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    improper integral problem...

    how to evaluate the improper integral using polar coordinates?
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    Hello,
    Quote Originally Posted by iwonder View Post
    how to evaluate the improper integral using polar coordinates?
    x=r \cos \theta
    y=r \sin \theta

    Then x^2+y^2=r^2 (\cos^2 \theta+ \sin^2 \theta)=r^2

    and the Jacobian of the transformation is r.
    So dx ~  dy=r ~ dr ~ d \theta

    Since (x,y) \in \mathbb{R} \times \mathbb{R}, and since r is positive, we have r \in [0,+ \infty) and  \theta \in [0,2 \pi], so that it covers all the plane.



    ---> =\int_0^{2 \pi} \int_0^\infty e^{-3r^2} ~ r ~ dr ~ d \theta

    =\left(\int_0^{2 \pi} ~ d \theta \right) \cdot \left(\int_0^\infty e^{-3r^2} ~ r ~ dr\right)

    In order to calculate the second one, make a substitution : u=r^2
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