Chapter 15 - Section 15.5 - Surfaces and Area - Exercises - Page 872: 19

$8 \pi \sqrt 5$

Here, we have $z=2; z=6$ and $z=2 \sqrt {x^2+y^2}$ Then $r_r=( \cos \theta) i+(\sin \theta) j+2k$ and $r_{\theta}=(-r \sin \theta) i+(r \cos \theta) j$ Also, $|r_r \times r_{\theta}|=r \sqrt 5$ Thus, the area is given by $A=\int_0^{2 \pi} \int_1^{3} r \sqrt 5 dr d \theta$ or, $A=\int_0^{2 \pi} 4 \sqrt 5 d \theta=8 \pi \sqrt 5$