Change of order of integration in polar coordinates.

Question

Change the order of integration. $$\int_{r=0}^{r = a\exp(\pi/4)}\int_{\theta=2\log(r/a)}^{\theta=\pi/2}\alpha(r,\theta)\ d\theta dr$$

From $2\log(r/a) < \theta < \pi/2$ and $0 < r < a \exp(\pi/4)$, I get $0 < r < a\exp(\theta /2)$ and $ -\infty < \theta <\pi/2$

So the integration should be $$\int_{\theta=-\infty}^{\theta=\pi/2}\int_{r =0}^{r=a\exp(\theta/2)}\alpha(r,\theta)\ d\theta dr$$

I think my answer is not correct since I am getting lower limit of $\theta < 0$ but in polar coordinates $\theta \in [0, 2\pi]$, is it correct ?

Answer 1

Your result is fine.

You don’t have to think of the polar coordinates. Instead, think of $x=r$ and $y=\theta$, and change the order in the familiar $xy$-coordinates.

Answer 2

The phase $\theta$ can be negative. You often don't want it to be negative however when you're interpreting your integral as an area -- but this is not the only use we have for integrals. However, even in this case, we may have $\theta$ out of the range $[0,2π]$ with some polar regions. Think, for example, of the region enclosed by two spirals.

So you're fine.