8. Let $g(x)=\sqrt{x(1-x)}$ be a function defined on the interval $[0,1]$, then the area of the figure enclosed by the graph of the function $x g(x)$ and the $x$-axis is $\qquad$ .
As shown in the figure, the graph of $g(x)$ is a semicircle with a radius of $\frac{1}{2}$.
Let $h(x)=x g(x)$, and note that
$$
\begin{array}{l}
h(x)+h(1-x)=x g(x)+(1-x) g(1-x) \\
=x g(x)+(1-x) g(x)=g(x),
\end{array}
$$

which means taking two points $x$ and $1-x$ on the $x$-axis that are symmetric about the center $C$, and drawing perpendicular segments from these points to the $x$-axis, the lengths of the segments between the $x$-axis and the graph of $h(x)$ and the segments between the graph of $h(x)$ and the semicircle are correspondingly equal. Thus, by the principle of Zu Chongzhi, the areas enclosed by the $x$-axis and the graph of $h(x)$ on both sides of $CD$ are equal to the areas enclosed by the graph of $h(x)$ and the semicircle.

Therefore, the area enclosed by the graph of $x g(x)$ and the $x$-axis is $\frac{1}{2} \cdot \frac{1}{2}\left(\frac{1}{2}\right)^{2} \pi=\frac{\pi}{16}$.