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 $y=x g(x)$ and the $x$-axis is $\frac{\pi}{16}$.
Obviously, the graph of $g(x)$ and the $x$-axis enclose a semicircle; we use $A$ to denote the figure enclosed by $x g(x)$ and the $x$-axis. The line $2 x=1$ is the axis of symmetry of the semicircle, dividing $A$ into two left and right parts. We know that
$$
x g(x)+(1-x) g(1-x)=x g(x)+(1-x) g(x)=g(x), 0 \leqslant x \leqslant \frac{1}{2} .
$$

The geometric meaning of this formula is shown in the following figure:
According to the two-dimensional form of Zu's principle, the sum of the areas of the left and right halves of $A$ is exactly one quarter of the area of the circle, i.e., the area we are looking for is $\frac{1}{4} \pi\left(\frac{1}{2}\right)^{2}=\frac{\pi}{16}$.