Answer
$$\pi[ \sqrt 2 +\ln (1+\sqrt 2)]$$
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=(\cos \theta) i+\sin \theta j; r_{\theta}=(-r\sin \theta) i+(r \cos \theta) j+k $
Now, $|r_r \times r_{\theta}|=\sqrt {1+r^2}$
$$ Surface \space Area =\iint_S |r_u \times r_v| du dv \\=\int_0^{2 \pi} \int_0^1 \sqrt {1+r^2} dr d\theta \\ = \int_0^{2 \pi} \dfrac{\sqrt 2}{2}+\dfrac{\ln (1+\sqrt 2)}{2} ] d\theta \\=\pi[ \sqrt 2 +\ln (1+\sqrt 2)]$$
