Answer
$$\dfrac{13\pi}{3} $$
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$
Now, $ 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, $|r_r \times r_{\theta}|=r\sqrt {4r^2+1}$
$$ Area=\int_0^{2 \pi} \int_0^{\sqrt 2} (r\sqrt {4r^2+1}) \space dr \space d \theta \\=\int_0^{2 \pi} (\dfrac{1}{12}) \times ({4r^2+1})^{3/2} \space d\theta \\= \dfrac{13\pi}{3} $$
