Answer
$$\left( {\overline x ,\overline y } \right) = \left( {\frac{{14}}{5},\frac{4}{5}} \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{4}{x},{\text{ }}y = 0,{\text{ }}x = 1,{\text{ }}x = 4,{\text{ }}\rho \left( {x,y} \right) = k{x^2} \cr
& {\text{The region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant y \leqslant \frac{4}{x},{\text{ }}1 \leqslant x \leqslant 4} \right\} \cr
& {\text{The mass }}m{\text{ of the lamina is given by}} \cr
& m = \iint\limits_R {\rho \left( {x,y} \right)dA} \cr
& m = \int_1^4 {\int_0^{4/x} {k{x^2}} } dydx \cr
& m = k\int_1^4 {{x^2}\left[ {\int_0^{4/x} {dy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& m = k\int_1^4 {{x^2}\left[ y \right]_0^{4/x}} dx \cr
& m = k\int_1^4 {{x^2}\left( {\frac{4}{x}} \right)} dx \cr
& m = 4k\int_1^4 x dx \cr
& {\text{Integrate}} \cr
& m = 2k\left[ {{x^2}} \right]_1^4 \cr
& m = 2k\left( {16 - 1} \right) \cr
& m = 30k \cr
& \cr
& {\text{*The moment of mass with respect to the }}x{\text{ axis is:}} \cr
& {M_x} = \iint\limits_R {y\rho \left( {x,y} \right)dA},{\text{ then}} \cr
& {M_x} = \int_1^4 {\int_0^{4/x} {y\left( {k{x^2}} \right)} } dydx \cr
& {M_x} = k\int_1^4 {{x^2}\left[ {\int_0^{4/x} {ydy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_x} = k\int_1^4 {{x^2}\left[ {\frac{1}{2}{y^2}} \right]_0^{4/x}} dx \cr
& {M_x} = \frac{k}{2}\int_1^4 {{x^2}\left[ {{{\left( {\frac{4}{x}} \right)}^2} - {0^2}} \right]} dx \cr
& {M_x} = \frac{k}{2}\int_1^4 {16} dx \cr
& {\text{Integrate}} \cr
& {M_x} = 8k\left[ x \right]_1^4 \cr
& {M_x} = 24k \cr
& \cr
& {\text{*The moment of mass with respect to the }}y{\text{ axis is:}} \cr
& {M_y} = \iint\limits_R {x\rho \left( {x,y} \right)dA},{\text{ then}} \cr
& {M_y} = \int_1^4 {\int_0^{4/x} {x\left( {k{x^2}} \right)} } dydx \cr
& {M_y} = k\int_1^4 {{x^3}\left[ {\int_0^{4/x} {dy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_y} = k\int_1^4 {{x^3}\left[ y \right]_0^{4/x}} dx \cr
& {M_y} = k\int_1^4 {{x^3}\left( {\frac{4}{x}} \right)} dx \cr
& {M_y} = 4k\int_1^4 {{x^2}} dx \cr
& {\text{Integrate}} \cr
& {M_y} = \frac{{4k}}{3}\left[ {{x^3}} \right]_1^4 \cr
& {M_y} = \frac{{4k}}{3}\left( {64 - 1} \right) \cr
& {M_y} = 84k \cr
& \cr
& {\text{The center of mass is}} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{{{M_y}}}{m},\frac{{{M_x}}}{m}} \right) \cr
& \frac{{{M_y}}}{m} = \frac{{84k}}{{30k}} = \frac{{14}}{5} \cr
& \frac{{{M_x}}}{m} = \frac{{24k}}{{30k}} = \frac{4}{5} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{{14}}{5},\frac{4}{5}} \right) \cr} $$
