converting improper double integrals to polar form: what do I do with infinity limits

Question

I need to convert

$$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}-e^{\frac{x^2+y^2}{5}}dA$$

To polar form.

I know $x^2+y^2 = r^2, $ and $dA = rdrd\theta$

But what do I do with the $\infty$ limits?

Answer 1

First notice that, in the original integral, the region over which you are integrating, is the whole of $\mathbb{R^2}$ (the $xy$-plane).

In polar coordinates, we must then have that both $r$ and $\theta$ also has its "maximum domain".

Please Note: My choice of wording in the above is VERY rough, as I want to explain it as easily as possible.

We will have the following limits: $$0\leq r\leq \infty\ , 0 \leq\theta \leq 2\pi$$ This is a rectangular region within the $r \theta$-plane (we will use this fact to interchange order of integration later).

This ensures that ALL circular regions within the original $xy$-plane gets included.

Our integral then becomes \begin{align*}\int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty-e^{\frac{x^2 + y^2}{5}}dA &= \int\limits_{\theta=0}^{2\pi} \int \limits_{r=0}^{\infty} -e^\frac{r^2}{5}rdrd\theta \\ \\ &= \int \limits_{r=0}^\infty\int \limits_{\theta =0}^{2\pi}-e^\frac{r^2}{5}rd\theta dr\\&= \int \limits_{r=0}^\infty -2\pi e^{\frac{r^2}{5}}rdr \\ &= -2 \pi\int \limits_{r=0}^\infty re^\frac{r^2}{5}dr\end{align*}

Notice, however that the above answer is in it's simplest form, since the improper integral given in the last step, does not converge on $r\in[0,\infty)$.

Answer 2

In this case, to cover the whole plane, $r$ goes from $0$ to $\infty$ (and $\theta$ goes from $0$ to $2\pi$).