[Math] Converting unit square domain in (x,y) to polar coordinates

Question

I have the following double integral

$\int_{0}^{1}\int_{0}^{1}\frac{x}{\sqrt{x^2+y^2}}dxdy$

The integrand is fairly simple: $\frac{x}{\sqrt{x^2+y^2}}dxdy=\frac{r\cos(\theta )}{\sqrt{r^2}}rd\theta{}dr=r\cos{(\theta)}d\theta{}dr$

My trouble is with the limits of integration. I've tried:

$0 \leq y \leq 1$ means $0 \leq{} r\sin(\theta{}) \leq 1$ so $0 \leq \theta \leq arcsin(1/r)$

But why isn't it just $0<\theta < \pi{}/2$ since the unit square is in the first quadrant?

My hunch for $r$ is that it varies within $1 \leq r \leq \sqrt{2}$

So we have $\int_{1}^{\sqrt{2}}\int_{0}^{\arcsin(1/r)}r\cos{(\theta)}d\theta{}dr=\int_{1}^{\sqrt{2}}1dr=\sqrt{2}-1$

My book gives
$2 (\ln(\sqrt{2} + 1) + \sqrt{2} − 1)$

Any advice on how to visualize this is much appreciated.

Answer 1

$$\int_{0}^{\pi /4}\int_{0}^{csc\theta }r\cos\theta drd\theta +\int_{\pi /4}^{\pi /2}\int_{0}^{sec\theta }r\cos\theta drd\theta$$