Answer
$$\dfrac{(13\sqrt {13}-1) \pi}{6} $$
Work Step by Step
Apply cylindrical coordinates. $ x=r \cos \theta ;\\ y= r \sin \theta ;\\ z \gt 0$
We know that $ r(r, \theta)=xi+yj+zk $ or, $ r^2=x^2+y^2+z^2$
$ r_r=\lt \cos \theta,\sin \theta, 2r \gt ; r_{\theta}=\lt -r\sin \theta, r\cos \theta,0 \gt $
Also, $|r_r \times r_{\theta}|=r\sqrt {4r^2+1}$
Now, $ Area=\int_0^{2 \pi} \int_1^{\sqrt 3} (r\sqrt {4r^2+1}) \space dr \space d \theta \\=\int_0^{2 \pi} (\dfrac{13\sqrt {13}}{12}-1) \space d \theta \\=\dfrac{(13\sqrt {13}-1) \pi}{6} $$
