Answer
$\dfrac{13 \pi}{3}$
Work Step by Step
Here, we have $z=2$ and $x^2+y^2=2$
We know that $x=r \cos \theta, y=r \sin \theta $
The surface area is given by:
$S=\iint_{R} \dfrac{|\nabla f|}{|\nabla f \cdot p|} dA=\iint_{R} \dfrac{\sqrt {4x^2+4y^2+1}}{1} dx dy$
This implies that
$S=\iint_{R} \sqrt {4(r \cos \theta)^2+4(r \sin \theta)^2+1} dx dy$
or, $S=\int_0^{2 \pi} \int_0^{\sqrt 2} \sqrt {4r^2+1} dr d \theta$
Thus, $S=\int_0^{2 \pi} (\dfrac{13}{16}) d \theta=\dfrac{13 \pi}{3}$
