So, lets say I have a pretty regular 2 units radius circle:
$$ x^2+y^2=4 $$
If I want to calculate the area:
$$ \int_{-2}^{2}dy \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} dx = 4\pi $$
So let's say I want to switch it to polars:
$$ x=r\cos(\theta) $$$$ y=r\sin(\theta) $$$$ J = \begin{vmatrix}x_r & y_r\\x_\theta & y_\theta\end{vmatrix} = r $$
So we normally would just do it like this:
$$ \int_{0}^{2\pi} d\theta \int_{0}^{2} rdr = 4\pi $$
But here is the problem, is there a way to obtain the integration limits $ 0, 2\pi $ and $ 0, 2 $ without actually knowing how the integration region looks like?
I tried the following
$$ y = 2 $$$$ r\sin(\theta)=2 $$$$ \theta = \arcsin(2/r) $$
And with $r$
$$ x=\sqrt{4-y^2} $$$$ r\cos(\theta)=\sqrt{4-r^2\sin^2(\theta)} $$$$ r=2 $$
I'm not entirely sure how to find $r$ for the lower limit (or the upper for that case), because I actually end up with the same absolute value. So I just tried with $r=\pm2$:
$$ \int_{-2}^{2} dr \int_{\arcsin(-2/r)}^{\arcsin(2/r)} rd\theta $$
And if I plug that into Wolfram Alpha I get $2\pi$ which is not exactly what I was looking for, but at least it's coherent.
But the point is, how do I get to $ 0, 2\pi $ and $ 0, 2 $ without actually looking at the plot?