Evaluating Improper Integrals in Polar Coordinates

[karush]

15.3.65 Improper integral arise in polar coordinates

$\textsf{Improper integral arise in polar coordinates when the radial coordinate r becomes arbitrarily large.}$
$\textsf{Under certain conditions, these integrals are treated in the usual way shown below.}$
\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
I&=\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*}
ok just want to make sure this is correct before the high dive..
also tried to plot this on desmos but

 

[jvicens]

Hello there,

I can confirm that your approach is correct. When dealing with improper integrals in polar coordinates, we can rewrite them with limits as shown in your example. The limit as b approaches infinity ensures that the radial coordinate r becomes arbitrarily large, which is necessary for the integral to be evaluated properly.

I'm not sure why you had trouble plotting this on Desmos, but the graph should look like a curve that approaches 0 as r approaches infinity, with the limits of integration being from 2 to infinity. I hope this helps and good luck with your high dive!