15.3.65 Rewriting double integral to infnty

[karush]

\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

 

[MarkFL]

Re: 15.3.65 rewriting dbl int to infnty

What is the problem, exactly as given?

 

[karush]

Re: 15.3.65 rewriting dbl int to infnty

What is the problem, exactly as given?

what is b

frankly I don't know how to finish this

 

[MarkFL]

Re: 15.3.65 rewriting dbl int to infnty

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?

 

[HallsofIvy]

Re: 15.3.65 rewriting dbl int to infnty

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