Is it fine this change of variables? Integral in polar coordinates

Question

$$\iint_{R} \frac{x}{y\sqrt{x^2+y^2}} dA$$ Where $R$ it's $\{x^2+y^2\leq 4,\ y\geq1\}$ I have changed to $x=r\cos{\theta}, y=r\sin{\theta}$. Seen the graph, we have $$\int_{-\pi/6}^{\pi/6} \int_{0}^2 \frac{r\cos\theta}{r\sin\theta\sqrt{r^2}}r\ dr \theta= \int_{-\pi/6}^{\pi/6} \int_{0}^2 \cot\theta\ dr \theta$$ is it ok?

Answer 1

Let's obtain it formally from inequalities $$x^2+y^2 \leqslant 4 \Leftrightarrow 0\leqslant r \leqslant 2$$ and $$y=r\sin \theta \geqslant 1 \Leftrightarrow r \geqslant \frac{1}{\sin \theta}$$ so for $r$ we have $\frac{1}{\sin \theta} \leqslant r \leqslant 2$. This last also give restrictions on $\theta$ i.e. we have $$\int\limits_{\frac{\pi}{6}}^{\pi-\frac{\pi}{6}}\int\limits_{\frac{1}{\sin \theta} }^{2}$$

Answer 2

The integral is rather straightforward without any computation. The domain is symmetric about the line $x=0$ and the integrand is an odd function of $x$. Therefore

$$I = \iint_R \frac{x}{y\sqrt{x^2+y^2}}dxdy = 0$$