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Thread: Double integral..ln

  1. #1
    Junior Member
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    Double integral..ln

    Hi all,

    I need help with the following task:

    Calculate
    ∫∫ ln(1+x^2+y^2) dxdy
    D

    where D is: (x,y); 1 <=x^2+y^2<=2

    (<= means bigger than or equal to)

    I think I should use polar coordinates where

    x= r cos
    y= r sin

    Then I get:

    ∫∫ ln(1+r^2)

    ∫ dr goes from 1 to 2
    ∫ d goes from 0 to 2 pi

    But what is the integral of ln(1+r^2)?


    I really need help,

    thanks!
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  2. #2
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    Remember that the transformation from cartesians to polars is $\displaystyle \displaystyle dy\,dx \to r\,dr\,d\theta$.

    So your double integral should be

    $\displaystyle \displaystyle \int_0^{2\pi}{\int_1^2{r\ln{(1 + r^2)}\,dr}\,d\theta}$.

    You can solve this using a $\displaystyle \displaystyle u$ substitution.
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  3. #3
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    Thank you, but I dont understand how to use u substitution.

    The thing that troubles me is that ln(1+r^2)*r has two "steps". one is that the anti-derivative is 1/(1+r^2) but then there is the inner r^2 to consider and also the r from the Jacobian...
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Prove It View Post
    $\displaystyle \displaystyle \int_0^{2\pi}{\int_1^2{r\ln{(1 + r^2)}\,dr}\,d\theta}$.
    look at the radius of the second circle.
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  5. #5
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    Quote Originally Posted by tinyone View Post
    Thank you, but I dont understand how to use u substitution.

    The thing that troubles me is that ln(1+r^2)*r has two "steps". one is that the anti-derivative is 1/(1+r^2) but then there is the inner r^2 to consider and also the r from the Jacobian...
    How could you possibly be doing double integrals in polar coordinates and not know how to use "substitution"?

    Let $\displaystyle u= 1+ r^2$. What is "du"?

    (And, as Krizalid suggests, the radii of the two circles are 1 and $\displaystyle \sqrt{2}$, not 1 and 2.)
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