Answer
The mass of the lamina is given by
$$
\frac{8}{15} k .
$$
The center of mass is at the point
$$
\begin{aligned}(\overline{x}, \overline{y}) &=\left(\frac{M_{y}}{m}, \frac{M_{x}}{m}\right)=\left(\frac{0}{\frac{8}{15} k }, \frac{\frac{32}{105} k}{\frac{8}{15} k }\right) =\left(0, \frac{4}{7}\right)
\end{aligned}
$$
Work Step by Step
$$
D=\left\{(x, y) | y=1-x^{2} , \quad y=0 \right\}.
$$
The mass of the lamina is is given by
$$
\begin{aligned} m &=\int_{-1}^{1} \int_{0}^{1-x^{2}} k y d y d x \\
&=k \int_{-1}^{1}\left[\frac{1}{2} y^{2}\right]_{y=0}^{y=1-x^{2}} d x \\
&=\frac{1}{2} k \int_{-1}^{1}\left(1-x^{2}\right)^{2} d x \\
&=\frac{1}{2} k \int_{-1}^{1}\left(1-2 x^{2}+x^{4}\right) d x \\
&=\frac{1}{2} k\left[x-\frac{2}{3} x^{3}+\frac{1}{5} x^{5}\right]_{-1}^{1} \\
&=\frac{1}{2} k\left(1-\frac{2}{3}+\frac{1}{5}+1-\frac{2}{3}+\frac{1}{5}\right) \\ &=\frac{8}{15} k \end{aligned}
$$
The center of mass of the lamina is given by
$$
\begin{aligned} M_{y} &=\int_{-1}^{1} \int_{0}^{1-x^{2}} k x y d y d x \\
&=k \int_{-1}^{1}\left[\frac{1}{2} x y^{2}\right]_{y=0}^{y=1-x^{2}} d x \\
&=\frac{1}{2} k \int_{-1}^{1} x\left(1-x^{2}\right)^{2} d x \\
&=\frac{1}{2} k \int_{-1}^{1}\left(x-2 x^{3}+x^{5}\right) d x \\
&=\frac{1}{2} k\left[\frac{1}{2} x^{2}-\frac{1}{2} x^{4}+\frac{1}{6} x^{5}\right]_{-1}^{1} \\
&=\frac{1}{2} k\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{6}-\frac{1}{2}+\frac{1}{2}-\frac{1}{6}\right)\\
&=0 ,
\end{aligned}
$$
$$
\begin{aligned} M_{x} &=\int_{-1}^{1} \int_{0}^{1-x^{2}} k y^{2} d y d x \\
&=k \int_{-1}^{1}\left[\frac{1}{3} y^{3}\right]_{y=0}^{y=1-x^{2}} d x \\
&=\frac{1}{3} k \int_{-1}^{1}\left(1-x^{2}\right)^{3} d x\\
&=\frac{1}{3} k \int_{-1}^{1}\left(1-3 x^{2}+3 x^{4}-x^{6}\right) d x \\ &=\frac{1}{3} k\left[x-x^{3}+\frac{3}{5} x^{5}-\frac{1}{7} x^{7}\right]_{-1}^{1} \\
&=\frac{1}{3} k\left(1-1+\frac{3}{5}-\frac{1}{7}+1-1+\frac{3}{5}-\frac{1}{7}\right) \\
&=\frac{32}{105} k.
\end{aligned}
$$
Hence, the center of mass is at the point
$$
\begin{aligned}(\overline{x}, \overline{y}) &=\left(\frac{M_{y}}{m}, \frac{M_{x}}{m}\right)=\left(\frac{0}{\frac{8}{15} k }, \frac{\frac{32}{105} k}{\frac{8}{15} k }\right) =\left(0, \frac{4}{7}\right)
\end{aligned}
$$
