Answer
$2 \ln 2+3$
Work Step by Step
Here, we have $r_x=\lt 1,0,2x-(2/x)\gt$
and
$r_{\theta}=\lt 0, 1, \sqrt {15} \gt$
Also, $|r_x\times r_{y}|=\dfrac{2}{x}+2x$
Thus, the area is given by $A=\int_1^{2} \int_0^{1} (\dfrac{2}{x}+2x) dy dx$
or, $A=[2 \ln x+x^2]_1^{2}= 2 \ln 2+3$
