Answer
$\dfrac{\pi}{6}(17\sqrt {17}-5\sqrt {5}) $
Work Step by Step
Here, we have $r_r=( \cos \theta) i+(\sin \theta) j+2k$
and
$r_{\theta}=(-r\sin \theta) i+( r\cos \theta) j$
Also, $|r_r \times r_{\theta}|=r\sqrt {4r^2+1}$
Thus, the area is given by $A=\int_0^{2 \pi} \int_1^{2} (r\sqrt {4r^2+1}) dr d \theta$
or, $A=\int_0^{2 \pi} (\dfrac{17\sqrt {17}}{12}- \dfrac{5\sqrt {5}}{12}-)d \theta= \dfrac{\pi}{6}(17\sqrt {17}-5\sqrt {5}) $
