Answer
$D$ is bounded by $ y=x^{2}, y=x+2$; $\rho(x, y)=kx^{2}$
The mass of the lamina is $m=\frac{63}{20} k,$
and the center of mass is
$$
(\bar{x}, \bar{y})=\left(\frac{8}{7}, \frac{118}{49}\right).
$$
Work Step by Step
$D$ is bounded by $ y=x^{2}, y=x+2$; $\rho(x, y)=kx^{2}$
The curves $ y=x^{2}, y=x+2$ intersect when
$$
x^{2}=x+2 \quad \Leftrightarrow \quad x^{2}-x-2=0 \quad \Leftrightarrow \quad x=-1, x=2
$$
Thus, the region $D$ is defined by:
$$ D=\left\{(x, y ) | \quad -1 \leq x \leq 2, \quad x^{2} \leq y \leq x+2 \right\} $$
is shown in the Figure. (Note that the equation of the upper boundary is $y=x+2$ and lower boundary is $y=x^{2}$ .) The mass of the lamina is
$$
\begin{aligned}
m&=\iint_{D} \rho(x, y) d A\\
&=\int_{-1}^{2} \int_{x^{2}}^{x+2} k x^{2} d y d x \\
&=k \int_{-1}^{2} x^{2} [y]_{y_{y}-x^{2}}^{y=x+2} d x\\
&=k \int_{-1}^{2}\left(x^{3}+2 x^{2}-x^{4}\right) d x \\
&=k\left[\frac{1}{4} x^{4}+\frac{2}{3}x^{3}-\frac{1}{5} x^{5}\right]_{-1}^{2} \\
&=k\left(\frac{44}{15}+\frac{13}{60}\right)\\
&=\frac{63}{20} k,\\
\\
M_{y}& =\int_{-1}^{2} \int_{x^{2}}^{x+2} k x^{3} d y d x\\
&=k \int_{-1}^{2} x^{3}[y]_{y=x^{2}}^{y=x+2} d x\\
&=k \int_{-1}^{2}\left(x^{4}+2 x^{3}-x^{5}\right) d x\\
&=k\left[\frac{1}{5} x^{5}+\frac{2}{4} x^{4}-\frac{1}{6} x^{6}\right]_{-1}^{2} \\
&=k\left(\frac{56}{15}-\frac{2}{15}\right)\\
&=\frac{18}{5} k,\\
M_{x} &=\int_{-1}^{2} \int_{x^{2}}^{x+2} k x^{2} y d y d x \\
&=k \int_{-1}^{2} x^{2}\left[\frac{1}{2} y^{2}\right]_{y=x^{2}}^{y=x+2} d x \\
&=\frac{1}{2} k \int_{-1}^{2} x^{2}\left(x^{2}+4 x+4-x^{4}\right) d x \\
&=\frac{1}{2} k \int_{-1}^{2}\left(x^{4}+4 x^{3}+4 x^{2}-x^{6}\right) d x \\
& =\frac{1}{2} k\left[\frac{1}{5} x^{5}+ x^{4}+\frac{4}{3} x^{3}-\frac{1}{7} x^{7}\right]_{-1}^{2} \\
&=k\left(\frac{776}{105}+\frac{41}{210}\right)\\
&=\frac{531}{70} k.
\end{aligned}
$$
Hence the mass of the lamina is $m=\frac{63}{20} k,$
and the center of mass is
$$
(\bar{x}, \bar{y})=\left(\frac{M_{y}}{m}, \frac{M_{x}}{m}\right) =\left(\frac{\frac{18}{5} k}{\frac{63}{20} k}, \frac{\frac{531}{70} k}{\frac{63}{20} k}\right)=\left(\frac{8}{7}, \frac{118}{49}\right)
$$
