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$
We have $ z=2$ and $ x^2+y^2=2$
$$ Surface \space Area =\iint_{R} \dfrac{|\nabla f|}{|\nabla f \cdot p|} dA=\iint_{R} \dfrac{\sqrt {4x^2+4y^2+1}}{1} \space dx \space dy\\=\iint_{R} \sqrt {4(r \cos \theta)^2+4(r \sin \theta)^2+1} \space dx \space dy \\=\int_0^{2 \pi} \int_0^{\sqrt 2} \sqrt {4r^2+1} \space dr \space d \theta\\=\int_0^{2 \pi} \space (\dfrac{13}{16}) \space d \theta \\=\dfrac{13 \pi}{3}$$
