Thomas' Calculus 13th Edition

$\overline{x}=\overline{y}=\dfrac{1}{2-\ln 4}$
$M= \int_1^{2} \int_{2/x}^{2} y \ dy \ dx \\= \int_{1}^{2} (2-\dfrac{2}{x}) \ dx \\ =2 -\ln 4$ $M_x=\int_1^{2} \int_{(2/x)}^{2} y \ dy \ dx \\= \int_{1}^{2} (2-\dfrac{2}{x^2}) \ dx \\= 1$ $M_y=\int_1^{2} \int_{(2/x)}^{2} x \ dy \ dx \\= \int_{1}^{2} x(2-\dfrac{2}{x}) \ dx \\= 1$ So, by symmetry, $\overline{x}=\overline{y}=\dfrac{1}{2-\ln 4}$