Answer
$$ \dfrac{\pi (17\sqrt {17}-5 \space\sqrt {5})}{6}$$
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 $ r_r= \cos \theta \space i+\sin \theta \space j+2 \space k ;\\ r_{\theta}=-r\sin \theta \space i+ r\cos \theta j $
Also, $|r_r \times r_{\theta}|=r\sqrt {4r^2+1}$
Now, $$ Area=\int_0^{2 \pi} \int_1^{2} (r\sqrt {4r^2+1}) \space dr \space d \theta \\=\int_0^{2 \pi} (\dfrac{17\sqrt {17}}{12}- \dfrac{5\sqrt {5}}{12}-) \space d \theta \\= \dfrac{\pi (17\sqrt {17}-5 \space\sqrt {5})}{6}$$
