Answer
$\dfrac{8\pi}{3}$
Work Step by Step
Here, $r_r=(\cos \theta) i+\sin \theta j; r_{\theta}=(-r\sin \theta) i+(r \cos \theta) j+k$
Then
$|r_r \times r_{\theta}|=\sqrt {1+r^2}$
The surface ares is given by: $S=\iint_S |r_u \times r_v| du dv$
Thus, $S=\int_0^{2 \pi} \int_0^1 (1+r^2) dr d\theta$
$\int_0^{2 \pi} (r+\dfrac{r^3}{3}]_0^1 d \theta=\int_0^{2 \pi} \dfrac{4}{3} d\theta=\dfrac{8\pi}{3}$
