Answer
$$\left( {\overline x ,\overline y } \right) = \left( {\frac{{1 - 4{e^{ - 3}}}}{{3\left( {1 - {e^3}} \right)}},\frac{{9\left( {1 - {e^{ - 4}}} \right)}}{{16\left( {1 - {e^3}} \right)}}} \right)$$
Work Step by Step
\leq$$\eqalign{
& y = {e^{ - x}},{\text{ }}y = 0,{\text{ }}x = 0,{\text{ }}x = 1,{\text{ }}\rho \left( {x,y} \right) = k{y^2} \cr
& {\text{The region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant y \leqslant {e^{ - x}},{\text{ 0}} \leqslant x \leqslant 1} \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_0^1 {\int_0^{{e^{ - x}}} {k{y^2}} } dydx \cr
& m = k\int_0^1 {\left[ {\int_0^{{e^{ - x}}} {{y^2}dy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& m = k\int_0^1 {\left[ {\frac{1}{3}{y^3}} \right]_0^{{e^{ - x}}}} dx \cr
& m = \frac{k}{3}\int_0^1 {{e^{ - 3x}}} dx \cr
& {\text{Integrate}} \cr
& m = \frac{k}{3}\left[ { - \frac{1}{3}{e^{ - 3x}}} \right]_0^1 \cr
& m = - \frac{k}{9}\left( {{e^{ - 3}} - {e^0}} \right) \cr
& m = \frac{k}{9}\left( {1 - {e^3}} \right) \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_0^1 {\int_0^{{e^{ - x}}} {y\left( {k{y^2}} \right)} } dydx \cr
& {M_x} = k\int_0^1 {\left[ {\int_0^{{e^{ - x}}} {{y^3}dy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_x} = k\int_0^1 {\left[ {\frac{1}{4}{y^4}} \right]} _0^{{e^{ - x}}}dx \cr
& {M_x} = \frac{1}{4}k\int_0^1 {\left[ {{{\left( {{e^{ - x}}} \right)}^4} - {{\left( 0 \right)}^4}} \right]} dx \cr
& {M_x} = \frac{1}{4}k\int_0^1 {\left( {{e^{ - 4x}} - 0} \right)} dx \cr
& {\text{Integrate}} \cr
& {M_x} = \frac{k}{{16}}\left[ { - {e^{ - 4x}}} \right]_0^1 \cr
& {M_x} = - \frac{k}{{16}}\left( {{e^{ - 4}} - {e^0}} \right) \cr
& {M_x} = \frac{k}{{16}}\left( {1 - {e^{ - 4}}} \right) \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_0^1 {\int_0^{{e^{ - x}}} {x\left( {k{y^2}} \right)} } dydx \cr
& {M_y} = k\int_0^1 {x\left[ {\int_0^{{e^{ - x}}} {{y^2}dy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_y} = k\int_0^1 {x\left[ {\frac{1}{3}{y^3}} \right]_0^{{e^{ - x}}}} dx \cr
& {M_y} = \frac{k}{3}\int_0^1 {x{e^{ - 3x}}} dx \cr
& {\text{Integrating by parts we obtain}} \cr
& {M_y} = \frac{k}{3}\left[ { - \frac{1}{3}x{e^{ - 3x}} - \frac{1}{9}{e^{ - 3x}}} \right]_0^1 \cr
& {M_y} = \frac{k}{3}\left[ { - \frac{1}{3}{e^{ - 3}} - \frac{1}{9}{e^{ - 3}}} \right]_0^1 - \frac{k}{3}\left[ { - 0 - \frac{1}{9}} \right] \cr
& {M_y} = - \frac{4}{{27}}k{e^{ - 3}} + \frac{k}{{27}} \cr
& {M_y} = \frac{k}{{27}}\left( {1 - 4{e^{ - 3}}} \right) \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{{\frac{k}{{27}}\left( {1 - 4{e^{ - 3}}} \right)}}{{\frac{k}{9}\left( {1 - {e^3}} \right)}} = \frac{{1 - 4{e^{ - 3}}}}{{3\left( {1 - {e^3}} \right)}} \cr
& \frac{{{M_x}}}{m} = \frac{{\frac{k}{{16}}\left( {1 - {e^{ - 4}}} \right)}}{{\frac{k}{9}\left( {1 - {e^3}} \right)}} = \frac{{9\left( {1 - {e^{ - 4}}} \right)}}{{16\left( {1 - {e^3}} \right)}} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{{1 - 4{e^{ - 3}}}}{{3\left( {1 - {e^3}} \right)}},\frac{{9\left( {1 - {e^{ - 4}}} \right)}}{{16\left( {1 - {e^3}} \right)}}} \right) \cr} $$
