Answer
$$6 \space \pi $$
Work Step by Step
Apply cylindrical coordinates. $ x=r \cos \theta ;\\ y= r \sin \theta ;\\ z \gt 0$
We know that $ r(r, \theta)=xi+yj+zk $ or, $ r^2=x^2+y^2+z^2$
We have $ z=1; z=4$ and $ x^2+y^2=1$
Now, $ r_z=\cos \theta \space i+ \sin \theta \space j+2 \space k; \\ r_{\theta}=-\sin \theta \space i+ \cos \theta \space j $
Also, $|r_z \times r_{\theta}|=1$
Now, $$ Area =\int_0^{2 \pi} \int_1^{4} (1) \space dr d \theta \\=\int_0^{2 \pi} 3 \space d \theta=6 \space \pi $$
