Answer
$$8 \space \pi \sqrt 5$$
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=2; z=6$ and $ z=2 \sqrt {x^2+y^2}$
Now, $ r_r= \cos \theta \space i+\sin \theta \space j+2k \\ r_{\theta}=-r \sin \theta \space i+r \cos \theta \space j $
Also, $|r_r \times r_{\theta}|=r \sqrt 5$
Now, $$ Area=\int_0^{2 \pi} \int_1^{3} r \sqrt 5 \space dr \space d \theta \\=\int_0^{2 \pi} 4 \space \sqrt 5\space d \theta \\=8 \space \pi \sqrt 5$$
