Answer
$$\dfrac{8\pi}{3}$$
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) i+\sin \theta j $ and $ r_{\theta}=(-r\sin \theta) i+(r \cos \theta) j+k $
so, $|r_r \times r_{\theta}|=\sqrt {1+r^2}$
$$ Surface \space Area =\int_0^{2 \pi} \int_0^1 (1+r^2) \space dr \space d\theta \\= \int_0^{2 \pi} [r+\dfrac{r^3}{3}]_0^1 \space d \theta \\=\int_0^{2 \pi} \dfrac{4}{3} \space d\theta\\=\dfrac{8\pi}{3}$$
