Chapter 15 - Section 15.5 - Surfaces and Area - Exercises - Page 873: 38

$\dfrac{49\pi }{3}$

Here, we have $x^2+y^2=z$ and $ z=r^2$ $r_r=( \cos \theta) i+(\sin \theta) j+2k$ 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 $A=\int_0^{2 \pi} \int_\sqrt 2^{\sqrt 6} (r\sqrt {4r^2+1}) dr d \theta$ or, $A=\int_0^{2 \pi} (\dfrac{125}{12}- \dfrac{27}{12}-)d \theta= \dfrac{49\pi }{3}$