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Math Help - Difficult double integral

  1. #1
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    Difficult double integral

    Hi!
    I'm stuck at this problem:
    Let D be the region given by |x| + |y| >= 1. Determine if the improper integral
    \int\limits_D {\int {\frac{{dxdy}}{{(x^2 + y^2 )^2 }}} }
    exists, and if it does, evaluate it.

    My attempt:
    Looking at the first quadrant and changing to polar coordinates I get r\cos \theta + r\sin \theta \ge 1 \Leftrightarrow r \ge \frac{1}{{\cos \theta + \sin \theta }}

    This gives me the integral

     <br />
4\int\limits_0^{\pi /2} {d\theta \int\limits_{\frac{1}{{\cos \theta + \sin \theta }}}^\infty {1/r^3 dr = } } 2\int\limits_0^{\pi /2} {\frac{{d\theta }}{{(\cos \theta + \sin \theta )^2 }}} <br />

    I don't know how to do that last iteration.
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by kloda View Post

     <br />
4\int\limits_0^{\pi /2} {d\theta \int\limits_{\frac{1}{{\cos \theta + \sin \theta }}}^\infty {1/r^3 dr }}<br />

    I don't know how to do that last iteration.
    It's ok until here

    An antiderivative of \frac{1}{r^3} is indeed \frac{-2}{r^2}

    But when substituting r by \frac{1}{\cos \theta+\sin \theta}, it yields :

    -2 (\cos \theta+ \sin \theta)^2


    So the latter integral should be 2 \int_0^{\frac{\pi}{2}} (\cos \theta+\sin \theta)^2 d \theta

    Note that (\cos \theta+\sin \theta)^2=\cos^2 \theta + \sin^2 \theta+2 \cos \theta \sin \theta=1+\sin 2 \theta

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  3. #3
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    Quote Originally Posted by Moo View Post
    But when substituting r by \frac{1}{\cos \theta+\sin \theta}, it yields :

    -2 (\cos \theta+ \sin \theta)^2
    Oops!
    Thanks for the quick reply!
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