8. Planar Region
$$
\begin{aligned}
S= & \left\{(x, y) \mid x, y \in\left[0, \frac{\pi}{2}\right],\right. \\
& \left.\sin ^{2} x+\sin x \cdot \sin y+\sin ^{2} y \leqslant \frac{3}{4}\right\}
\end{aligned}
$$

The area of the region equals $\qquad$
8. $\frac{\pi^{2}}{6}$.
$$
\begin{array}{l} 
\text { By } 2\left(\sin ^{2} x-\sin x \cdot \sin y+\sin ^{2} y\right) \\
= 2-2 \cos (x+y) \cdot \cos (x-y)+ \\
\cos (x+y)-\cos (x-y) \\
= \frac{3}{2}-2\left[\cos (x+y)+\frac{1}{2}\right] . \\
{\left[\cos (x-y)-\frac{1}{2}\right] } \\
\leqslant \frac{3}{2},
\end{array}
$$

we get $\left[\cos (x+y)+\frac{1}{2}\right]\left[\cos (x-y)-\frac{1}{2}\right] \geqslant 0$,
which means $\left\{\begin{array}{l}x+y \leqslant \frac{2 \pi}{3}, \\ |x-y| \leqslant \frac{\pi}{3}\end{array}\right.$ or $\left\{\begin{array}{l}x+y \geqslant \frac{2 \pi}{3}, \\ |x-y| \geqslant \frac{\pi}{3} \text {. }\end{array}\right.$
Combining $x, y \in\left[0, \frac{\pi}{2}\right]$, we can obtain the planar region as shown in Figure 2, whose area is
$$
\left(\frac{\pi}{2}\right)^{2}-\left(\frac{\pi}{6}\right)^{2}-\frac{1}{2}\left(\frac{\pi}{3}\right)^{2}=\frac{\pi^{2}}{6} \text {. }
$$