3. Given the regions $x^{2}+y^{2} \leqslant 1$ and $[x]^{2}+[y]^{2}=1$ with areas $S_{1}$ and $S_{2}$ respectively (where $[x]$ denotes the greatest integer less than or equal to $x$). Then ( ).
(A) $S_{1}>S_{2}$
(B) $S_{1}=S_{2}$
(C) $S_{1}<S_{2}$
(D) Cannot compare the sizes of $S_{1}$ and $S_{2}$
3.C.

From the problem, we know $S_{1}=\pi \times 1^{2}=\pi$.
Since $[x], [y] \in \mathbf{Z}, [x]^{2}+[y]^{2}=1$, therefore, $\left\{\begin{array}{l}[x]=0, \\ [y]= \pm 1\end{array}\right.$ or $\left\{\begin{array}{l}[x]= \pm 1, \\ [y]=0 . .\end{array}\right.$
Solving, we get $\left\{\begin{array}{l}0 \leqslant x<1, \\ -1 \leqslant y<0 \text { or } 1 \leqslant y<2\end{array}\right.$
or $\left\{\begin{array}{l}-1 \leqslant x<0 \text { or } 1 \leqslant x<2, \\ 0 \leqslant y<1 .\end{array}\right.$
Drawing the graph, we find $S_{2}=4$.
Thus, $S_{1}<S_{2}$.