Answer
$\dfrac{1}{2-\ln 4}$
Work Step by Step
$M= \int_1^{2} \int_{2/x}^{2} y \ dy \ dx = \int_{1}^{2} (2-\dfrac{2}{x}) \ dx =2 -\ln 4$
Now, $M_x=\int_1^{2} \int_{2/x}^{2} y \ dy \ dx = \int_{1}^{2} (2-\dfrac{2}{x^2}) \ dx = 1$
and $M_y=\int_1^{2} \int_{2/x}^{2} x \ dy \ dx = \int_{1}^{2} x(2-\dfrac{2}{x}) \ dx = 1$
Thus, $\overline{x}=\overline{y}=\dfrac{1}{2-\ln 4}$
