Answer
$$ \frac{4}{15}$$
Work Step by Step
By using Eq. (2), we get
\begin{align*}
M_{x}&= \rho \int_{a}^{b} y\left(g_{1}(y)-g_{2}(y)\right) d y\\
&=\int_{0}^{1} y \sqrt{1-y} d y
\end{align*}
Let $$ u^2=1-y\ \ \ \ \ \ \ \ \ 2udu=-dy$$
at $y=1$, $u=0$, $y=0\to u=1$, so
\begin{align*}
M_{x} &=\int_{1}^{0} y \sqrt{1-y} d y\\
&= 2\int_1^0 ( u^2-1)u^2du\\
&= 2\int_1^0 ( u^4-u^2)du\\
&=2\left(\frac{1}{5}u^5-\frac{1}{3}u^3\right)\bigg|_0^1\\
&=\frac{4}{15}
\end{align*}
By using Eq. (3), we get
\begin{aligned}
M_{x}&=\frac{1}{2} \rho \int_{a}^{b} f(x)^{2} d x \\
&=\frac{1}{2} \int_{0}^{1}\left(1-x^{2}\right)^{2} d x \\
&=\frac{1}{2} \int_{0}^{1}\left(1-2 x^{2}+x^{4}\right) d x \\
&=\left.\frac{1}{2}\left(x-\frac{2}{3} x^{3}+\frac{x^{5}}{5}\right)\right|_{0} ^{1} \\
&=\frac{1}{2}\left(\frac{8}{15}\right)= \frac{4}{15}
\end{aligned}
