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  1. MHB Master
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    #1
    \begin{align*}\displaystyle
    \int_{\alpha}^{\beta}\int_{a}^{\infty}
    g(r,\theta) \, rdr\theta
    =\lim_{b \to \infty}
    \int_{\alpha}^{\beta}\int_{a}^{b}g(r,\theta)rdrd\theta
    \end{align*}
    $\textit{Evaluate the Given}$
    \begin{align*}\displaystyle
    &=\iint\limits_{R} e^{-x^2-y^2} \, dA \\
    (r,\theta) \, 2 \le r \le \infty \\
    &\, 0 \le \theta \le \pi/2
    \end{align*}$\textit{Rewrite with limits}$
    \begin{align*}\displaystyle
    &\lim_{b \to \infty}\int_{0}^{\pi/2}\int_2^{\infty} e^{-x^2-y^2} rdrd\theta
    \end{align*}

    just seeing if I'm going in the right direction☕
    Last edited by karush; September 14th, 2017 at 00:53.

  2. Pessimist Singularitarian
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    #2
    What is the problem, exactly as given?

  3. MHB Master
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    #3 Thread Author
    Quote Originally Posted by MarkFL View Post
    What is the problem, exactly as given?
    what is b

    frankly I don't know how to finish this

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    #4
    Quote Originally Posted by karush View Post
    what is b

    frankly I don't know how to finish this
    You've got an integrand in terms of $x$ and $y$, and differentials in terms of $r$ and $\theta$...can you state the problem exactly as it was given to you?

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    #5
    Of course x2+ y2= r2 so that integral is the same as $ \displaystyle \int_0^{2\pi}\int_2^\infty e^{-r^2}rdrd\theta$. Letting $ \displaystyle u= r^2$, that is easy to integrate.

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