I have to calculate:
$$\iint _{R} \frac{x}{\sqrt{x^{2}+y^{2}}}dA$$
and R is this region:
$$x^{2}+y^{2}=16; x^{2}+y^{2}=4; y = \sqrt{3}x; y=\frac{x}{\sqrt{3}}$$
so I used substitution:
$$x = r\cos \theta \: \: \: \wedge \: \: \: y = r\sin \theta$$
therefore I have:
$$\int_{\frac{\pi }{6}}^{\frac{\pi }{3}}\int_{2}^{4}\frac{rcos\theta }{r}rdrd\theta$$
$$=6\left [ \frac{\sqrt{3}}{2}-\frac{1}{2} \right ]$$
is this right way to approach this double integral?