Chapter 15 - Section 15.5 - Surfaces and Area - Exercises - Page 872: 21

$6 \pi $

Here, we have $z=1; z=4$ and $x^2+y^2=1$ Then $r_z=( \cos \theta) i+(\sin \theta) j+2k$ and $r_{\theta}=(-\sin \theta) i+( \cos \theta) j$ Also, $|r_z \times r_{\theta}|=1$ Thus, the area is given by $A=\int_0^{2 \pi} \int_1^{4} (1) dr d \theta$ or, $A=\int_0^{2 \pi} 3 d \theta=6 \pi $