Answer
Mass of lamina: $42k$
Center of lamina: $(2,\frac{85}{28})$
Work Step by Step
Find the mass of the lamina:
$m=\iint_D\sigma(x,y)dA=\int_1^3\int_1^4ky^2dydx=\int_1^3\frac{ky^3}{3}]_1^4dx=\int_1^321k dx=21kx]_1^3=42k$
Find the moment of the lamina about the x-axis:
$M_x=\iint_D y\sigma(x,y)dydx=\int_1^3\int_1^4ky^3dydx=\int_1^3\frac{ky^4}{4}]_1^4dx=\int_1^3\frac{255k}{4}dx=\frac{255kx}{4}]_1^3=\frac{255k}{2}$
Find the moment of the lamina about the y-axis:
$M_y=\iint_Dx\sigma(x,y)dydx=\int_1^3\int_1^4kxy^2dydx=\int_1^3\frac{kxy^3}{3}]_1^4dx=\int_1^321kxdx=\frac{21kx^2}{2}]_1^3=84k$
Then,
$\bar{x}=\frac{M_y}{m}=\frac{84k}{42k}=2$ and $\bar{y}=\frac{M_x}{y}=\frac{255k/2}{42k}=\frac{85}{28}$
So, the center of the lamina is $(\bar{x},\bar{y})=(2,\frac{85}{28})$.
