Chapter 15 - Section 15.5 - Surfaces and Area - Exercises - Page 873: 46

$\dfrac{13\pi}{3} $

Here, we have $r_r=( \cos \theta) i+(\sin \theta) j+2k$ and $r_{\theta}=(-r\sin \theta) i+( r\cos \theta) j$ Also, $|r_r \times r_{\theta}|=r\sqrt {4r^2+1}$ Thus, the area is given by $A=\int_0^{2 \pi} \int_0^{\sqrt 2} (r\sqrt {4r^2+1}) dr d \theta$ or, $A=\int_0^{2 \pi} (\dfrac{1}{12})[{4r^2+1})^{3/2} d\theta= \dfrac{13\pi}{3} $