Answer
The mass of the lamina is $42 k$ and the center of mass is at the point $(\overline{x} ,\overline{y} )=(2,\frac{85}{28})$.
Work Step by Step
$$
D=\left\{(x, y) | 1\leq x \leq3 , \quad 1 \leq y \leq4 \right\}.
$$
The mass of the lamina is is given by
$$
\begin{aligned}
m&=\iint_{D} \rho(x, y) d A \\
&=\int_{1}^{3} \int_{1}^{4} k y^{2} d y d x \\
&=k \int_{1}^{3} d x \int_{1}^{4} y^{2} d y \\
&=k[x]_{1}^{3}\left[\frac{1}{3} y^{3}\right]_{1}^{4}\\
&=k(2)(21) \\
&=42 k
\end{aligned}
$$
The center of mass of the lamina is given by
$$
\begin{aligned}
\overline{x} &=\frac{1}{m} \iint_{D} x \rho(x, y) d A \\
&=\frac{1}{42 k} \int_{1}^{3} \int_{1}^{4} k x y^{2} d y d x \\
&=\frac{1}{42} \int_{1}^{3} x d x \int_{1}^{4} y^{2} d y\\
&=\frac{1}{42}\left[\frac{1}{2} x^{2}\right]_{1}^{3}\left[\frac{1}{3} y^{3}\right]_{1}^{4} \\
&=\frac{1}{42}(4)(21)\\
&=2,
\end{aligned}
$$
and
$$
\begin{aligned}
\overline{y}&=\frac{1}{m} \iint_{D} y \rho(x, y) d A \\
&=\frac{1}{42 k} \int_{1}^{3} \int_{1}^{4} k y^{3} d y d x \\
&=\frac{1}{42} \int_{1}^{3} d x \int_{1}^{4} y^{3} d y \\
&=\frac{1}{42}[x]_{1}^{3}\left[\frac{1}{4} y^{4}\right]_{1}^{4}\\
&=\frac{1}{42}(2)\left(\frac{255}{4}\right) \\
&=\frac{85}{28}.
\end{aligned}
$$
Hence, the mass of the lamina is $42 k$ and the center of mass is at the point $(\overline{x} ,\overline{y} )=(2,\frac{85}{28})$.
