I'm studying in preparation for a Mathematical Analysis II examination and I'm solving past exam exercises. If it's any indicator of difficulty, the exercise is Exercise 4 of 4, part $d$ and graded for 10%.
Calculate the integral $$\iint_R\dfrac{dxdy}{\sqrt{x^2+y^2}}$$ (by changing polar coordinates) where $R=\left\{(x,y):1\leq x^2+y^2\leq 2, x\leq0, y\geq0\right\}$.
I was able to come to an answer, but I'm not sure about its correctness. I started by noting that the given $R$ is a part of a circular ring with centre O and area of the circle with centre O and radius 2 minus the circle with centre O and radius 1.
I was then able to substitute $x=r\cos\theta$ and $y=r\sin\theta$, and make some assumptions based on the thought above: By replacing the substitutes of x and y in $1\leq x^2 + y^2\leq 2 \Rightarrow 1\leq r^2(\cos^2\theta + \sin^2\theta)\leq 2 \Rightarrow 1\leq r \leq \sqrt{2}$. Additionally, based on the $x\leq 0, y\geq 0$ from the exercise itself (and the visualisation of the top-left half of the circular ring that is $R$), we've got that $\dfrac{\pi}{2}\leq\theta\leq\pi$.
With these two assumptions in mind, I move on and calculate the integral as follows: $$\iint_R\dfrac{1}{\sqrt{x^2+y^2}}dx dy = \int^{\pi}_{\frac{\pi}{2}}\int^{\sqrt{2}}_1\dfrac{1}{\sqrt{r^2\cos^2\theta + r^2\sin^2\theta }}rdrd\theta = \int^{\pi}_{\frac{\pi}{2}}\int^{\sqrt{2}}_1 (\dfrac{1}{r})rdrd\theta = \frac{\pi}{2} (\sqrt{2} - 1 ) $$
Does my answer look correct? I feel I made bad assumptions, or too many and a mistake in my process leads to the weird result. Like I missed crucial steps. In any case, any validation or help would be extremely appreciated.
By rotational symmetry, one can also compute $1/4$ of the integral over the whole angle $2\pi$: since $dx\,dy=r\,dr\,d\varphi$, $$ \int_{R} \frac{1}{\sqrt{x^2+y^2}}d(x,y) = \frac{1}{4}\int_{1<r<\sqrt 2}\frac{1}{\sqrt{x^2+y^2}}d(x,y)=\frac{2\pi}{4}\int_1^{\sqrt{2}}dr=\frac{\pi}{2}(\sqrt{2}-1). $$ But it is pretty much the same as you did; it was just to remind that symmetry in some cases can be a powerful ally.