Answer
$$\dfrac{49\pi }{3}$$
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$
We have $ x^2+y^2=z $ and $ z=r^2$
$ r_r= \cos \theta \space i+\sin \theta \space j+2 \space k ;
\\r_{\theta}=-r\sin \theta \space i+ r\cos \theta \space j $.
Also, $|r_r \times r_{\theta}|=r\sqrt {4r^2+1}$
Now, $$ Area=\int_0^{2 \pi} \int_\sqrt 2^{\sqrt 6} (r\sqrt {4r^2+1}) \space dr \space d \theta\\=\int_0^{2 \pi} (\dfrac{125}{12}- \dfrac{27}{12})\space d \theta \\= \space \dfrac{49\pi }{3}$$
