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Math Help - The double integral of this problem.

  1. #1
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    The double integral of this problem.



    Evaluate the iterated integral. Ignore the answer. I want to know how to integrate this. I know you first integrate from respect with x first. But how do you integrate the first part? Trig Substitution? Its been awhile since I done integrals.
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  2. #2
    Super Member Random Variable's Avatar
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    EDIT:  \int^{3}_{1} \int_{0}^{y} \frac{4}{x^{2}+y^{2}} \ dx \  dy = 4 \int^{3}_{1} \int^{y}_{0} \frac{1}{y^{2}(1+\frac{x^{2}}{y^{2}})} \ dx \ dy

     = 4 \int^{3}_{1} \frac{1}{y^{2}} \int^{y}_{0} \frac{1}{1+(\frac{x}{y})^2} \ dx dy = 4 \int^{3}_{1} \frac{1}{y} \int_{0}^{1} \frac{1}{1+u^{2}} \ du \ dy =  =  4 \int^{3}_{1} \frac{1}{y} \arctan u \Big|^{u=1}_{u=0} \ dy

     = \pi \int^{3}_{1} \frac{dy}{y} = \pi \ln y \Big|^{3}_{1} = \pi \ln 3
    Last edited by Random Variable; March 23rd 2010 at 06:24 PM.
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  3. #3
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    Why did the y change to 3 then to 1?
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  4. #4
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    or use polar coordinates:

    \int_{1}^{3}{\int_{0}^{y}{\frac{4}{x^{2}+y^{2}}\,d  x}\,dy}=4\int_{0}^{\frac{\pi }{4}}{\int_{\sec \varphi }^{3\sec \varphi }{\frac{dr\,d\varphi}{r}}}=4\ln (3)\int_{0}^{\frac{\pi }{4}}{d\varphi }=\pi \ln 3.
    Last edited by Krizalid; March 23rd 2010 at 06:55 PM.
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by asdf122345 View Post
    Why did the y change to 3 then to 1?
    Because I made a mistake.
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