #1
    1. The problem statement, all variables and given/known data
    find ∫∫ sin(x+y) dxdy in the domain D where D=(x,y) where y≥√x and y≤2x and y≥(1/x) and y≤2/x


    2. Relevant equations
    i

    3. The attempt at a solution
    took y as variable so i have two domains D1 where x is between 0.5 and 1 and y between 1/x and √x
    D2 where x is between 1 and 4 power 1/3 and y is between 2x and 2/x
    i will reach to one variable in terms of x but it is very hard to find them cos(x+√x) cos(3x) cos(1/x +x) so??????
     
  2. jcsd
  3. Jan 4, 2012 #2
    any one?????????????? please????????????????????
     
  4. Jan 4, 2012 #3

    I like Serena

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    Hey queenstudy! :smile:

    I agree something like cos(x+√x) or cos(x+1/x) is pretty hard to integrate.
    They do not have anti-derivatives that can be expressed in standard functions, so either you should leave them as is, or you should approximate them numerically.

    Is it possible that your problem statement only asked to write down the boundaries of your integral?

    As it is, the boundaries you found are not the right boundaries...
     
  5. Jan 5, 2012 #4
    they were in the exam so what the hell should i do??
    about the boundaries i know they arent true
     
  6. Jan 5, 2012 #5

    ehild

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    Ask your teacher. Maybe you copied something wrong or he wrote something wrong. Perhaps the problem can be solved with some transformation of the coordinates, but I do not know how.
    As for the boundaries, see attached picture.

    ehild
     

    Attached Files:

  7. Jan 6, 2012 #6
    he gave us the domain which is exactly the same , but how can i solve it is it like impossible or what???
     
  8. Jan 6, 2012 #7
    so there is no way the problem can be solved??
     
  9. Jan 6, 2012 #8

    I like Serena

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    I'm going to go out on a limb here and say: no, it can't be solved.
    There is no formula for it. It can only be approximated numerically or geometrically.

    Oh, and of course it is possible to write down the integral with proper boundaries (if you can find them).
     
  10. Jan 6, 2012 #9
    thank you very much , i guess i am a bit better , i hope you are right
     
  11. Jan 6, 2012 #10

    SammyS

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    Do you know how to handle a coordinate transform for a double integral?

    I suggest trying:
    u=x2-y2

    v=2xy​

    I haven't tried it, so I can't guarantee that it will help.
     
  12. Jan 6, 2012 #11
    [itex]\sin(x+y)=\sin x\cos y+\cos x\sin y[/itex] Should help.
     
  13. Jan 6, 2012 #12

    SammyS

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    After working with this a bit (not done yet) I suggest transforming to Hyperbolic coordinates. (see Wikipedia)

    [itex]\displaystyle u = -\frac{1}{2} \ln \left( \frac{y}{x} \right)[/itex]

    [itex]\displaystyle v = \sqrt{xy}[/itex]

    The inverse transformation is:

    [itex]\displaystyle x = v\, e^u ,\quad y = v\, e^{-u}[/itex]

    The limits of integration work out nicely. I haven't looked at the integrand or integration in detail.
     
  14. Jan 7, 2012 #13
    so you are saying it is possible to solve this double integral??
    another question: is i have cos(p(x)) what is its derivative??
     
  15. Jan 7, 2012 #14
    it made it much harder ,
     
  16. Jan 7, 2012 #15

    SammyS

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    I'm not saying it's possible to solve. In fact, after trying these transformations plus a couple more, I suspect that it can't be done in closed form.

    The transformations may make it possible to describe the region more simply, but all that I tried made the anti-derivative impossible to express in closed form.

    Use the chain rule to get the derivative of cos(p(x)) .
    It's the derivative of cosine --evaluated at p(x) -- times the derivative of p(x).​
     
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