Answer
$ \dfrac{\pi}{6}(13\sqrt {13}-1) $
Work Step by Step
Here, we have $r_r=\lt \cos \theta,\sin \theta, 2r$
and
$r_{\theta}=\lt -r\sin \theta, r\cos \theta,0 \gt $
Also, $|r_r \times r_{\theta}|=r\sqrt {4r^2+1}$
Thus, the area is given by $A=\int_0^{2 \pi} \int_1^{\sqrt 3} (r\sqrt {4r^2+1}) dr d \theta$
or, $A=\int_0^{2 \pi} (\dfrac{13\sqrt {13}}{12}-1)d \theta= \dfrac{\pi}{6}(13\sqrt {13}-1) $
