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Math Help - improper double integral

  1. #1
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    improper double integral

    Compute this integral.
    I've tried using polar coordinates (it seems the obvious thing to do) but the domain pretty much messed everthing up. since D is the upper right square [0,1]X[0,1].
    I got 2 possible boundaries to choose from:
    1) 0<=R<=sqrt(2) , 0<=Theta<=sin(1/R)
    2) 0<=R<=1/sin(Theta) , 0<=Theta<=pie/2

    I couldn't integrate either of them.. your assistance is welcome :-)
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  2. #2
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    Quote Originally Posted by zokomoko View Post
    Compute this integral.
    I've tried using polar coordinates (it seems the obvious thing to do) but the domain pretty much messed everthing up. since D is the upper right square [0,1]X[0,1].
    I got 2 possible boundaries to choose from:
    1) 0<=R<=sqrt(2) , 0<=Theta<=sin(1/R)
    2) 0<=R<=1/sin(Theta) , 0<=Theta<=pie/2

    I couldn't integrate either of them.. your assistance is welcome :-)
    If you switch to polar coordinates won't you just have

    \int_{\theta = 0}^{\pi/4} \int_{r = 0}^{r = 1/\cos \theta} dr \, d \theta + \int^{\pi/2}_{\theta = \pi/4} \int_{r = 0}^{r = 1/\sin \theta} dr \, d \theta

    and the resulting integrals in \theta are not nasty.
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  3. #3
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    thanks

    thank you for your answer, it was helpful :-)

    I got stuck with another problem, it says "compute the following integral. hint: use the rotational transformation to get new coordinates (u,v) so the u = ..."

    I didn't understand how the substitution they suggested is a rotational transformation, and consequently what v(x,y) should be.

    thank you in advance.
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  4. #4
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    Quote Originally Posted by zokomoko View Post

    I've tried using polar coordinates
    To do that, split the original square into two triangles:

    \begin{aligned}<br />
   \int_{0}^{1}{\int_{0}^{1}{\frac{dx\,dy}{\sqrt{x^{2  }+y^{2}}}}}&=2\int_{0}^{1}{\int_{0}^{x}{\frac{dy\,  dx}{\sqrt{x^{2}+y^{2}}}}} \\ <br />
 & =2\int_{0}^{\frac{\pi }{4}}{\int_{0}^{\sec \varphi }{dr}\,d\varphi }=2\int_{0}^{\frac{\pi }{4}}{\sec \varphi \,d\varphi } \\ <br />
 & =2\ln \left| \sec \varphi +\tan \varphi  \right|\bigg|_{0}^{\frac{\pi }{4}}=2\ln \left( \sqrt{2}+1 \right).<br />
\end{aligned}
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